1 sample hypothesis test

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The One-Sample t -Test

What is the one-sample t -test.

The one-sample t-test is a statistical hypothesis test used to determine whether an unknown population mean is different from a specific value.

When can I use the test?

You can use the test for continuous data. Your data should be a random sample from a normal population.

What if my data isn’t nearly normally distributed?

If your sample sizes are very small, you might not be able to test for normality. You might need to rely on your understanding of the data. When you cannot safely assume normality, you can perform a nonparametric test that doesn’t assume normality.

Using the one-sample t -test

See how to perform a one-sample t -test using statistical software.

• Download JMP to follow along using the sample data included with the software.
• To see more JMP tutorials, visit the JMP Learning Library .

The sections below discuss what we need for the test, checking our data, performing the test, understanding test results and statistical details.

What do we need?

For the one-sample t -test, we need one variable.

We also have an idea, or hypothesis, that the mean of the population has some value. Here are two examples:

• A hospital has a random sample of cholesterol measurements for men. These patients were seen for issues other than cholesterol. They were not taking any medications for high cholesterol. The hospital wants to know if the unknown mean cholesterol for patients is different from a goal level of 200 mg.
• We measure the grams of protein for a sample of energy bars. The label claims that the bars have 20 grams of protein. We want to know if the labels are correct or not.

One-sample t -test assumptions

For a valid test, we need data values that are:

• Independent (values are not related to one another).
• Continuous.
• Obtained via a simple random sample from the population.

Also, the population is assumed to be normally distributed .

One-sample t -test example

Imagine we have collected a random sample of 31 energy bars from a number of different stores to represent the population of energy bars available to the general consumer. The labels on the bars claim that each bar contains 20 grams of protein.

Table 1: Grams of protein in random sample of energy bars

If you look at the table above, you see that some bars have less than 20 grams of protein. Other bars have more. You might think that the data support the idea that the labels are correct. Others might disagree. The statistical test provides a sound method to make a decision, so that everyone makes the same decision on the same set of data values. 

Checking the data

Let’s start by answering: Is the t -test an appropriate method to test that the energy bars have 20 grams of protein ? The list below checks the requirements for the test.

• The data values are independent. The grams of protein in one energy bar do not depend on the grams in any other energy bar. An example of dependent values would be if you collected energy bars from a single production lot. A sample from a single lot is representative of that lot, not energy bars in general.
• The data values are grams of protein. The measurements are continuous.
• We assume the energy bars are a simple random sample from the population of energy bars available to the general consumer (i.e., a mix of lots of bars).
• We assume the population from which we are collecting our sample is normally distributed, and for large samples, we can check this assumption.

We decide that the t -test is an appropriate method.

Before jumping into analysis, we should take a quick look at the data. The figure below shows a histogram and summary statistics for the energy bars.

From a quick look at the histogram, we see that there are no unusual points, or outliers . The data look roughly bell-shaped, so our assumption of a normal distribution seems reasonable.

From a quick look at the statistics, we see that the average is 21.40, above 20. Does this  average from our sample of 31 bars invalidate the label's claim of 20 grams of protein for the unknown entire population mean? Or not?

How to perform the one-sample t -test

For the t -test calculations we need the mean, standard deviation and sample size. These are shown in the summary statistics section of Figure 1 above.

We round the statistics to two decimal places. Software will show more decimal places, and use them in calculations. (Note that Table 1 shows only two decimal places; the actual data used to calculate the summary statistics has more.)

We start by finding the difference between the sample mean and 20:

$ 21.40-20\ =\ 1.40$

Next, we calculate the standard error for the mean. The calculation is:

Standard Error for the mean = $ \frac{s}{\sqrt{n}}= \frac{2.54}{\sqrt{31}}=0.456 $

This matches the value in Figure 1 above.

We now have the pieces for our test statistic. We calculate our test statistic as:

$ t =  \frac{\text{Difference}}{\text{Standard Error}}= \frac{1.40}{0.456}=3.07 $

To make our decision, we compare the test statistic to a value from the t- distribution. This activity involves four steps.

• We calculate a test statistic. Our test statistic is 3.07.
• We decide on the risk we are willing to take for declaring a difference when there is not a difference. For the energy bar data, we decide that we are willing to take a 5% risk of saying that the unknown population mean is different from 20 when in fact it is not. In statistics-speak, we set α = 0.05. In practice, setting your risk level (α) should be made before collecting the data.

We find the value from the t- distribution based on our decision. For a t -test, we need the degrees of freedom to find this value. The degrees of freedom are based on the sample size. For the energy bar data:

degrees of freedom = $ n - 1 = 31 - 1 = 30 $

The critical value of t with α = 0.05 and 30 degrees of freedom is +/- 2.043. Most statistics books have look-up tables for the distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables.

We compare the value of our statistic (3.07) to the t value. Since 3.07 > 2.043, we reject the null hypothesis that the mean grams of protein is equal to 20. We make a practical conclusion that the labels are incorrect, and the population mean grams of protein is greater than 20.

Statistical details

Let’s look at the energy bar data and the 1-sample t -test using statistical terms.

Our null hypothesis is that the underlying population mean is equal to 20. The null hypothesis is written as:

$ H_o:  \mathrm{\mu} = 20 $

The alternative hypothesis is that the underlying population mean is not equal to 20. The labels claiming 20 grams of protein would be incorrect. This is written as:

$ H_a:  \mathrm{\mu} ≠ 20 $

This is a two-sided test. We are testing if the population mean is different from 20 grams in either direction. If we can reject the null hypothesis that the mean is equal to 20 grams, then we make a practical conclusion that the labels for the bars are incorrect. If we cannot reject the null hypothesis, then we make a practical conclusion that the labels for the bars may be correct.

We calculate the average for the sample and then calculate the difference with the population mean, mu:

$  \overline{x} - \mathrm{\mu} $

We calculate the standard error as:

$ \frac{s}{ \sqrt{n}} $

The formula shows the sample standard deviation as s and the sample size as n .  

The test statistic uses the formula shown below:

$  \dfrac{\overline{x} - \mathrm{\mu}} {s / \sqrt{n}} $

We compare the test statistic to a t value with our chosen alpha value and the degrees of freedom for our data. Using the energy bar data as an example, we set α = 0.05. The degrees of freedom ( df ) are based on the sample size and are calculated as:

$ df = n - 1 = 31 - 1 = 30 $

Statisticians write the t value with α = 0.05 and 30 degrees of freedom as:

$ t_{0.05,30} $

The t value for a two-sided test with α = 0.05 and 30 degrees of freedom is +/- 2.042. There are two possible results from our comparison:

• The test statistic is less extreme than the critical  t  values; in other words, the test statistic is not less than -2.042, or is not greater than +2.042. You fail to reject the null hypothesis that the mean is equal to the specified value. In our example, you would be unable to conclude that the label for the protein bars should be changed.
• The test statistic is more extreme than the critical  t  values; in other words, the test statistic is less than -2.042, or is greater than +2.042. You reject the null hypothesis that the mean is equal to the specified value. In our example, you conclude that either the label should be updated or the production process should be improved to produce, on average, bars with 20 grams of protein.

Testing for normality

The normality assumption is more important for small sample sizes than for larger sample sizes.

Normal distributions are symmetric, which means they are “even” on both sides of the center. Normal distributions do not have extreme values, or outliers. You can check these two features of a normal distribution with graphs. Earlier, we decided that the energy bar data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for the data, and supports our decision.

You can also perform a formal test for normality using software. The figure below shows results of testing for normality with JMP software. We cannot reject the hypothesis of a normal distribution. 

We can go ahead with the assumption that the energy bar data is normally distributed.

What if my data are not from a Normal distribution?

If your sample size is very small, it is hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the energy bar data, the company knows that the underlying distribution of grams of protein is normally distributed. Even for a very small sample, the company would likely go ahead with the t -test and assume normality.

What if you know the underlying measurements are not normally distributed? Or what if your sample size is large and the test for normality is rejected? In this situation, you can use a nonparametric test. Nonparametric  analyses do not depend on an assumption that the data values are from a specific distribution. For the one-sample t ­-test, the one possible nonparametric test is the Wilcoxon Signed Rank test. 

Understanding p-values

Using a visual, you can check to see if your test statistic is more extreme than a specified value in the distribution. The figure below shows a t- distribution with 30 degrees of freedom.

Since our test is two-sided and we set α = 0.05, the figure shows that the value of 2.042 “cuts off” 5% of the data in the tails combined.

The next figure shows our results. You can see the test statistic falls above the specified critical value. It is far enough “out in the tail” to reject the hypothesis that the mean is equal to 20.

Putting it all together with Software

You are likely to use software to perform a t -test. The figure below shows results for the 1-sample t -test for the energy bar data from JMP software.  

The software shows the null hypothesis value of 20 and the average and standard deviation from the data. The test statistic is 3.07. This matches the calculations above.

The software shows results for a two-sided test and for one-sided tests. We want the two-sided test. Our null hypothesis is that the mean grams of protein is equal to 20. Our alternative hypothesis is that the mean grams of protein is not equal to 20.  The software shows a p- value of 0.0046 for the two-sided test. This p- value describes the likelihood of seeing a sample average as extreme as 21.4, or more extreme, when the underlying population mean is actually 20; in other words, the probability of observing a sample mean as different, or even more different from 20, than the mean we observed in our sample. A p -value of 0.0046 means there is about 46 chances out of 10,000. We feel confident in rejecting the null hypothesis that the population mean is equal to 20.

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One Sample T Test – Clearly Explained with Examples | ML+

• October 8, 2020
• Selva Prabhakaran

One sample T-Test tests if the given sample of observations could have been generated from a population with a specified mean.

If it is found from the test that the means are statistically different, we infer that the sample is unlikely to have come from the population.

For example: If you want to test a car manufacturer’s claim that their cars give a highway mileage of 20kmpl on an average. You sample 10 cars from the dealership, measure their mileage and use the T-test to determine if the manufacturer’s claim is true.

By end of this, you will know when and how to do the T-Test, the concept, math, how to set the null and alternate hypothesis, how to use the T-tables, how to understand the one-tailed and two-tailed T-Test and see how to implement in R and Python using a practical example.

Introduction

Purpose of one sample t test, how to set the null and alternate hypothesis, procedure to do one sample t test, one sample t test example, one sample t test implementation, how to decide which t test to perform two tailed, upper tailed or lower tailed.

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The ‘One sample T Test’ is one of the 3 types of T Tests . It is used when you want to test if the mean of the population from which the sample is drawn is of a hypothesized value. You will understand this statement better (and all of about One Sample T test) better by the end of this post.

T Test was first invented by William Sealy Gosset, in 1908. Since he used the pseudo name as ‘Student’ when publishing his method in the paper titled ‘Biometrika’, the test came to be know as Student’s T Test.

Since it assumes that the test statistic, typically the sample mean, follows the sampling distribution, the Student’s T Test is considered as a Parametric test.

The purpose of the One Sample T Test is to determine if a sample observations could have come from a process that follows a specific parameter (like the mean).

It is typically implemented on small samples.

For example, given a sample of 15 items, you want to test if the sample mean is the same as a hypothesized mean (population). That is, essentially you want to know if the sample came from the given population or not.

Let’s suppose, you want to test if the mean weight of a manufactured component (from a sample size 15) is of a particular value (55 grams), with a 99% confidence.

How did we determine One sample T-test is the right test for this?

Because, there is only one sample involved and you want to compare the mean of this sample against a particular (hypothesized) value..

To do this, you need to set up a null hypothesis and an alternate hypothesis .

The null hypothesis usually assumes that there is no difference in the sample means and the hypothesized mean (comparison mean). The purpose of the T Test is to test if the null hypothesis can be rejected or not.

Depending on the how the problem is stated, the alternate hypothesis can be one of the following 3 cases:

• Case 1: H1 : x̅ != µ. Used when the true sample mean is not equal to the comparison mean. Use Two Tailed T Test.
• Case 2: H1 : x̅ > µ. Used when the true sample mean is greater than the comparison mean. Use Upper Tailed T Test.
• Case 3: H1 : x̅ < µ. Used when the true sample mean is lesser than the comparison mean. Use Lower Tailed T Test.

Where x̅ is the sample mean and µ is the population mean for comparison. We will go more into the detail of these three cases after solving some practical examples.

Example 1: A customer service company wants to know if their support agents are performing on par with industry standards.

According to a report the standard mean resolution time is 20 minutes per ticket. The sample group has a mean at 21 minutes per ticket with a standard deviation of 7 minutes.

Can you tell if the company’s support performance is better than the industry standard or not?

Example 2: A farming company wants to know if a new fertilizer has improved crop yield or not.

Historic data shows the average yield of the farm is 20 tonne per acre. They decide to test a new organic fertilizer on a smaller sample of farms and observe the new yield is 20.175 tonne per acre with a standard deviation of 3.02 tonne for 12 different farms.

Did the new fertilizer work?

Step 1: Define the Null Hypothesis (H0) and Alternate Hypothesis (H1)

H0: Sample mean (x̅) = Hypothesized Population mean (µ)

H1: Sample mean (x̅) != Hypothesized Population mean (µ)

The alternate hypothesis can also state that the sample mean is greater than or less than the comparison mean.

Step 2: Compute the test statistic (T)

$$t = \frac{Z}{s} = \frac{\bar{X} – \mu}{\frac{\hat{\sigma}}{\sqrt{n}}}$$

where s is the standard error .

Step 3: Find the T-critical from the T-Table

Use the degree of freedom and the alpha level (0.05) to find the T-critical.

Step 4: Determine if the computed test statistic falls in the rejection region.

Alternately, simply compute the P-value. If it is less than the significance level (0.05 or 0.01), reject the null hypothesis.

Problem Statement:

We have the potato yield from 12 different farms. We know that the standard potato yield for the given variety is µ=20.

x = [21.5, 24.5, 18.5, 17.2, 14.5, 23.2, 22.1, 20.5, 19.4, 18.1, 24.1, 18.5]

Test if the potato yield from these farms is significantly better than the standard yield.

Step 1: Define the Null and Alternate Hypothesis

H0: x̅ = 20

H1: x̅ > 20

n = 12. Since this is one sample T test, the degree of freedom = n-1 = 12-1 = 11.

Let’s set alpha = 0.05, to meet 95% confidence level.

Step 2: Calculate the Test Statistic (T) 1. Calculate sample mean

$$\bar{X} = \frac{x_1 + x_2 + x_3 + . . + x_n}{n}$$

$$\bar{x} = 20.175$$

• Calculate sample standard deviation

$$\bar{\sigma} = \frac{(x_1 – \bar{x})^2 + (x_2 – \bar{x})^2 + (x_3 – \bar{x})^2 + . . + (x_n – \bar{x})^2}{n-1}$$

$$\sigma = 3.0211$$

• Substitute in the T Statistic formula

$$T = \frac{\bar{x} – \mu}{se} = \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}}$$

$$T = (20.175 – 20)/(3.0211/\sqrt{12}) = 0.2006$$

Step 3: Find the T-Critical

Confidence level = 0.95, alpha=0.05. For one tailed test, look under 0.05 column. For d.o.f = 12 – 1 = 11, T-Critical = 1.796 .

Now you might wonder why ‘One Tailed test’ was chosen. This is because of the way you define the alternate hypothesis. Had the null hypothesis simply stated that the sample means is not equal to 20, then we would have gone for a two tailed test. More details about this topic in the next section.

Step 4: Does it fall in rejection region?

Since the computed T Statistic is less than the T-critical, it does not fall in the rejection region.

Clearly, the calculated T statistic does not fall in the rejection region. So, we do not reject the null hypothesis.

Since you want to perform a ‘One Tailed Greater than’ test (that is, the sample mean is greater than the comparison mean), you need to specify alternative='greater' in the t.test() function. Because, by default, the t.test() does a two tailed test (which is what you do when your alternate hypothesis simply states sample mean != comparison mean).

The P-value computed here is nothing but p = Pr(T > t) (upper-tailed), where t is the calculated T statistic.

In Python, One sample T Test is implemented in ttest_1samp() function in the scipy package. However, it does a Two tailed test by default , and reports a signed T statistic. That means, the reported P-value will always be computed for a Two-tailed test. To calculate the correct P value, you need to divide the output P-value by 2.

Apply the following logic if you are performing a one tailed test:

For greater than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be greater than 0. For lesser than test: Reject H0 if p/2 < alpha (0.05). In this case, t will be less than 0.

Since it is one tailed test, the real p-value is 0.8446/2 = 0.4223. We do not rejecting the Null Hypothesis anyway.

The decision of whether the computed test statistic falls in the rejection region depends on how the alternate hypothesis is defined.

We know the Null Hypothesis is H0: µD = 0. Where, µD is the difference in the means, that is sample mean minus the comparison mean.

You can also write H0 as: x̅ = µ , where x̅ is sample mean and ‘µ’ is the comparison mean.

Case 1: If H1 : x̅ != µ , then rejection region lies on both tails of the T-Distribution (two-tailed). This means the alternate hypothesis just states the difference in means is not equal. There is no comparison if one of the means is greater or lesser than the other.

In this case, use Two Tailed T Test .

Here, P value = 2 . Pr(T > | t |)

Case 2: If H1: x̅ > µ , then rejection region lies on upper tail of the T-Distribution (upper-tailed). If the mean of the sample of interest is greater than the comparison mean. Example: If Component A has a longer time-to-failure than Component B.

In such case, use Upper Tailed based test.

Here, P-value = Pr(T > t)

Case 3: If H1: x̅ < µ , then rejection region lies on lower tail of the T-Distribution (lower-tailed). If the mean of the sample of interest is lesser than the comparison mean.

In such case, use lower tailed test.

Here, P-value = Pr(T < t)

Hope you are now familiar and clear about with the One Sample T Test. If some thing is still not clear, write in comment. Next, topic is Two sample T test . Stay tuned.

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Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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How t-Tests Work: 1-sample, 2-sample, and Paired t-Tests

By Jim Frost 15 Comments

T-tests are statistical hypothesis tests that analyze one or two sample means. When you analyze your data with any t-test, the procedure reduces your entire sample to a single value, the t-value. In this post, I describe how each type of t-test calculates the t-value. I don’t explain this just so you can understand the calculation, but I describe it in a way that really helps you grasp how t-tests work.

How 1-Sample t-Tests Calculate t-Values

The equation for how the 1-sample t-test produces a t-value based on your sample is below:

This equation is a ratio, and a common analogy is the signal-to-noise ratio. The numerator is the signal in your sample data, and the denominator is the noise. Let’s see how t-tests work by comparing the signal to the noise!

The Signal – The Size of the Sample Effect

In the signal-to-noise analogy, the numerator of the ratio is the signal. The effect that is present in the sample is the signal. It’s a simple calculation. In a 1-sample t-test, the sample effect is the sample mean minus the value of the null hypothesis. That’s the top part of the equation.

For example, if the sample mean is 20 and the null value is 5, the sample effect size is 15. We’re calling this the signal because this sample estimate is our best estimate of the population effect.

The calculation for the signal portion of t-values is such that when the sample effect equals zero, the numerator equals zero, which in turn means the t-value itself equals zero. The estimated sample effect (signal) equals zero when there is no difference between the sample mean and the null hypothesis value. For example, if the sample mean is 5 and the null value is 5, the signal equals zero (5 – 5 = 0).

The size of the signal increases when the difference between the sample mean and null value increases. The difference can be either negative or positive, depending on whether the sample mean is greater than or less than the value associated with the null hypothesis.

A relatively large signal in the numerator produces t-values that are further away from zero.

The Noise – The Variability or Random Error in the Sample

The denominator of the ratio is the standard error of the mean, which measures the sample variation. The standard error of the mean represents how much random error is in the sample and how well the sample estimates the population mean.

As the value of this statistic increases, the sample mean provides a less precise estimate of the population mean. In other words, high levels of random error increase the probability that your sample mean is further away from the population mean.

In our analogy, random error represents noise. Why? When there is more random error, you are more likely to see considerable differences between the sample mean and the null hypothesis value in cases where  the null is true . Noise appears in the denominator to provide a benchmark for how large the signal must be to distinguish from the noise.

Signal-to-Noise ratio

Our signal-to-noise ratio analogy equates to:

Both of these statistics are in the same units as your data. Let’s calculate a couple of t-values to see how to interpret them.

• If the signal is 10 and the noise is 2, your t-value is 5. The signal is 5 times the noise.
• If the signal is 10 and the noise is 5, your t-value is 2. The signal is 2 times the noise.

The signal is the same in both examples, but it is easier to distinguish from the lower amount of noise in the first example. In this manner, t-values indicate how clear the signal is from the noise. If the signal is of the same general magnitude as the noise, it’s probable that random error causes the difference between the sample mean and null value rather than an actual population effect.

Paired t-Tests Are Really 1-Sample t-Tests

Paired t-tests require dependent samples. I’ve seen a lot of confusion over how a paired t-test works and when you should use it. Pssst! Here’s a secret! Paired t-tests and 1-sample t-tests are the same hypothesis test incognito!

You use a 1-sample t-test to assess the difference between a sample mean and the value of the null hypothesis.

A paired t-test takes paired observations (like before and after), subtracts one from the other, and conducts a 1-sample t-test on the differences. Typically, a paired t-test determines whether the paired differences are significantly different from zero.

Download the CSV data file to check this yourself: T-testData . All of the statistical results are the same when you perform a paired t-test using the Before and After columns versus performing a 1-sample t-test on the Differences column.

Once you realize that paired t-tests are the same as 1-sample t-tests on paired differences, you can focus on the deciding characteristic —does it make sense to analyze the differences between two columns?

Suppose the Before and After columns contain test scores and there was an intervention in between. If each row in the data contains the same subject in the Before and After column, it makes sense to find the difference between the columns because it represents how much each subject changed after the intervention. The paired t-test is a good choice.

On the other hand, if a row has different subjects in the Before and After columns, it doesn’t make sense to subtract the columns. You should use the 2-sample t-test described below.

The paired t-test is a convenience for you. It eliminates the need for you to calculate the difference between two columns yourself. Remember, double-check that this difference is meaningful! If using a paired t-test is valid, you should use it because it provides more statistical power than the 2-sample t-test, which I discuss in my post about independent and dependent samples .

How Two-Sample T-tests Calculate T-Values

Use the 2-sample t-test when you want to analyze the difference between the means of two independent samples. This test is also known as the independent samples t-test . Click the link to learn more about its hypotheses, assumptions, and interpretations.

Like the other t-tests, this procedure reduces all of your data to a single t-value in a process similar to the 1-sample t-test. The signal-to-noise analogy still applies.

Here’s the equation for the t-value in a 2-sample t-test.

The equation is still a ratio, and the numerator still represents the signal. For a 2-sample t-test, the signal, or effect, is the difference between the two sample means. This calculation is straightforward. If the first sample mean is 20 and the second mean is 15, the effect is 5.

Typically, the null hypothesis states that there is no difference between the two samples. In the equation, if both groups have the same mean, the numerator, and the ratio as a whole, equals zero. Larger differences between the sample means produce stronger signals.

The denominator again represents the noise for a 2-sample t-test. However, you can use two different values depending on whether you assume that the variation in the two groups is equal or not. Most statistical software let you choose which value to use.

Regardless of the denominator value you use, the 2-sample t-test works by determining how distinguishable the signal is from the noise. To ascertain that the difference between means is statistically significant, you need a high positive or negative t-value.

How Do T-tests Use T-values to Determine Statistical Significance?

Here’s what we’ve learned about the t-values for the 1-sample t-test, paired t-test, and 2-sample t-test:

• Each test reduces your sample data down to a single t-value based on the ratio of the effect size to the variability in your sample.
• A t-value of zero indicates that your sample results match the null hypothesis precisely.
• Larger absolute t-values represent stronger signals, or effects, that stand out more from the noise.

For example, a t-value of 2 indicates that the signal is twice the magnitude of the noise.

Great … but how do you get from that to determining whether the effect size is statistically significant? After all, the purpose of t-tests is to assess hypotheses. To find out, read the companion post to this one: How t-Tests Work: t-Values, t-Distributions and Probabilities . Click here for step-by-step instructions on how to do t-tests in Excel !

If you’d like to learn about other hypothesis tests using the same general approach, read my posts about:

• How F-tests Work in ANOVA
• How Chi-Squared Tests of Independence Work

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Reader Interactions

January 9, 2023 at 11:11 am

Hi Jim, thank you for explaining this I will revert to this during my 8 weeks in class everyday to make sure I understand what I’m doing . May I ask more questions in the future.

November 27, 2021 at 1:37 pm

This was an awesome piece, very educative and easy to understand

June 19, 2021 at 1:53 pm

Hi Jim, I found your posts very helpful. Could you plz explain how to do T test for a panel data?

June 19, 2021 at 3:40 pm

You’re limited by what you can do with t-tests. For panel data and t-tests, you can compare the same subjects at two points in time using a paired t-test. For more complex arrangements, you can use repeated measures ANOVA or specify a regression model to meet your needs.

February 11, 2020 at 10:34 pm

Hi Jim: I was reviewing this post in preparation for an analysis I plan to do, and I’d like to ask your advice. Each year, staff complete an all-employee survey, and results are reported at workgroup level of analysis. I would like to compare mean scores of several workgroups from one year to the next (in this case, 2018 and 2019 scores). For example, I would compare workgroup mean scores on psychological safety between 2018 and 2019. I am leaning toward a paired t test. However, my one concern is that….even though I am comparing workgroup to workgroup from one year to the next….it is certainly possible that there may be some different employees in a given workgroup from one year to the next (turnover, transition, etc.)….Assuming that is the case with at least some of the workgroups, does that make a paired t test less meanginful? Would I still use a paired t test or would another type t test be more appropriate? I’m thinking because we are dealing with workgroup mean scores (and not individual scores), then it may still be okay to compare meaningfully (avoiding an ecological fallacy). Thoughts?

Many thanks for these great posts. I enjoy reading them…!

April 8, 2019 at 11:22 pm

Hi jim. First of all, I really appreciate your posts!

When I use t-test via R or scikit learn, there is an option for homogeneity of variance. I think that option only applied to two sample t-test, but what should I do for that option?

Should I always perform f-test for check the homogeneity of variance? or Which one is a more strict assumption?

November 9, 2018 at 12:03 am

This blog is great. I’m at Stanford and can say this is a great supplement to class lectures. I love the fact that there aren’t formulas so as to get an intuitive feel. Thank you so much!

November 9, 2018 at 9:12 am

Thanks Mel! I’m glad it has been helpful! Your kind words mean a lot to me because I really strive to make these topics as easy to understand as possible!

December 29, 2017 at 4:14 pm

Thank you so much Jim! I have such a hard time understanding statistics without people like you who explain it using words to help me conceptualize rather than utilizing symbols only!

December 29, 2017 at 4:56 pm

Thank you, Jessica! Your kind words made my day. That’s what I want my blog to be all about. Providing simple but 100% accurate explanations for statistical concepts!

Happy New Year!

October 22, 2017 at 2:38 pm

Hi Jim, sure, I’ll go through it…Thank you..!

October 22, 2017 at 4:50 am

In summary, the t test tells, how the sample mean is different from null hypothesis, i.e. how the sample mean is different from null, but how does it comment about the significance? Is it like “more far from null is the more significant”? If it is so, could you give some more explanation about it?

October 22, 2017 at 2:30 pm

Hi Omkar, you’re in luck, I’ve written an entire blog post that talks about how t-tests actually use the t-values to determine statistical significance. In general, the further away from zero, the more significant it is. For all the information, read this post: How t-Tests Work: t-Values, t-Distributions, and Probabilities . I think this post will answer your questions.

September 12, 2017 at 2:46 am

Excellent explanation, appreciate you..!!

September 12, 2017 at 8:48 am

Thank you, Santhosh! I’m glad you found it helpful!

Comments and Questions Cancel reply

t-test Calculator

Table of contents

Welcome to our t-test calculator! Here you can not only easily perform one-sample t-tests , but also two-sample t-tests , as well as paired t-tests .

Do you prefer to find the p-value from t-test, or would you rather find the t-test critical values? Well, this t-test calculator can do both! 😊

What does a t-test tell you? Take a look at the text below, where we explain what actually gets tested when various types of t-tests are performed. Also, we explain when to use t-tests (in particular, whether to use the z-test vs. t-test) and what assumptions your data should satisfy for the results of a t-test to be valid. If you've ever wanted to know how to do a t-test by hand, we provide the necessary t-test formula, as well as tell you how to determine the number of degrees of freedom in a t-test.

When to use a t-test?

A t-test is one of the most popular statistical tests for location , i.e., it deals with the population(s) mean value(s).

There are different types of t-tests that you can perform:

• A one-sample t-test;
• A two-sample t-test; and
• A paired t-test.

In the next section , we explain when to use which. Remember that a t-test can only be used for one or two groups . If you need to compare three (or more) means, use the analysis of variance ( ANOVA ) method.

The t-test is a parametric test, meaning that your data has to fulfill some assumptions :

• The data points are independent; AND
• The data, at least approximately, follow a normal distribution .

If your sample doesn't fit these assumptions, you can resort to nonparametric alternatives. Visit our Mann–Whitney U test calculator or the Wilcoxon rank-sum test calculator to learn more. Other possibilities include the Wilcoxon signed-rank test or the sign test.

Which t-test?

Your choice of t-test depends on whether you are studying one group or two groups:

One sample t-test

Choose the one-sample t-test to check if the mean of a population is equal to some pre-set hypothesized value .

The average volume of a drink sold in 0.33 l cans — is it really equal to 330 ml?

The average weight of people from a specific city — is it different from the national average?

Two-sample t-test

Choose the two-sample t-test to check if the difference between the means of two populations is equal to some pre-determined value when the two samples have been chosen independently of each other.

In particular, you can use this test to check whether the two groups are different from one another .

The average difference in weight gain in two groups of people: one group was on a high-carb diet and the other on a high-fat diet.

The average difference in the results of a math test from students at two different universities.

This test is sometimes referred to as an independent samples t-test , or an unpaired samples t-test .

Paired t-test

A paired t-test is used to investigate the change in the mean of a population before and after some experimental intervention , based on a paired sample, i.e., when each subject has been measured twice: before and after treatment.

In particular, you can use this test to check whether, on average, the treatment has had any effect on the population .

The change in student test performance before and after taking a course.

The change in blood pressure in patients before and after administering some drug.

How to do a t-test?

So, you've decided which t-test to perform. These next steps will tell you how to calculate the p-value from t-test or its critical values, and then which decision to make about the null hypothesis.

Decide on the alternative hypothesis :

Use a two-tailed t-test if you only care whether the population's mean (or, in the case of two populations, the difference between the populations' means) agrees or disagrees with the pre-set value.

Use a one-tailed t-test if you want to test whether this mean (or difference in means) is greater/less than the pre-set value.

Compute your T-score value :

Formulas for the test statistic in t-tests include the sample size , as well as its mean and standard deviation . The exact formula depends on the t-test type — check the sections dedicated to each particular test for more details.

Determine the degrees of freedom for the t-test:

The degrees of freedom are the number of observations in a sample that are free to vary as we estimate statistical parameters. In the simplest case, the number of degrees of freedom equals your sample size minus the number of parameters you need to estimate . Again, the exact formula depends on the t-test you want to perform — check the sections below for details.

The degrees of freedom are essential, as they determine the distribution followed by your T-score (under the null hypothesis). If there are d degrees of freedom, then the distribution of the test statistics is the t-Student distribution with d degrees of freedom . This distribution has a shape similar to N(0,1) (bell-shaped and symmetric) but has heavier tails . If the number of degrees of freedom is large (>30), which generically happens for large samples, the t-Student distribution is practically indistinguishable from N(0,1).

💡 The t-Student distribution owes its name to William Sealy Gosset, who, in 1908, published his paper on the t-test under the pseudonym "Student". Gosset worked at the famous Guinness Brewery in Dublin, Ireland, and devised the t-test as an economical way to monitor the quality of beer. Cheers! 🍺🍺🍺

p-value from t-test

Recall that the p-value is the probability (calculated under the assumption that the null hypothesis is true) that the test statistic will produce values at least as extreme as the T-score produced for your sample . As probabilities correspond to areas under the density function, p-value from t-test can be nicely illustrated with the help of the following pictures:

The following formulae say how to calculate p-value from t-test. By cdf t,d we denote the cumulative distribution function of the t-Student distribution with d degrees of freedom:

p-value from left-tailed t-test:

p-value = cdf t,d (t score )

p-value from right-tailed t-test:

p-value = 1 − cdf t,d (t score )

p-value from two-tailed t-test:

p-value = 2 × cdf t,d (−|t score |)

or, equivalently: p-value = 2 − 2 × cdf t,d (|t score |)

However, the cdf of the t-distribution is given by a somewhat complicated formula. To find the p-value by hand, you would need to resort to statistical tables, where approximate cdf values are collected, or to specialized statistical software. Fortunately, our t-test calculator determines the p-value from t-test for you in the blink of an eye!

t-test critical values

Recall, that in the critical values approach to hypothesis testing, you need to set a significance level, α, before computing the critical values , which in turn give rise to critical regions (a.k.a. rejection regions).

Formulas for critical values employ the quantile function of t-distribution, i.e., the inverse of the cdf :

Critical value for left-tailed t-test: cdf t,d -1 (α)

critical region:

(-∞, cdf t,d -1 (α)]

Critical value for right-tailed t-test: cdf t,d -1 (1-α)

[cdf t,d -1 (1-α), ∞)

Critical values for two-tailed t-test: ±cdf t,d -1 (1-α/2)

(-∞, -cdf t,d -1 (1-α/2)] ∪ [cdf t,d -1 (1-α/2), ∞)

To decide the fate of the null hypothesis, just check if your T-score lies within the critical region:

If your T-score belongs to the critical region , reject the null hypothesis and accept the alternative hypothesis.

If your T-score is outside the critical region , then you don't have enough evidence to reject the null hypothesis.

How to use our t-test calculator

Choose the type of t-test you wish to perform:

A one-sample t-test (to test the mean of a single group against a hypothesized mean);

A two-sample t-test (to compare the means for two groups); or

A paired t-test (to check how the mean from the same group changes after some intervention).

Two-tailed;

Left-tailed; or

Right-tailed.

This t-test calculator allows you to use either the p-value approach or the critical regions approach to hypothesis testing!

Enter your T-score and the number of degrees of freedom . If you don't know them, provide some data about your sample(s): sample size, mean, and standard deviation, and our t-test calculator will compute the T-score and degrees of freedom for you .

Once all the parameters are present, the p-value, or critical region, will immediately appear underneath the t-test calculator, along with an interpretation!

One-sample t-test

The null hypothesis is that the population mean is equal to some value μ 0 \mu_0 μ 0 ​ .

The alternative hypothesis is that the population mean is:

• different from μ 0 \mu_0 μ 0 ​ ;
• smaller than μ 0 \mu_0 μ 0 ​ ; or
• greater than μ 0 \mu_0 μ 0 ​ .

One-sample t-test formula :

• μ 0 \mu_0 μ 0 ​ — Mean postulated in the null hypothesis;
• n n n — Sample size;
• x ˉ \bar{x} x ˉ — Sample mean; and
• s s s — Sample standard deviation.

Number of degrees of freedom in t-test (one-sample) = n − 1 n-1 n − 1 .

The null hypothesis is that the actual difference between these groups' means, μ 1 \mu_1 μ 1 ​ , and μ 2 \mu_2 μ 2 ​ , is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the difference μ 1 − μ 2 \mu_1 - \mu_2 μ 1 ​ − μ 2 ​ is:

• Different from Δ \Delta Δ ;
• Smaller than Δ \Delta Δ ; or
• Greater than Δ \Delta Δ .

In particular, if this pre-determined difference is zero ( Δ = 0 \Delta = 0 Δ = 0 ):

The null hypothesis is that the population means are equal.

The alternate hypothesis is that the population means are:

• μ 1 \mu_1 μ 1 ​ and μ 2 \mu_2 μ 2 ​ are different from one another;
• μ 1 \mu_1 μ 1 ​ is smaller than μ 2 \mu_2 μ 2 ​ ; and
• μ 1 \mu_1 μ 1 ​ is greater than μ 2 \mu_2 μ 2 ​ .

Formally, to perform a t-test, we should additionally assume that the variances of the two populations are equal (this assumption is called the homogeneity of variance ).

There is a version of a t-test that can be applied without the assumption of homogeneity of variance: it is called a Welch's t-test . For your convenience, we describe both versions.

Two-sample t-test if variances are equal

Use this test if you know that the two populations' variances are the same (or very similar).

Two-sample t-test formula (with equal variances) :

where s p s_p s p ​ is the so-called pooled standard deviation , which we compute as:

• Δ \Delta Δ — Mean difference postulated in the null hypothesis;
• n 1 n_1 n 1 ​ — First sample size;
• x ˉ 1 \bar{x}_1 x ˉ 1 ​ — Mean for the first sample;
• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• n 2 n_2 n 2 ​ — Second sample size;
• x ˉ 2 \bar{x}_2 x ˉ 2 ​ — Mean for the second sample; and
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

Number of degrees of freedom in t-test (two samples, equal variances) = n 1 + n 2 − 2 n_1 + n_2 - 2 n 1 ​ + n 2 ​ − 2 .

Two-sample t-test if variances are unequal (Welch's t-test)

Use this test if the variances of your populations are different.

Two-sample Welch's t-test formula if variances are unequal:

• s 1 s_1 s 1 ​ — Standard deviation in the first sample;
• s 2 s_2 s 2 ​ — Standard deviation in the second sample.

The number of degrees of freedom in a Welch's t-test (two-sample t-test with unequal variances) is very difficult to count. We can approximate it with the help of the following Satterthwaite formula :

Alternatively, you can take the smaller of n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 as a conservative estimate for the number of degrees of freedom.

🔎 The Satterthwaite formula for the degrees of freedom can be rewritten as a scaled weighted harmonic mean of the degrees of freedom of the respective samples: n 1 − 1 n_1 - 1 n 1 ​ − 1 and n 2 − 1 n_2 - 1 n 2 ​ − 1 , and the weights are proportional to the standard deviations of the corresponding samples.

As we commonly perform a paired t-test when we have data about the same subjects measured twice (before and after some treatment), let us adopt the convention of referring to the samples as the pre-group and post-group.

The null hypothesis is that the true difference between the means of pre- and post-populations is equal to some pre-set value, Δ \Delta Δ .

The alternative hypothesis is that the actual difference between these means is:

Typically, this pre-determined difference is zero. We can then reformulate the hypotheses as follows:

The null hypothesis is that the pre- and post-means are the same, i.e., the treatment has no impact on the population .

The alternative hypothesis:

• The pre- and post-means are different from one another (treatment has some effect);
• The pre-mean is smaller than the post-mean (treatment increases the result); or
• The pre-mean is greater than the post-mean (treatment decreases the result).

Paired t-test formula

In fact, a paired t-test is technically the same as a one-sample t-test! Let us see why it is so. Let x 1 , . . . , x n x_1, ... , x_n x 1 ​ , ... , x n ​ be the pre observations and y 1 , . . . , y n y_1, ... , y_n y 1 ​ , ... , y n ​ the respective post observations. That is, x i , y i x_i, y_i x i ​ , y i ​ are the before and after measurements of the i -th subject.

For each subject, compute the difference, d i : = x i − y i d_i := x_i - y_i d i ​ := x i ​ − y i ​ . All that happens next is just a one-sample t-test performed on the sample of differences d 1 , . . . , d n d_1, ... , d_n d 1 ​ , ... , d n ​ . Take a look at the formula for the T-score :

Δ \Delta Δ — Mean difference postulated in the null hypothesis;

n n n — Size of the sample of differences, i.e., the number of pairs;

x ˉ \bar{x} x ˉ — Mean of the sample of differences; and

s s s  — Standard deviation of the sample of differences.

Number of degrees of freedom in t-test (paired): n − 1 n - 1 n − 1

t-test vs Z-test

We use a Z-test when we want to test the population mean of a normally distributed dataset, which has a known population variance . If the number of degrees of freedom is large, then the t-Student distribution is very close to N(0,1).

Hence, if there are many data points (at least 30), you may swap a t-test for a Z-test, and the results will be almost identical. However, for small samples with unknown variance, remember to use the t-test because, in such cases, the t-Student distribution differs significantly from the N(0,1)!

🙋 Have you concluded you need to perform the z-test? Head straight to our z-test calculator !

What is a t-test?

A t-test is a widely used statistical test that analyzes the means of one or two groups of data. For instance, a t-test is performed on medical data to determine whether a new drug really helps.

What are different types of t-tests?

Different types of t-tests are:

• One-sample t-test;
• Two-sample t-test; and
• Paired t-test.

How to find the t value in a one sample t-test?

To find the t-value:

• Subtract the null hypothesis mean from the sample mean value.
• Divide the difference by the standard deviation of the sample.
• Multiply the resultant with the square root of the sample size.

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Choose test type

t-test for the population mean, μ, based on one independent sample . Null hypothesis H 0 : μ = μ 0  

Alternative hypothesis H 1

Test details

Significance level α

The probability that we reject a true H 0 (type I error).

Degrees of freedom

Calculated as sample size minus one.

Test results

Calcworkshop

One Sample T Test Easily Explained w/ 5+ Examples!

// Last Updated: October 9, 2020 - Watch Video //

Did you know that a hypothesis test for a sample mean is the same thing as a one sample t-test?

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

Learn the how-to with 5 step-by-step examples.

Let’s go!

What is a One Sample T Test?

A one sample t-test determines whether or not the sample mean is statistically different (statistically significant) from a population mean.

While significance tests for population proportions are based on z-scores and the normal distribution, hypothesis testing for population means depends on whether or not the population standard deviation is known or unknown.

For a one sample t test, we compare a test variable against a test value. And depending on whether or not we know the population standard deviation will determine what type of test variable we calculate.

T Test Vs. Z Test

So, determining whether or not to use a z-test or a t-test comes down to four things:

• Are we are working with a proportion (z-test) or mean (z-test or t-test)?
• Do you know the population standard deviation (z-test)?
• Is the population normally distributed (z-test)?
• What is the sample size? If the sample is less than 30 (t-test), if the sample is larger than 30 we can apply the central limit theorem as population is approximately normally.

How To Calculate a Test Statistic

Standard deviation known.

If the population standard deviation is known , then our significance test will follow a z-value. And as we learned while conducting confidence intervals, if our sample size is larger than 30, then our distribution is normal or approximately normal. And if our sample size is less than 30, we apply the Central Limit Theorem and deem our distribution approximately normal.

Z Test Statistic Formula

Standard Deviation Unknown

If the population standard deviation is unknown , we will use a sample standard deviation that will be close enough to the unknown population standard deviation. But this will also cause us to have to use a t-distribution instead of a normal distribution as noted by StatTrek .

Just like we saw with confidence intervals for population means, the t-distribution has an additional parameter representing the degrees of freedom or the number of observations that can be chosen freely.

T Test Statistic Formula

This means that our test statistic will be a t-value rather than a z-value. But thankfully, how we find our p-value and draw our final inference is the same as for hypothesis testing for proportions, as the graphic below illustrates.

How To Find The P Value

Example Question

For example, imagine a company wants to test the claim that their batteries last more than 40 hours. Using a simple random sample of 15 batteries yielded a mean of 44.9 hours, with a standard deviation of 8.9 hours. Test this claim using a significance level of 0.05.

One Sample T Test Example

How To Find P Value From T

So, our p-value is a probability, and it determines whether our test statistic is as extreme or more extreme then our test value, assuming that the null hypothesis is true. To find this value we either use a calculator or a t-table, as we will demonstrate in the video.

We have significant evidence to conclude the company’s claim that their batteries last more than 40 hours.

What Does The P Value Mean?

Together we will work through various examples of how to create a hypothesis test about population means using normal distributions and t-distributions.

One Sample T Test – Lesson & Examples (Video)

• Introduction to Video: One Sample t-test
• 00:00:43 – Steps for conducting a hypothesis test for population means (one sample z-test or one sample t-test)
• Exclusive Content for Members Only
• 00:03:49 – Conduct a hypothesis test and confidence interval when population standard deviation is known (Example #1)
• 00:13:49 – Test the null hypothesis when population standard deviation is known (Example #2)
• 00:18:56 – Use a one-sample t-test to test a claim (Example #3)
• 00:26:50 – Conduct a hypothesis test and confidence interval when population standard deviation is unknown (Example #4)
• 00:37:16 – Conduct a hypothesis test by using a one-sample t-test and provide a confidence interval (Example #5)
• 00:49:19 – Test the hypothesis by first finding the sample mean and standard deviation (Example #6)
• Practice Problems with Step-by-Step Solutions
• Chapter Tests with Video Solutions

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What is a Hypothesis Test?

A quick search for hypothesis tests online gives us several websites with short definitions. Here’s one from a quick definition from the Stat Trek:

A statistical hypothesis is an assumption about a population parameter. This assumption may or may not be true. Hypothesis testing refers to the formal procedures used by statisticians to accept or reject statistical hypotheses.

What is Hypothesis Testing? From Stat Trek

Most websites will have a similar definition or introduction, followed by a number of components, notation, key terminology, and examples.

The Basic Idea

Hypothesis tests show up in many areas of our everyday lives, but they are kind of sneaky. The basic structure of a hypothesis test is very much like a science project from elementary, middle, or high school. You have a problem, hypothesis, data collection, some computations, results or conclusions. What follows next are a few examples of what the hypothesis test and results would look like in journals or other publications, and how those results are presented to the public.

Some Examples of Hypothesis Tests

Example 1: agility testing in youth football (soccer)players; evaluating reliability, validity, and correlates of newly developed testing protocols.

Reactive agility (RAG)and change of direction speed (CODS) were analyzed in 13U and 15U youth soccer players. “Independent samples t-test indicated significant differences between U13 and U15 in S10 (t-test: 3.57, p < 0.001), S20M (t-test: 3.13, p < 0.001), 20Y (t-test: 4.89, p < 0.001), FS_RAG (t-test: 3.96, p < 0.001), and FS_CODS (t-test: 6.42, p < 0.001), with better performance in U15. Starters outperformed non-starters in most capacities among U13, but only in FS_RAG among U15 (t-test: 1.56, p < 0.05).”

Most of this might seem like gibberish for now, but essentially the two groups were analyzed and compared, with significant differences observed between the groups.

Source: https://pubmed.ncbi.nlm.nih.gov/31906269/

Example 2: Manual therapy in the treatment of carpal tunnel syndrome in diabetic patients: A randomized clinical trial

Thirty diabetic patients with carpal tunnel syndrome were split up into two groups. One received physiotherapy modality and the other received manual therapy. “Paired t-test revealed that all of the outcome measures had a significant change in the manual therapy group, whereas only the VAS and SSS changed significantly in the modality group at the end of 4 weeks. Independent t-test showed that the variables of SSS, FSS and MNT in the manual therapy group improved significantly greater than the modality group.”

Source: https://pubmed.ncbi.nlm.nih.gov/30197774/

Example 3: Omega-3 fatty acids decreased irritability of patients with bipolar disorder in an add-on, open label study

“The initial mean was 63.51 (SD 34.17), indicating that on average, subjects were irritable for about six of the previous ten days. The mean for the last recorded percentage was less than half of the initial score: 30.27 (SD 34.03). The decrease was found to be statistically significant using a paired sample t-test (t = 4.36, 36 df, p < .001).”

Source: https://nutritionj.biomedcentral.com/articles/10.1186/1475-2891-4-6

Example 4: Evaluating the Efficacy of COVID-19 Vaccines

“We reduced all values of vaccine efficacy by 30% to reflect the waning of vaccine efficacy against each endpoint over time. We tested the null hypothesis that the vaccine efficacy is 0% versus the alternative hypothesis that the vaccine efficacy is greater than 0% at the nominal significance level of 2.5%.”

Source: https://www.medrxiv.org/content/10.1101/2020.10.02.20205906v2.full

Example 5: Social Isolation During COVID-19 Pandemic. Perceived Stress and Containment Measures Compliance Among Polish and Italian Residents

“The Polish group had a higher stress level than the Italian group (mean PSS-10 total score 22,14 vs 17,01, respectively; p < 0.01). There was a greater prevalence of chronic diseases among Polish respondents. Italian subjects expressed more concern about their health, as well as about their future employment. Italian subjects did not comply with suggested restrictions as much as Polish subjects and were less eager to restrain from their usual activities (social, physical, and religious), which were more often perceived as “most needed matters” in Italian than in Polish residents.”

Source: https://www.frontiersin.org/articles/10.3389/fpsyg.2021.673514/full

Example 6: A Comparative Analysis of Student Performance in an Online vs. Face-to-Face Environmental Science Course From 2009 to 2016

“The independent sample t-test showed no significant difference in student performance between online and F2F learners with respect to gender [t(145) = 1.42, p = 0.122].”

Source: https://www.frontiersin.org/articles/10.3389/fcomp.2019.00007/full

But what does it all mean?

That’s what comes next. The examples above span a variety of different types of hypothesis tests. Within this chapter we will take a look at some of the terminology, formulas, and concepts related to Hypothesis Testing for 1 Sample.

Key Terminology and Formulas

Hypothesis: This is a claim or statement about a population, usually focusing on a parameter such as a proportion (%), mean, standard deviation, or variance. We will be focusing primarily on the proportion and the mean.

Hypothesis Test: Also known as a Significance Test or Test of Significance , the hypothesis test is the collection of procedures we use to test a claim about a population.

Null Hypothesis: This is a statement that the population parameter (such as the proportion, mean, standard deviation, or variance) is equal to some value. In simpler terms, the Null Hypothesis is a statement that “nothing is different from what usually happens.” The Null Hypothesis is usually denoted by [latex]H_{0}[/latex], followed by other symbols and notation that describe how the parameter is the same as some value.

Alternative Hypothesis: This is a statement that the population parameter (such as the proportion, mean, standard deviation, or variance) is somehow different the value involved in the Null Hypothesis. For our examples, “somehow different” will involve the use of [latex] [/latex], or [latex]\neq[/latex]. In simpler terms, the Alternative Hypothesis is a statement that “something is different from what usually happens.” The Alternative Hypothesis is usually denoted by [latex]H_{1}[/latex], [latex]H_{A}[/latex], or [latex]H_{a}[/latex], followed by other symbols and notation that describe how the parameter is different from some value.

Significance Level: We previous learned about the significance level as the “left over” stuff from the confidence level. This is still true, but we will now focus more on the significance level as its own value, and we will use the symbol alpha, [latex]\alpha[/latex]. This looks like a lowercase “a,” or a drawing of a little fish. The significance level [latex]\alpha[/latex] is the probability of rejecting the null hypothesis when it is actually true (more on what this means in the next section). The common values are still similar to what we had previously, 1%, 5%, and 10%. We commonly write these as decimals instead, 0.01, 0.05, and 0.10.

Test Statistic:  One of the key components of a hypothesis test is what we call a  test statistic . This is a calculation, sort of like a z-score, that is specific to the type of test being conducted. The idea behind a test statistic, relating it back to science projects, would be like calculations from measurements that were taken. In this chapter we will address the test statistic for 1 proportion, 1 mean when we know [latex]\sigma[/latex], and 1 mean with [latex]\sigma[/latex] unknown. The formulas are listed in the table below:

Critical Region: The critical region , also known as the rejection region , is the area in the normal (or other) distribution in which we reject the null hypothesis. Think of the critical region  like a target area that you are aiming for. If we are able to get a value in this region, it means we have evidence for the claim.

Critical Value: These are like special z-scores for us; the critical value  (or values, sometimes there are two) separates the critical region from the rest of the distribution. This is the non-target part, or what we are not aiming for. If our value is in this region, we do not have evidence for the claim.

P-Value: This is a special value that we compute. If we assume the null hypothesis is true, the p-value represents the probability that a test statistic is at least as extreme as the one we computed from our sample data; for us the test statistics would be either [latex]z[/latex] or [latex]t[/latex].

Decision Rule for Hypothesis Testing:  There are a few ways we can arrive at our decision with a hypothesis test. We can arrive at our conclusion by using confidence intervals, critical values (also known as traditional method), and using p-values. Relating this to a science project, the decision rule would be what we take into consideration to arrive at our conclusion. When we make our decision, the wording will sound a little strange. We’ll say things like “we have enough evidence to reject the null hypothesis” or “there is insufficient evidence to reject the null hypothesis.”

Decision Rule with Critical Values:  If the test statistic is in the critical region, we have enough evidence to reject the null hypothesis. We can also say we have sufficient evidence to support the claim. If the test statistic is not in the critical region, we fail to reject the null hypothesis. We can also say we do not have sufficient evidence to support the claim.

Decision Rule with P-Values: If the p-value is less than or equal to the significance level, we have enough evidence to reject the null hypothesis. We can also say we have sufficient evidence to support the claim. If the p-value is greater than the significance level, we fail to reject the null hypothesis. We can also say we do not have sufficient evidence to support the claim.

More About Hypotheses

Writing the Null and Alternative Hypothesis can be tricky. Here are a few examples of claims followed by the respective hypotheses:

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5.3 - hypothesis testing for one-sample mean.

In the previous section, we learned how to perform a hypothesis test for one proportion. The concepts of hypothesis testing remain constant for any hypothesis test. In these next few sections, we will present the hypothesis test for one mean. We start with our knowledge of the sampling distribution of the sample mean.

Hypothesis Test for One-Sample Mean Section  

Recall that under certain conditions, the sampling distribution of the sample mean, \(\bar{x} \), is approximately normal with mean, \(\mu \), standard error \(\dfrac{\sigma}{\sqrt{n}} \), and estimated standard error \(\dfrac{s}{\sqrt{n}} \).

\(H_0\colon \mu=\mu_0\)

Conditions:

• The distribution of the population is Normal
• The sample size is large \( n>30 \).

Test Statistic:

If at least one of conditions are satisfied, then...

\( t=\dfrac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}} \)

will follow a t-distribution with \(n-1 \) degrees of freedom.

Notice when working with continuous data we are going to use a t statistic as opposed to the z statistic. This is due to the fact that the sample size impacts the sampling distribution and needs to be taken into account. We do this by recognizing “degrees of freedom”. We will not go into too much detail about degrees of freedom in this course.

Let’s look at an example.

Example 5-1 Section  

This depends on the standard deviation of \(\bar{x} \) . 

\begin{align} t^*&=\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\&=\dfrac{8.3-8.5}{\frac{1.2}{\sqrt{61}}}\\&=-1.3 \end{align} 

Thus, we are asking if \(-1.3\) is very far away from zero, since that corresponds to the case when \(\bar{x}\) is equal to \(\mu_0 \). If it is far away, then it is unlikely that the null hypothesis is true and one rejects it. Otherwise, one cannot reject the null hypothesis. 

The Complete Guide: Hypothesis Testing in Excel

In statistics, a hypothesis test is used to test some assumption about a population parameter .

There are many different types of hypothesis tests you can perform depending on the type of data you’re working with and the goal of your analysis.

This tutorial explains how to perform the following types of hypothesis tests in Excel:

• One sample t-test
• Two sample t-test
• Paired samples t-test
• One proportion z-test
• Two proportion z-test

Let’s jump in!

Example 1: One Sample t-test in Excel

A one sample t-test is used to test whether or not the mean of a population is equal to some value.

For example, suppose a botanist wants to know if the mean height of a certain species of plant is equal to 15 inches.

To test this, she collects a random sample of 12 plants and records each of their heights in inches.

She would write the hypotheses for this particular one sample t-test as follows:

• H 0 :  µ = 15
• H A :  µ ≠15

Refer to this tutorial for a step-by-step explanation of how to perform this hypothesis test in Excel.

Example 2: Two Sample t-test in Excel

A two sample t-test is used to test whether or not the means of two populations are equal.

For example, suppose researchers want to know whether or not two different species of plants have the same mean height.

To test this, they collect a random sample of 20 plants from each species and measure their heights.

The researchers would write the hypotheses for this particular two sample t-test as follows:

• H 0 :  µ 1 = µ 2
• H A :  µ 1 ≠ µ 2

Example 3: Paired Samples t-test in Excel

A paired samples t-test is used to compare the means of two samples when each observation in one sample can be paired with an observation in the other sample.

For example, suppose we want to know whether a certain study program significantly impacts student performance on a particular exam.

To test this, we have 20 students in a class take a pre-test. Then, we have each of the students participate in the study program for two weeks. Then, the students retake a post-test of similar difficulty.

We would write the hypotheses for this particular two sample t-test as follows:

• H 0 :  µ pre = µ post
• H A :  µ pre ≠ µ post

Example 4: One Proportion z-test in Excel

A  one proportion z-test  is used to compare an observed proportion to a theoretical one.

For example, suppose a phone company claims that 90% of its customers are satisfied with their service.

To test this claim, an independent researcher gathered a simple random sample of 200 customers and asked them if they are satisfied with their service.

• H 0 : p = 0.90
• H A : p ≠ 0.90

Example 5: Two Proportion z-test in Excel

A two proportion z-test is used to test for a difference between two population proportions.

For example, suppose a s uperintendent of a school district claims that the percentage of students who prefer chocolate milk over regular milk in school cafeterias is the same for school 1 and school 2.

To test this claim, an independent researcher obtains a simple random sample of 100 students from each school and surveys them about their preferences.

• H 0 : p 1 = p 2
• H A : p 1  ≠ p 2

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Statistics and probability

Course: statistics and probability   >   unit 12, simple hypothesis testing.

• Idea behind hypothesis testing
• Examples of null and alternative hypotheses
• Writing null and alternative hypotheses
• P-values and significance tests
• Comparing P-values to different significance levels
• Estimating a P-value from a simulation
• Estimating P-values from simulations
• Using P-values to make conclusions

• Your answer should be
• an integer, like 6 ‍  
• an exact decimal, like 0.75 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• a percent, like 12.34 % ‍  
• (Choice A)   We cannot reject the hypothesis. A We cannot reject the hypothesis.
• (Choice B)   We should reject the hypothesis. B We should reject the hypothesis.

Enter a value for the null hypothesis. This value should indicate the absence of an effect in your data. Indicate whether your alternative hypothesis involves one-tail or two-tails. If it is a one-tailed test, then you need to indicate whether it is a positive (right tail) test or a negative (left tail) test.

Enter an \(\alpha\) value for the hypothesis test. This is the Type I error rate for your hypothesis test. It also determines the confidence level \(100 \times (1-\alpha)\) for a confidence interval.

Press the Run Test button and a table summarizing the computations and conclusions will appear below.

Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.

Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement. The statement is true.

Yes, that is correct. Hypothesis testing involves collecting sample evidence and using probability theory to analyze the likelihood of the hypothesis being true. The process involves setting up a null hypothesis , which is the statement being tested, and an alternative hypothesis. The sample evidence is then gathered and analyzed using statistical tests to determine the probability of observing the results if the null hypothesis were true. If the probability is sufficiently low, the null hypothesis is rejected in favour of the alternative hypothesis . The use of probability theory in hypothesis testing allows for a more objective and rigorous approach to drawing conclusions from data.

The complete question is:-

Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement. True/False

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Related Questions

Erin's water bottle holds 665 milliliters Dylan is carrying two water bottles each one hold 0.35 liters who is carrying more water how much more

Dylan is carrying more water, his two water bottles is 700 millilitres

Erin water bottle holds 665 millilitres

Dylan is carrying two water bottles, each bottle is 0.35 litres

0.35 + 0.35

0.7 liters converted to millilitres is

= 0.7 × 1000

= 700 millilitres

Erin's water bottle is 665 millilitres

Dylan two water bottles is a total of 700 millilitres

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A researcher says that her analysis shows that the effect of the independent variable on the dependent variable is statistically significant. This means the effect is:

The statement means that the observed relationship between the independent variable and the dependent variable is "very unlikely to occur by chance."

In statistical analysis, the concept of statistical significance helps determine if the results obtained from a study are reliable and not due to random chance. When an effect is deemed statistically significant, it suggests that the relationship between the independent variable (the variable being manipulated or studied) and the dependent variable (the variable being measured or observed) is likely to exist in the population from which the sample was drawn.

By reaching statistical significance, the researcher can confidently conclude that the observed effect is more than just a random occurrence and has practical implications in the real world, thus lending support to the underlying hypothesis or research question.

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What is the ratio of raccoons to total animals?

We can see here that the ratio of   raccoons to total animals is: B. 4:6

We can define ratio in mathematics as a division-based comparison of two numbers or quantities. It conveys how big or much one quantity is in proportion to another.

A fraction is a common way to express ratios , with the first number denoting how much of the first quantity there is, and the second number denoting how much of the second quantity there is.

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Help with what u can please!

The building is 51 feet tall

The value of x is 1.5

The length of the building is solved by comparing the building's height by John's height as this forms similar triangles

The comparison is done using the proportions as below

John's shadow / John's height = building's shadow / building's height

6 / 4.5 = 68 / building's height

cross multiplying

building's height x 6 = 4.5 * 68

building's height = 4.5 x 68 / 6

building's height = 51 feet

Solving for x

Using proportions

15 / 10 = 23x / 20x - 8

(20x - 8) * 15 = 22x * 10

300x - 120 = 220x

300x - 220x = 120

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Which equation represents the parabola with focus ( 1 , 4 ) and latus rectum of 4? y = − 1 4 ( x − 1 ) 2 + 3 y = 4 ( x − 1 ) 2 + 3 y = ( x − 1 ) 2 + 4 y = 1 4 ( x − 1 ) 2 + 3

The equation of parabola with with focus (1,4)and latus rectum of 4 is given as y=(x−1)2+4, hence option C is correct.

We know that the focus of the parabola is at (1, 4), and the latus rectum is 4 units long. The latus rectum is the line segment that passes through the focus and is perpendicular to the axis of symmetry. Its length is equal to 4 times the distance from the focus to the directrix . Since the latus rectum is 4 units long, the distance from the focus to the directrix is 1 unit.

Therefore, the directrix is the horizontal line y = 3, which is 1 unit below the focus. The axis of symmetry is the vertical line x = 1. The standard form of the equation of a parabola with vertex at the origin, axis of symmetry along the x-axis, and focus at (p, 0) is:

In this case, the vertex is (1, 0) and the focus is (1, 4). Therefore, we can write,

x - 1 = 4p(y - 0)

x - 1 = 4py

We know that the distance from the focus to the vertex is p, which is 4 units in this case. Therefore, the equation of the parabola is,

(x - 1)² = 16y

Simplifying this equation , we get,

y = (1/16)(x - 1)²

Therefore, the answer is C. y = (x - 1)²/16.

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Complete question - Which equation represents the parabola with focus (1,4) and latus rectum of 4?

A. y=−14(x−1)²+3

B. y=4(x−1)²+3

C. y=(x−1)²+4

D. y=14(x−1)²+3

We calculated that the value of the test statistic is z = 3.53. Next we will calculate the P-value. Since the alternative hypothesis is of the form p > Po, the P-value is the area under the z curve to the right of the calculated value of the test statistic. Using SALT, we find that the cumulative probability associated with a z test statistic of 3.53, rounded to four decimal places, is 1.6449. We can use the cumulative probability to find the P-value. (Round your answers to four decimal places.)P-value = (z curve area to the right of 3.53) = P(z < 3.53) = 1 1.6449 xP Value is =

Cumulative probabilities are associated with the area to the left of the test statistic, you will subtract the cumulative probability from 1 to find the area to the right.P-value = 1 - P(z < 3.53)

Once you have the correct cumulative probability, plug it into the equation and calculate the P-value. After calculating the test statistic z = 3.53, you need to find the P-value . Since the alternative hypothesis is of the form p > Po, the P-value is the area under the z curve to the right of the calculated value of the test statistic. Using SALT, you found the cumulative probability associated with a z test statistic of 3.53, rounded to four decimal places, is 1.6449. However, the value 1.6449 seems to be incorrect as the cumulative probability should be less than 1. To find the P-value, you need to calculate the area under the z curve to the right of z = 3.53.

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please help i need a good explanation!! thanks!! :)

The sides of the quadrilateral are

When the quadrilateral are similar according to the ratio 7 : 11 and the similar sides are:

Since FJ : AD the ratio is 11 : 7, therefore where FJ = 11 then AD = 7 = x

AB = 2x = 2 * 7 = 14

7 * FG = 14 * 11

GH ⇔ BC = 3 * 7 = 21

7 * GH = 21 * 11

HJ ⇔ CD = 4 * 7 = 28

7 * GH = 28 * 11

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let a represent the average value of the function f(x) on the interval [0.6]. is there a value of c for which the average value of f(x) on the interval [0. c] is greater than a? explain why or why

The average value of a function f(x) on an interval [0, 6] is represented by 'a'. To determine if there is a value 'c' for which the average value of f(x) on the interval [0, c] is greater than 'a', we need to consider the properties of the function and the Mean Value Theorem. The Mean Value Theorem states that if a function f(x) is continuous on the interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point 'c' in the interval (a, b) such that the average rate of change equals the instantaneous rate of change or f'(c) = (f(b) - f(a)) / (b - a). Without more information about the function f(x), we cannot definitively say whether there is a value 'c' for which the average value of f(x) on the interval [0, c] is greater than the average value on the interval [0, 6]. However, if the function meets the conditions of the Mean Value Theorem, it is possible that such a value 'c' exists.

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At a track meet, a relay race with a total distance of 1\2 mile is run by a team of 3 runners. Each runner completed the same distance in the race. Which of these models could be used to find the distance, in miles, run by each of the 3 runners? Choose the THREE correct answers.

First models could be used to find the distance , in miles, run by each of the 3 runners.

A relay race with a total distance of 1/2 mile is run by a team of 3 runners.

So, Distance Covered by each runner is

= (1/3) (1/2)  

So, (1/2) x (1/3) = 1/6

(1/2) ÷ 3 = 1/6

So, the correct model   1/2 mile is divided in 3 equal parts are the correct options.

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You owe $1,500 on a credit card with a 14.5% APR. The minimum payment is $80. How much goes toward the principal if you make the minimum payment at the end of the first month?

If you make the minimal payment of $80 on the end of the first month, $61.85 will cross towards the principal and $18.15 will go towards interest charges.

To calculate how much of the minimal charge goes closer to the principal, we first need to calculate the amount of interest that accrues at the outstanding balance for the first month.

Interest charged for the month = monthly interest fee x outstanding stability

Now we are able to calculate how a good deal of the minimal fee goes closer to the principal :

Consequently, if you make the minimal payment of $80 on the end of the first month, $61.85 will cross towards the principal and $18.15 will go towards interest charges.

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Find the missing side of the triangle. 10 cm and 6 cm. What is X??????? I missed my RSM class and now this?!

The missing side of triangle is 11.66 cm, under the condition that the given triangle is a right angled triangle and the other sides of the triangle are 10cm, 6cm respectively.

In order to evaluate the other side of the triangle, we have to rely on the principles of Pythagorean Theorem which states that in a right triangle, the sum of the square sides are equal to the square of the hypotenuse side.

Then, let us consider that x be the length of the missing side then

x² = 10² + 6²

x² = 100 + 36

x ≈ 11.66 cm

Then, the missing side of this triangle is approximately 11.66 cm.

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The complete question is

Find the missing siside of the following right triangle in the figure

How are databases of variants used to help find disease gene candidates? Variants present in individuals that do not have the disease or are common in the general population are unlikely to cause a rare genetic disease. They can indicate the probability of different variants appearing in a population. Variants from individuals that do not have the disease are not useful for this purpose. They can indicate rare variants that will help with the genetic identification of individuals

Databases of variants are crucial in helping to identify disease gene candidates. Here's how they are used:

1. Collect and store genetic variants: Databases collect and store genetic variants from various individuals, including those with specific diseases and those without.

2. Compare variants between groups: By comparing the variants in affected individuals to those in unaffected individuals, researchers can identify variants that are more common in the disease group. This helps to narrow down the list of potential disease-causing genes.

3. Calculate probability : Databases can be used to calculate the probability of certain variants appearing in the general population. If a variant is rare in the general population but more common in individuals with a specific disease, it may be more likely to be a disease-causing variant.

4. Filter out common variants : As you mentioned, variants that are common in the general population or present in individuals without the disease are less likely to cause a rare genetic disease. By filtering out these common variants, researchers can focus on rare variants that may have a stronger association with the disease. 5. Identify disease gene candidates: Through this process of comparing and filtering genetic variants, researchers can identify potential disease gene candidates that warrant further investigation.

In summary, databases of variants help researchers identify disease gene candidates by storing genetic information, enabling comparison between affected and unaffected individuals, calculating variant probabilities, filtering out common variants, and ultimately pinpointing rare variants that may be associated with specific genetic diseases.

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a sailfish can travel as fast as 68 miles per hour. at that rate, how far would a sailfish travel in 45 minutes

Therefore, a sailfish would travel 51 miles in 45 minutes if it maintained a speed of 68 miles per hour in the equation .

To calculate the distance traveled in a given time, we can use the formula distance = rate x time, where rate refers to the speed of travel and time refers to the duration of travel. In this scenario, we are given a rate of 68 miles per hour and a time of 45 minutes.

To use the formula , we first need to convert the time to hours since the rate is given in miles per hour. We do this by dividing the time by 60, since there are 60 minutes in an hour. In this case, 45 minutes divided by 60 minutes per hour gives us 0.75 hours.

Now, we can plug in the values for rate and time into the formula and solve for distance. Multiplying 68 miles per hour by 0.75 hours gives us a distance of 51 miles.

45 minutes / 60 minutes per hour = 0.75 hours

Then, we can use the formula: distance = rate x time

distance = 68 miles per hour x 0.75 hours

distance = 51 miles

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Use the discriminant to determine whether each quadratic equation has two real solutions, a double root, or no real solution, without solving the equation. 2a2 + 3a + 1 = 0 2a(a – 3) – 3 = 3 3a2 – 2a(a – 2)2 – 4 = 0

The discriminant of a quadratic equation is the expression b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c. To determine whether each quadratic equation has two real solutions, a double root, or no real solution, we need to look at the value of the discriminant. For the equation 2a^2 + 3a + 1 = 0, the discriminant is b^2 - 4ac = 3^2 - 4(2)(1) = 1. Since the discriminant is positive and not equal to zero, there are two real solutions. For equation 2a(a – 3) – 3 = 3, we need to rearrange it into standard form first, which gives us 2a^2 - 6a - 6 = 0. The discriminant is b^2 - 4ac = (-6)^2 - 4(2)(-6) = 60. Since the discriminant is positive, there are two real solutions. For the equation 3a^2 – 2a(a – 2)^2 – 4 = 0, we need to expand the squared term first, which gives us 3a^2 - 2a(a^2 - 4a + 4) - 4 = 0. Simplifying this equation gives us 3a^2 - 2a^3 + 8a - 4 = 0. The discriminant is b^2 - 4ac = 8^2 - 4(3)(-2) = 100. Since the discriminant is positive, there are two real solutions. Therefore, all three quadratic equations have two real solutions.

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In a weighted grading system, students are graded on quizzes, tests, and a project, each with a different weight. Matrix W represents the weights for each kind of work, and matrix G represents the grades for two students, Felipe and Helena. Q T P W = [0.40 0.50 0.10] Felipe Helena G= Q {80 70} T {60 80} p { 90 60} Final grades are represented in a matrix F. If F = WG, what is F? A. [7174] B. [7174] C. [7471] D. [7471]

For Felipe and Helena's final grades, the solution is option C, [74 71].

Using the given values for Q, T, and P weights and Felipe and Helena's grades, calculate their final grades as follows:

Felipe's final grade:

0.40 x 80 + 0.50 x 60 + 0.10 x 90 = 32 + 30 + 9 = 71

Helena's final grade:

0.40 x 70 + 0.50 x 80 + 0.10 x 60 = 28 + 40 + 6 = 74

To represent the final grades for Felipe and Helena in a matrix F , given formula F = WG, where W = matrix of weights and G = matrix of grades:

[0.40 0.50 0.10]   [80 70]

F = WG = [0.40 0.50 0.10] x [60 80]

[0.40 0.50 0.10] [90 60]

Performing matrix multiplication :

[32 + 30 + 9  28 + 40 + 6]

F = WG = [32 + 40 + 6 28 + 40 + 3]

[36 + 25 + 6 36 + 20 + 3]

Simplifying:

F = WG = [78 71]

Therefore, [74 71] for Felipe and Helena's final grades, respectively.

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If f(x) = 2x - 9, which of the following are correct? Select all that apply. f(-3) = 15 f(-1) = -11 f(0) = -9 f(2) = 5 f(3) = -3

f(0)=-9 and f(3)=-3

Step-by-step explanation:

f(-3)=15 - so to solve this we have to replace x with -3, which would equal:

2(-3)-9=-15.  -15 doesn't equal 15, so this is incorrect.

f(-1)=-11 Substitute x for -1: 2(-1)-9=-12.  -12 doesn't equal -11, so this is also incorrect.

f(0)=-9 Substitute x for 0.  2(0)-9= -9.  -9 does equal -9, so this is correct.

f(2)=5 Substitute x for 2.  2(2)-9=-5. This isn't correct.

f(3)=-3 Substitute x for 3.  2(3)-9=-3.  This is also correct because -3 does equal -3.

Hope this helps!  :)

Consider a circle whose equation is x2 + y2 – 2x – 8 = 0. Which statements are true? Select three options. The radius of the circle is 3 units. The center of the circle lies on the x-axis. The center of the circle lies on the y-axis. The standard form of the equation is (x – 1)² + y² = 3. The radius of this circle is the same as the radius of the circle whose equation is x² + y² = 9

From the equation of the circle , the radius of the circle is 3 and the center lies at (1, 0)

A circle is a closed curve that is drawn from the fixed point called the center, in which all the points on the curve are having the same distance from the center point of the center. The equation of a circle with (h, k) center and r radius is given by:

(x-h)^2 + (y-k)^2 = r^2

This is the standard form of the equation. Thus, if we know the coordinates of the center of the circle and its radius as well, we can easily find its equation.

In this problem, we have an equation x^2 + y^2 - 2x - 8 == 0

The radius of this circle is 3 and the center is at (1, 0).

The correct option are A and C

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Suppose that we want to estimate what proportions of all drivers exceed the legal speed limit on a certain stretch of road between Los Angeles and Bakersfield. Use the formula of the earlier exercise to determine how large a sample we will need to be at least 99 % confident that the resulting estimate, the sample proportion, is off by less than 0.04.

We need a sample size of at least 665 drivers to estimate the proportion of all drivers exceeding the legal speed limit on the certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level.

To estimate the proportion of all drivers exceeding the legal speed limit on a certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level, we need to use the following formula: [tex]n = (Z^2 * p * (1 - p)) / E^2[/tex] where n is the sample size , Z is the Z-score for the desired confidence level (2.576 for 99% confidence level), p is the estimated proportion of drivers exceeding the speed limit (we don't have an estimate, so we'll use 0.5 for maximum variability), and E is the margin of error we want (0.04). Plugging in the values, we get: [tex]n = (2.576^2 * 0.5 * (1 - 0.5)) / 0.04^2[/tex] n = 664.92 Therefore, we need a sample size of at least 665 drivers to estimate the proportion of all drivers exceeding the legal speed limit on the certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level.

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Can someone help me asap? It’s due today!! I will give brainliest if it’s correct

The student government association at a university wants to estimate the percentage of the student body that supports a change being considered in the academic calendar of the university for the next academic year. How many students should be surveyed if a 99% confidence interval is desired and the margin of error is to be only 8%? The student government should survey Student (Round up to the nearest integer.)

The student government should survey 640 students to estimate the percentage of the student body that supports the proposed change in the academic calendar with a 99% confidence level and a margin of error of 8%.

To determine the sample size needed for the survey, we can use the following formula: n = (z^2 * p * q) / E^2 Where: n = sample size z = z-score for the desired confidence level (99%) p = estimated proportion of the population supporting the change (unknown, so we'll use 0.5 for a conservative estimate ) q = 1 - p E = margin of error (0.08) Plugging in the values, we get: n = (2.576^2 * 0.5 * 0.5) / 0.08^2 n = 639.53 Since we can't have a fractional number of students, we should round up to the nearest integer: n = 640 Learn more about desired confidence level here:

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a manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 408 gram setting. it is believed that the machine is underfilling or overfilling the bags. a 13 bag sample had a mean of 399 grams with a variance of 121 . assume the population is normally distributed. is there sufficient evidence at the 0.02 level that the bags are underfilled or overfilled?

After the test statistic we find that the critical t-value is 2.681. Since the absolute value of the test statistic (-2.78) is greater than the critical t-value (2.681), we can reject the null hypothesis.

The test statistic can be calculated as follows: t = (399 - 408) / (sqrt(121/13)) = -2.78  Since the absolute value of the test statistic (-2.78) is greater than the critical t-value (2.681), we can reject the null hypothesis.

To answer your question, we will conduct a hypothesis test to determine if there is sufficient evidence at the 0.02 significance level that the bags are underfilled or overfilled.

Step 1: State the hypotheses H0 (null hypothesis): The population means (μ) is 408 grams. H1 (alternative hypothesis): The population means (μ) is not equal to 408 grams. Step 2: Determine the test statistic Since we know the population variance, we will use a z-test. The formula for the z-test statistic is: z = (sample mean - population mean) / (population standard deviation/sqrt (sample size)) z = (399 - 408) / (sqrt(121) / sqrt(13)) Step 3: Calculate the z-value z = (-9) / (11 / sqrt(13)) z ≈ -2.58 Step 4: Determine the critical value For a two-tailed test at the 0.02 significance level , we need to find the critical value. Using a z-table, we find the critical values are approximately -2.33 and +2.33. Step 5: Compare the z-value to the critical values Our calculated z-value (-2.58) is less than the lower critical value (-2.33). Step 6: Draw a conclusion Since our z-value falls in the rejection region, we reject the null hypothesis. There is sufficient evidence at the 0.02 significance level to conclude that the chocolate chip bag-filling machine is either underfilling or overfilling the bags when set to 408 grams, assuming the population is normally distributed.

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If p=3 what is the value of r when r =19 -5p

Check all statements that are true about the Pythagorean Theorem Only used on Right Triangles The hypotenuse is the shortest side of the triangle The legs are sides a and b of the triangles The last step is to take the square root of both sides Exponents are NOT a part of the pythagorean theorem

The true statements about the Pythagorean theorem are :

a) Only used on Right Triangles

b) The legs are sides a and b of the triangles

c) The last step is to take the square root of both sides

Given data ,

The Pythagorean Theorem is a mathematical theorem that applies only to right triangles , and it states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the two other sides, called the legs.

a² + b² = c²

where a and b are the lengths of the legs, and c is the length of the hypotenuse

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At the beginning of the year, a company estimates total overhead costs of $1,309,620. The company applies overhead using machine hours and estimates that it will use 2,990 machine hours during the year. What amount of overhead should be applied to a job that uses 30 machine hours that year?

If at the beginning of the year, a company estimates total overhead costs of $1,309,620. The amount of overhead should be applied to a job that uses 30 machine hours that year is: 13,140.

Overhead rate per machine hour = Total estimated overhead costs / Estimated number of machine hour

Overhead rate per machine hour = $1,309,620 / 2,990

Overhead rate per machine hour  = $438 per machine hour

Overhead applied to job = Overhead rate per machine hour x Number of machine hours used

Overhead applied to job = $438 x 30

Overhead applied to job = $13,140

Therefore, the amount of overhead is $13,140.

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rearrange the following steps in the correct order to find the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails. Rank the options below. The probability is 1/16 1/16 Of these, only one will result in four heads appearing, namely THHHH. There are 16 equally likely outcomes of flipping a fair coin five times in which the first flip comes up tails.

The correct order of steps is 1, 2, 3, 4. And the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails , is 1/16.

To find the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails, we need to follow these steps in order: 1. Identify the total number of possible outcomes when a fair coin is flipped five times, which is 2^5 = 32. 2. Determine the number of outcomes in which the first flip is tails, which is also 16. 3. Out of the 16 outcomes where the first flip is tails , identify the number of outcomes in which exactly four heads appear. There is only one such outcome: THHHH. 4. Calculate the conditional probability by dividing the number of favourable outcomes (i.e. THHHH) by the number of total outcomes given the condition (i.e. the first flip is tails), which is 1/16. Therefore, the correct order of steps is 1, 2, 3, 4. And the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails, is 1/16.

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Complete the following tasks on the plane. Using a blue pencil, shade the region that contains points that are more than 2 1/2 units and less than 3 1/4 units from the y-axis.

The points that are more than the fraction 2 1/2 units and less than 3 1/4 units contains the points from 2.5 to 3.25.

Given measurements are 2 1/2 and 3 1/4.

We have to shade the region in between these two points .

Both are written in mixed fractional form.

This can be written in decimal as,

2 1/2 = 2 + 1/2 = 2 +6 0.5 = 2.5

3 1/4 = 3 + 1/4 = 3 + 0.25 = 3.25

Hence the shaded region will contain the points from 2.5 to 3.25.

2.5 is at the exact middle of 2 and 3.

3.25 is at 1/4th distance of 3 and 4. That is if we divide the distance between 3 and 4 to 4 equal spaces, the first space is 3 1/4.

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How can we easily recognize when a system of linear equations is inconsistent or not?

To recognize linear equations write them in Ax = b form, Gaussian elimination, check in raw-echelon for all coefficients in zero, and if it exists its a contradiction and inconsistent .

To easily recognize when a system of linear equations is inconsistent, you can follow these steps: 1. Write the system of linear equations in the form Ax = b, where A is the matrix of coefficients, x is the vector of variables, and b is the constant vector. 2. Perform Gaussian elimination or row reduction on the augmented matrix [A | b] to obtain the row-echelon form. 3. Check for any row in the row-echelon form where all the coefficients of the variables are zero, but the constant term is nonzero (i.e., 0x + 0y + ... + 0z = k, where k ≠ 0). 4. If such a row exists, then the system of linear equations is inconsistent because it represents a contradiction (e.g., 0 = k, where k ≠ 0). If no such row is found, then the system is either consistent and has a unique solution or consistent and has infinitely many solutions, depending on the number of free variables.

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find the conditional probability of the indicated event when two fair dice (one red and one green) are rolled. hint [see example 1.] the sum is 4, given that the green one is either 3 or 2.

The conditional probability of the sum being 4, given that the green die shows either a 3 or a 2, is 1/6.

To find the conditional probability of the sum being 4, given that the green die is either 3 or 2, we need to use the formula: P(A|B) = P(A and B) / P(B) where A is the event of getting a sum of 4 and B is the event of getting either a 3 or 2 on the green die. First, let's calculate the probability of getting a 3 or 2 on the green die: P(B) = 1/3 + 1/3 = 2/3 since there are 3 possible outcomes for each die and the green die can either be 3 or 2. Next, we need to calculate the probability of getting a sum of 4 and a green die of either 3 or 2: P(A and B) = 2/36 since there are only 2 ways to get a sum of 4 with a green die of either 3 or 2: (1,3) and (2,2). Now we can plug in the values into the formula: P(A|B) = (2/36) / (2/3) = 1/18 Therefore, the conditional probability of getting a sum of 4, given that the green die is either 3 or 2, is 1/18. To find the conditional probability of the indicated event, we'll use the formula: P(A / B) = P(A / B) / P(B) Here, event A is the sum of the numbers on the two dice being 4, and event B is the green die showing either a 3 or a 2. First, let's find P(B). There are 6 possible outcomes for each die, so there are 6x6=36 total possible outcomes when rolling both dice. There are 2 favorable outcomes for event B: the green die showing a 3 or a 2. Therefore, P(B) = 2/6 = 1/3. Now, let's find P(A / B). This is the probability of both events A and B happening at the same time. For the sum to be 4 and the green die to show a 2, the red die must show a 2. For the sum to be 4 and the green die to show a 3, the red die must show a 1. There are 2 favorable outcomes for P(A /B) out of the 36 possible outcomes. Therefore, P(A ∩ B) = 2/36 = 1/18. Finally, we can find the conditional probability P(A | B) using the formula: P(A / B) = P(A / B) / P(B) = (1/18) / (1/3) = (1/18) * (3/1) = 3/18 = 1/6. So, the conditional probability of the sum being 4, given that the green die shows either a 3 or a 2, is 1/6.

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(1) data are obtained from a random sample of assets in millions of dollars of 30 credit unions and the sample mean turns out to be 11.091. assume that the sample standard deviation is 14.404. the population is assumed to be normally distributed. (a) find the 90% confidence interval for the mean. write the solution with two decimal places: for example: (x.xx, x.xx).

The 90% confidence interval for the mean is (6.63, 15.55). Lower limit = X - Margin of error = 11.091 - 4.459 ≈ 6.63 and Upper limit = X + Margin of error = 11.091 + 4.459 ≈ 15.55

To find the 90% confidence interval for the mean, we can use the formula: CI = X ± Zα/2 * (σ/√n) Where X is the sample mean (11.091), Zα/2 is the z-score corresponding to the desired confidence level (90% confidence interval = 1.645), σ is the sample standard deviation (14.404), and n is the sample size (30). Plugging in the values, we get: CI = 11.091 ± 1.645 * (14.404/√30) CI = (3.20, 18.98) Therefore, the 90% confidence interval for the mean is (3.20, 18.98).

The 90% confidence interval for the mean . Given the sample size (n = 30), the sample mean (X = 11.091), and the sample standard deviation (s = 14.404), we can calculate the confidence interval using the formula: X ± t * (s / √n) First, we need to find the t-value for a 90% confidence interval with 29 degrees of freedom (n - 1). You can use a t-table or a calculator to find this value. For a 90% confidence interval, the t-value is approximately 1.699. Next, we calculate the margin of error : Margin of error = t * (s / √n) = 1.699 * (14.404 / √30) = 4.459 Now, we can find the confidence interval: Lower limit = X - Margin of error = 11.091 - 4.459 ≈ 6.63 Upper limit = X + Margin of error = 11.091 + 4.459 ≈ 15.55 Thus, the 90% confidence interval for the mean is (6.63, 15.55).

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At a particular restaurant, 52% of all customers order an appetizer and 32% of all customers order dessert. If 27% of all customers order both an appetizer and dessert, what is the probability a randomly selected customer orders an appetizer or dessert or both? Write your answer as a decimal (not as a percentage).

The probability of randomly selected c ustomers ordering both an appetizer and dessert is 57%.

The probability is defined as the possibility of an event being equal to the ratio of the number of favorable outcomes and the total number of outcomes.

Given data as:

P(E₁) = 52%

P(E₂) = 32%

P(E₁ or E₂) = 27%.

Using the formula:

⇒ P(E₁ or E₂) = P(E₁) + P(E₂) - P(E₁ & E₂)

Substitute the values and solve for P(E₁ & E₂)

⇒ P(E₁ & E₂) = 52% + 32% - 27%

⇒ P(E₁ & E₂) = 57%

Therefore, the probability of randomly selected customers ordering both an appetizer and dessert is 57%.

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