two sample hypothesis testing assumes that the samples are

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Statistics By Jim

Making statistics intuitive

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

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A free online introduction to statistics

The Two-Sample t -Test

What is the two-sample t -test.

The two-sample t -test (also known as the independent samples t -test) is a method used to test whether the unknown population means of two groups are equal or not.

Is this the same as an A/B test?

Yes, a two-sample t -test is used to analyze the results from A/B tests.

When can I use the test?

You can use the test when your data values are independent, are randomly sampled from two normal populations and the two independent groups have equal variances.

What if I have more than two groups?

Use a multiple comparison method. Analysis of variance (ANOVA) is one such method. Other multiple comparison methods include the Tukey-Kramer test of all pairwise differences, analysis of means (ANOM) to compare group means to the overall mean or Dunnett’s test to compare each group mean to a control mean.

What if the variances for my two groups are not equal?

You can still use the two-sample t- test. You use a different estimate of the standard deviation. 

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.

See how to perform a two-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 .

Using the two-sample t -test

The sections below discuss what is needed to perform the test, checking our data, how to perform the test and statistical details.

What do we need?

For the two-sample t -test, we need two variables. One variable defines the two groups. The second variable is the measurement of interest.

We also have an idea, or hypothesis, that the means of the underlying populations for the two groups are different. Here are a couple of examples:

• We have students who speak English as their first language and students who do not. All students take a reading test. Our two groups are the native English speakers and the non-native speakers. Our measurements are the test scores. Our idea is that the mean test scores for the underlying populations of native and non-native English speakers are not the same. We want to know if the mean score for the population of native English speakers is different from the people who learned English as a second language.
• We measure the grams of protein in two different brands of energy bars. Our two groups are the two brands. Our measurement is the grams of protein for each energy bar. Our idea is that the mean grams of protein for the underlying populations for the two brands may be different. We want to know if we have evidence that the mean grams of protein for the two brands of energy bars is different or not.

Two-sample t -test assumptions

To conduct a valid test:

• Data values must be independent. Measurements for one observation do not affect measurements for any other observation.
• Data in each group must be obtained via a random sample from the population.
• Data in each group are normally distributed .
• Data values are continuous.
• The variances for the two independent groups are equal.

For very small groups of data, it can be hard to test these requirements. Below, we'll discuss how to check the requirements using software and what to do when a requirement isn’t met.

Two-sample t -test example

One way to measure a person’s fitness is to measure their body fat percentage. Average body fat percentages vary by age, but according to some guidelines, the normal range for men is 15-20% body fat, and the normal range for women is 20-25% body fat.

Our sample data is from a group of men and women who did workouts at a gym three times a week for a year. Then, their trainer measured the body fat. The table below shows the data.

Table 1: Body fat percentage data grouped by gender

You can clearly see some overlap in the body fat measurements for the men and women in our sample, but also some differences. Just by looking at the data, it's hard to draw any solid conclusions about whether the underlying populations of men and women at the gym have the same mean body fat. That is the value of statistical tests – they provide a common, statistically valid way to make decisions, so that everyone makes the same decision on the same set of data values.

Checking the data

Let’s start by answering: Is the two-sample t -test an appropriate method to evaluate the difference in body fat between men and women?

• The data values are independent. The body fat for any one person does not depend on the body fat for another person.
• We assume the people measured represent a simple random sample from the population of members of the gym.
• We assume the data are normally distributed, and we can check this assumption.
• The data values are body fat measurements. The measurements are continuous.
• We assume the variances for men and women are equal, and we can check this assumption.

Before jumping into analysis, we should always take a quick look at the data. The figure below shows histograms and summary statistics for the men and women.

The two histograms are on the same scale. From a quick look, we can see that there are no very unusual points, or outliers . The data look roughly bell-shaped, so our initial idea of a normal distribution seems reasonable.

Examining the summary statistics, we see that the standard deviations are similar. This supports the idea of equal variances. We can also check this using a test for variances.

Based on these observations, the two-sample t -test appears to be an appropriate method to test for a difference in means.

How to perform the two-sample t -test

For each group, we need the average, standard deviation and sample size. These are shown in the table below.

Table 2: Average, standard deviation and sample size statistics grouped by gender

Without doing any testing, we can see that the averages for men and women in our samples are not the same. But how different are they? Are the averages “close enough” for us to conclude that mean body fat is the same for the larger population of men and women at the gym? Or are the averages too different for us to make this conclusion?

We'll further explain the principles underlying the two sample t -test in the statistical details section below, but let's first proceed through the steps from beginning to end. We start by calculating our test statistic. This calculation begins with finding the difference between the two averages:

$ 22.29 - 14.95 = 7.34 $

This difference in our samples estimates the difference between the population means for the two groups.

Next, we calculate the pooled standard deviation. This builds a combined estimate of the overall standard deviation. The estimate adjusts for different group sizes. First, we calculate the pooled variance:

$ s_p^2 = \frac{((n_1 - 1)s_1^2) + ((n_2 - 1)s_2^2)} {n_1 + n_2 - 2} $

$ s_p^2 = \frac{((10 - 1)5.32^2) + ((13 - 1)6.84^2)}{(10 + 13 - 2)} $

$ = \frac{(9\times28.30) + (12\times46.82)}{21} $

$ = \frac{(254.7 + 561.85)}{21} $

$ =\frac{816.55}{21} = 38.88 $

Next, we take the square root of the pooled variance to get the pooled standard deviation. This is:

$ \sqrt{38.88} = 6.24 $

We now have all the pieces for our test statistic. We have the difference of the averages, the pooled standard deviation and the sample sizes.  We calculate our test statistic as follows:

$ t = \frac{\text{difference of group averages}}{\text{standard error of difference}} = \frac{7.34}{(6.24\times \sqrt{(1/10 + 1/13)})} = \frac{7.34}{2.62} = 2.80 $

To evaluate the difference between the means in order to make a decision about our gym programs, we compare the test statistic to a theoretical value from the t- distribution. This activity involves four steps:

• We decide on the risk we are willing to take for declaring a significant difference. For the body fat data, we decide that we are willing to take a 5% risk of saying that the unknown population means for men and women are not equal when they really are. In statistics-speak, the significance level, denoted by α, is set to 0.05. It is a good practice to make this decision before collecting the data and before calculating test statistics.
• We calculate a test statistic. Our test statistic is 2.80.
• We find the theoretical value from the t- distribution based on our null hypothesis which states that the means for men and women are equal. Most statistics books have look-up tables for the t- distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables. To find this value, we need the significance level (α = 0.05) and the degrees of freedom . The degrees of freedom ( df ) are based on the sample sizes of the two groups. For the body fat data, this is: $ df = n_1 + n_2 - 2 = 10 + 13 - 2 = 21 $ The t value with α = 0.05 and 21 degrees of freedom is 2.080.
• We compare the value of our statistic (2.80) to the t value. Since 2.80 > 2.080, we reject the null hypothesis that the mean body fat for men and women are equal, and conclude that we have evidence body fat in the population is different between men and women.

Statistical details

Let’s look at the body fat data and the two-sample t -test using statistical terms.

Our null hypothesis is that the underlying population means are the same. The null hypothesis is written as:

$ H_o:  \mathrm{\mu_1} =\mathrm{\mu_2} $

The alternative hypothesis is that the means are not equal. This is written as:

$ H_o:  \mathrm{\mu_1} \neq \mathrm{\mu_2} $

We calculate the average for each group, and then calculate the difference between the two averages. This is written as:

$\overline{x_1} -  \overline{x_2} $

We calculate the pooled standard deviation. This assumes that the underlying population variances are equal. The pooled variance formula is written as:

The formula shows the sample size for the first group as n 1 and the second group as n 2 . The standard deviations for the two groups are s 1 and s 2 . This estimate allows the two groups to have different numbers of observations. The pooled standard deviation is the square root of the variance and is written as s p .

What if your sample sizes for the two groups are the same? In this situation, the pooled estimate of variance is simply the average of the variances for the two groups:

$ s_p^2 = \frac{(s_1^2 + s_2^2)}{2} $

The test statistic is calculated as:

$ t = \frac{(\overline{x_1} -\overline{x_2})}{s_p\sqrt{1/n_1 + 1/n_2}} $

The numerator of the test statistic is the difference between the two group averages. It estimates the difference between the two unknown population means. The denominator is an estimate of the standard error of the difference between the two unknown population means. 

Technical Detail: For a single mean, the standard error is $ s/\sqrt{n} $  . The formula above extends this idea to two groups that use a pooled estimate for s (standard deviation), and that can have different group sizes.

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

$ df = n_1 + n_2 - 2 = 10 + 13 - 2 = 21 $

The formula shows the sample size for the first group as n 1 and the second group as n 2 .  Statisticians write the t value with α = 0.05 and 21 degrees of freedom as:

$ t_{0.05,21} $

The t value with α = 0.05 and 21 degrees of freedom is 2.080. There are two possible results from our comparison:

• The test statistic is lower than the t value. You fail to reject the hypothesis of equal means. You conclude that the data support the assumption that the men and women have the same average body fat.
• The test statistic is higher than the t value. You reject the hypothesis of equal means. You do not conclude that men and women have the same average body fat.

t -Test with unequal variances

When the variances for the two groups are not equal, we cannot use the pooled estimate of standard deviation. Instead, we take the standard error for each group separately. The test statistic is:

$ t = \frac{ (\overline{x_1} -  \overline{x_2})}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} $

The numerator of the test statistic is the same. It is the difference between the averages of the two groups. The denominator is an estimate of the overall standard error of the difference between means. It is based on the separate standard error for each group.

The degrees of freedom calculation for the t value is more complex with unequal variances than equal variances and is usually left up to statistical software packages. The key point to remember is that if you cannot use the pooled estimate of standard deviation, then you cannot use the simple formula for the degrees of freedom.

Testing for normality

The normality assumption is more important   when the two groups have 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 body fat data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for men and women, and supports our decision.

You can also perform a formal test for normality using software. The figure above shows results of testing for normality with JMP software. We test each group separately. Both the test for men and the test for women show that we cannot reject the hypothesis of a normal distribution. We can go ahead with the assumption that the body fat data for men and for women are normally distributed.

Testing for unequal variances

Testing for unequal variances is complex. We won’t show the calculations in detail, but will show the results from JMP software. The figure below shows results of a test for unequal variances for the body fat data.

Without diving into details of the different types of tests for unequal variances, we will use the F test. Before testing, we decide to accept a 10% risk of concluding the variances are equal when they are not. This means we have set α = 0.10.

Like most statistical software, JMP shows the p -value for a test. This is the likelihood of finding a more extreme value for the test statistic than the one observed. It’s difficult to calculate by hand. For the figure above, with the F test statistic of 1.654, the p- value is 0.4561. This is larger than our α value: 0.4561 > 0.10. We fail to reject the hypothesis of equal variances. In practical terms, we can go ahead with the two-sample t -test with the assumption of equal variances for the two groups.

Understanding p-values

Using a visual, you can check to see if your test statistic is a more extreme value in the distribution. The figure below shows a t- distribution with 21 degrees of freedom.

Since our test is two-sided and we have set α = .05, the figure shows that the value of 2.080 “cuts off” 2.5% of the data in each of the two tails. Only 5% of the data overall is further out in the tails than 2.080. Because our test statistic of 2.80 is beyond the cut-off point, we reject the null hypothesis of equal means.

Putting it all together with software

The figure below shows results for the two-sample t -test for the body fat data from JMP software.

The results for the two-sample t -test that assumes equal variances are the same as our calculations earlier. The test statistic is 2.79996. The software shows results for a two-sided test and for one-sided tests. The two-sided test is what we want (Prob > |t|). Our null hypothesis is that the mean body fat for men and women is equal. Our alternative hypothesis is that the mean body fat is not equal. The one-sided tests are for one-sided alternative hypotheses – for example, for a null hypothesis that mean body fat for men is less than that for women.

We can reject the hypothesis of equal mean body fat for the two groups and conclude that we have evidence body fat differs in the population between men and women. The software shows a p -value of 0.0107. We decided on a 5% risk of concluding the mean body fat for men and women are different, when they are not. It is important to make this decision before doing the statistical test.

The figure also shows the results for the t- test that does not assume equal variances. This test does not use the pooled estimate of the standard deviation. As was mentioned above, this test also has a complex formula for degrees of freedom. You can see that the degrees of freedom are 20.9888. The software shows a p- value of 0.0086. Again, with our decision of a 5% risk, we can reject the null hypothesis of equal mean body fat for men and women.

Other topics

If you have more than two independent groups, you cannot use the two-sample t- test. You should use a multiple comparison   method. ANOVA, or analysis of variance, is one such method. Other multiple comparison methods include the Tukey-Kramer test of all pairwise differences, analysis of means (ANOM) to compare group means to the overall mean or Dunnett’s test to compare each group mean to a control mean.

What if my data are not from normal distributions?

If your sample size is very small, it might be hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the body fat data, the trainer knows that the underlying distribution of body fat is normally distributed. Even for a very small sample, the trainer 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 nonparametric analyses. These types of analyses do not depend on an assumption that the data values are from a specific distribution. For the two-sample t ­-test, the Wilcoxon rank sum test is a nonparametric test that could be used.

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Two-Sample Hypothesis Tests A two-sample test compares samples with each other rather than comparing with a benchmark, as in a one-sample test. For two proportions, the samples may be pooled if the population proportions are assumed equal, and the test statistic is the difference of proportions divided by the standard error, the square root of the sum of the sample variances. For proportions, normality may be assumed if both samples are large, that is, if they each contain at least 10 successes and 10 failures. For independent samples, the comparison of means generally utilizes the Student's distribution, because the population variances are almost always unknown. If the unknown variances are assumed equal, we use a pooled variance estimate and add the degrees of freedom. If the unknown variances are assumed unequal, we do not pool the variances and we reduce the degrees of freedom by using Welch's formula. The test statistic is the difference of means divided by their standard error. For tests of means or proportions, equal sample sizes are desirable, but not necessary. The test for paired samples uses the differences of paired observations, thereby being one-sample test. The test for equality of two variances is named after Sir Ronald Fisher. Its test statistic is the ratio of the sample variances. We want to see if the ratio differs significantly from 1. The table shows critical values based on both numerator and denominator degrees of freedom. When you finish this chapter you should be able to test. distribution and why it is used. test and when they are violated. In This Chapter To learn more about the book this website supports, please visit its . Copyright Any use is subject to the and | You must be a registered user to view the in this website. If you already have a username and password, enter it below. If your textbook came with a card and this is your first visit to this site, you can to register. 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(optional) Select some text on the page (or do this before you open the "Notes" drawer). 3. Highlighter Color: 4. Search for: Search in: Course-wide Content Quizzes Instructor Resources Course-wide Content Introduction to Statistics and Data Science Chapter 15 hypothesis testing: two sample tests, 15.1 two sample t test. We can also use the t test command to conduct a hypothesis test on data where we have samples from two populations. To introduce this lets consider an example from sports analytics. In particular, let us consider the NBA draft and the value of a lottery pick in the draft. Teams which do make the playoffs are entered into a lottery to determine the order of the top picks in the draft for the following year. These top 14 picks are called lottery picks. Using historical data we might want to investigate the value of a lottery pick against those players who were selected outside the lottery. We can now make a boxplot comparing the career scoring averages of the lottery picks between these two pick levels. From this boxplot we notice that the lottery picks tend to have a higher point per game (PPG) average. However, we certainly see many exceptions to this rule. We can also compute the averages of the PTS column for these two groups: Lottery.Pick ppg NumberPlayers Lottery 11.236927 371 Not Lottery 7.107924 366 This table once again demonstrates that the lottery picks tend to average more points. However, we might like to test this trend to see if have sufficient evidence to conclude this trend is real (this could also just be a function of sampling error). 15.1.1 Regression analysis Our first technique for looking for a difference between our two categories is linear regression with a categorical explanatory variable. We fit a regression model of the form: \[PTS=\beta \delta_{\text{ not lottery}}+\alpha\] Where \(\delta_{\text{ not lottery}}\) is equal to one if the draft pick fell outside the lottery and zero otherwise. To see if this relationship is real we can form a confidence interval for the coefficients. From this we can see that Lottery picks to tend to average more point per game over their careers. The magnitude of this effect is somewhere between 3.5 and 4.7 points more for lottery picks. 15.1.2 Two Sample t test approach For this we can use the two-sample t-test to compare the means of these two distinct populations. Here the alternative hypothesis is that the lottery players score more points \[H_A: \mu_L > \mu_{NL}\] thus the null hypothesis is \[H_0: \mu_L \leq \mu_{NL}.\] We can now perform the test in R using the same t.test command as before. Notice that I used the magic tilde ~ to split the PTS column into the lottery/non-lottery pick subdivisions. I could also do this manually and get the same answer: The very small p-value here indicates that the population mean of the lottery picks is truly greater than the population mean of the non-lottery picks. The 95% confidence interval also tells us that this difference is rather large (at least 3.85 points). Conditions for using a two-sample t test: These are roughly the same as the conditions for using a one sample t test, although we now need to assume that BOTH samples satisfy the conditions. Must be looking for a difference in the population means (averages) 30 or greater samples in both groups (CLT) If you have less than 30 in one sample, you can use the t test must you must then assume that the population is roughly mound shaped. At this point you would probably like to know why we would ever want to do a two sample t test instead of a linear regression? My answer is that a two sample t test is more robust against a difference in variance between the two groups. Recall, that one of the assumptions of simple linear regression is that the variance of the residuals does not depend on the explanatory variable(s). By default R does a type of t test which does not assume equal variance between the two groups. This is the one advantage of using the t test command. 15.1.2.1 Paired t test Lets say we are trying to estimate the effect of a new training regiment on the 40 yard dash times for soccer players. Before implementing the training regime we measure the 40 yard dash times of the 30 players. First lets read this data set into R. First, we can compare the mean times before and after the training: Also we could make a side by side boxplot for the soccer players times before and after the training We could do a simple t test to examine whether mean of the players times after the training regime is implemented decrease (on average). Here we have the alternative hypothesis that \(H_a: \mu_b-\mu_a>0\) and thus the null hypothesis that \(H_0: \mu_b-\mu_a \leq 0\) . Using the two sample t test format in R we have: Here we cannot reject the null hypothesis that the training had no effect on the players sprinting performance. However, we haven’t used all of the information available to us in this scenario. The t test we have just run doesn’t know that we recorded the before and after for the same players more than once. As far as R knows the before and after times could be entirely different players as if we are comparing the results between one team which received the training and one who didn’t. Therefore, R has to be pretty conservative in its predictions. The differences between the two groups could be due to many reasons other than the training regime implemented. Maybe the second set of players just started off being a little bit faster, etc. The data we collected is actually more powerful because we know the performance of the same players before and after the test. This greatly reduces the number of variables which need to be accounted for in our statistical test. Luckily, we can easily let R know that our data points are paired . Setting the paired keyword to true lets R know that the two columns should be paired together during the test. We can see that running the a paired t test gives us a much smaller p value. Moreover, we can now safely conclude that the new training regiment is effective in at least modestly reducing the 40 yard dash times of the soccer players. This is our first example of the huge subject of experimental design which is the study of methods which can be used to create data sets which have more power to distinguish differences between groups. Where possible it is better to collect data for the same subjects under two conditions as this will allow for more powerful statistical analysis of the data (i.e a paired t test instead of a normal t test). Whenever the assumptions are met for a paired t test, you will be expected to perform a paired t test in this class. 15.2 Two Sample Proportion Tests We can also use statistical hypothesis testing to compare the proportion between two samples. For example, we might conduct a survey of 100 smokers and 50 non-smokers to see whether they buy organic foods. If we find that 30/100 smokers buy organic and only 11/50 non-smokers buy organic then can we conclude that more smokers buy organic foods that smokers? \(H_a: p_s > p_n\) and \(H_0: p_s \leq p_n\) . In this case we don’t have sufficient evidence to conclude that a larger fraction of smokers buy organic foods. It is common when analyzing survey data to want to compare proportions between populations. The key assumptions when performing a two-sample proportion test are that we have at least 5 successes and 5 failures in BOTH samples. 15.3 Extra Example: Birth Weights and Smoking For this example we are going to use a data from a study on the risk factors associated with giving birth to a low-weight baby (sometimes defined as less than 2,500 grams). This data set is another one which is build into R . To load this data for analysis type: You can view all a description of the data by typing ?birthwt once it is loaded. To begin we could look at the raw birth weight of mothers who were smokers versus non-smokers. We can do some EDA on this data using a boxplot: From the boxplot we can see that the median birth weight of babies whose mothers smoked was smaller. We can test the data for a difference in the means using a t.test command. Notice we can use the ~ shorthand to split the data into those two groups faster than filtering. Here we get a small p value meaning we have sufficient evidence to reject the null hypothesis that the mean weight of babies of women who smoked is greater than or equal to those of non-smokers. Within this data set we also have a column low which classifies whether the babies birth weight is considered low using the medical criterion (birth weight less than 2,500 grams): We can see that smoking gives a higher fraction of low-weight births. However, this could just be due to sampling error so let’s run a proportion test to find out. Once again we find we have sufficient evidence to reject the null hypothesis that smoking does not increase the risk of a low birth weight. 15.4 Homework 15.4.1 concept questions. What the assumptions behind using a two sample proportion test? Hint these will be the same as forming a confidence interval for for the fraction of a population, with two samples where this assumption needs to hold. What assumptions are required for a two sample t test with small \(N\leq 30\) sample sizes? A paired t test may be used for any two sample experiment (True/False) The power of any statistical test will increase with increasing sample sizes. (True/False) Where possible it is better to collect data on the same individuals when trying to distinguish a difference in the average response to a condition (True/False) The paired t test is a more powerful statistical test than a normal t test (True/ False) 15.4.2 Practice Problems For each of the scenarios below form the null and alternative hypothesis. We have conducted an educational study on two classrooms of 30 students using two different teaching methods. The first method had 50% of students pass a standardized test, and the classroom using the second teaching method had 60% of the students pass. A basketball coach is extremely superstitious and believes that when he wears his lucky tie the team has a greater chance of winning the game. He comes to you because he is looking to design an experiment to test this belief. If the team has 40 games in the upcoming season, design an experiment and the (null and alt) hypothesis to test the coaches claims. For the below question work out the number of errors in the data set. Before the Olympics all athletes are required to submit a urine sample to be tested for banned substances. This is done by estimating the concentration of certain compounds in the urine and is prone to some degree of laboratory error. In addition, the concentration of these compounds are known to vary with the individual (genetic, diet, etc). To weigh the evidence present in a drug test the laboratory conducts a statistical test. To ensure they don’t falsely convict athletes of doping they use a significance level of \(\alpha=0.01\) . If they test 3000 athletes, all of whom are clean about how many will be falsely accused of doping? Explain the issue with this procedure. 15.4.3 Advanced Problems Load the drug_use data set from the fivethirtyeight package. Run a hypothesis test to determine if a larger proportion of 22-23 year olds are using marijuana then 24-25 year olds. Interpret your results statistically and practically. Import the data set Cavaliers_Home_Away_2016 . Form a hypothesis on whether being home or away for the game had an effect on the proportion of games won by the Cavaliers during the 2016-2017 season, test this hypothesis using a hypothesis test. Load the data set animal_sleep and compare the average total sleep time (sleep_total column) between carnivores and herbivores (using the vore column) to divide the between the two categories. To begin make a boxplot to compare the total sleep time between these two categories. Do we have sufficient evidence to conclude the average total sleep time differs between these groups? Load the HR_Employee_Attrition data set. We wish to investigate whether the daily rate (pay) has anything to do with whether a employee has quit (the attrition column is “Yes”). To begin make a boxplot of the DailyRate column split into these Attrition categories. Use the boxplot to help form the null hypothesis for your test and decide on an alternative hypothesis. Conduct a statistical hypothesis test to determine if we have sufficient evidence to conclude that those employees who quit tended to be paid less. Report and interpret the p value for your test. Load the BirdCaptureData data set. Perform a hypothesis test to determine if the proportion of orange-crowned warblers (SpeciesCode==OCWA) caught at the station is truly less than the proportion of Yellow Warblers (SpeciesCode==YWAR). Report your p value and interpret the results statistically and practically. (All of Statistics Problem) In 1861, 10 essays appeared in the New Orleans Daily Crescent. They were signed “Quintus Curtius Snodgrass” and one hypothesis is that these essays were written by Mark Twain. One way to look for similarity between writing styles is to compare the proportion of three letter words found in two works. For 8 Mark Twain essays we have: From 10 Snodgrass essays we have that: Perform a two sample t test to examine these two data sets for a difference in the mean values. Report your p value and a 95% confidence interval for the results. What are the issues with using a t-test on this data? Consider the analysis of the kidiq data set again. Run a regression with kid_score as the response and mom_hs as the explanatory variable and look at the summary() of your results. Notice the p-value which is reported in the last line of the summary. This “F-test” is a hypothesis test with the null hypothesis that the explanatory variable tells us nothing about the value of the response variable. Perform a t test for the a difference in means in the kid_score values based on the mom_hs column. What is your conclusion? Repeat the t test again using the command: User Preferences Content preview. Arcu felis bibendum ut tristique et egestas quis: Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris Duis aute irure dolor in reprehenderit in voluptate Excepteur sint occaecat cupidatat non proident Keyboard Shortcuts 5.5 - hypothesis testing for two-sample proportions. We are now going to develop the hypothesis test for the difference of two proportions for independent samples. The hypothesis test follows the same steps as one group. These notes are going to go into a little bit of math and formulas to help demonstrate the logic behind hypothesis testing for two groups. If this starts to get a little confusion, just skim over it for a general understanding! Remember we can rely on the software to do the calculations for us, but it is good to have a basic understanding of the logic! We will use the sampling distribution of \(\hat{p}_1-\hat{p}_2\) as we did for the confidence interval. For a test for two proportions, we are interested in the difference between two groups. If the difference is zero, then they are not different (i.e., they are equal). Therefore, the null hypothesis will always be: \(H_0\colon p_1-p_2=0\) Another way to look at it is \(H_0\colon p_1=p_2\). This is worth stopping to think about. Remember, in hypothesis testing, we assume the null hypothesis is true. In this case, it means that \(p_1\) and \(p_2\) are equal. Under this assumption, then \(\hat{p}_1\) and \(\hat{p}_2\) are both estimating the same proportion. Think of this proportion as \(p^*\). Therefore, the sampling distribution of both proportions, \(\hat{p}_1\) and \(\hat{p}_2\), will, under certain conditions, be approximately normal centered around \(p^*\), with standard error \(\sqrt{\dfrac{p^*(1-p^*)}{n_i}}\), for \(i=1, 2\). We take this into account by finding an estimate for this \(p^*\) using the two-sample proportions. We can calculate an estimate of \(p^*\) using the following formula: \(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}\) This value is the total number in the desired categories \((x_1+x_2)\) from both samples over the total number of sampling units in the combined sample \((n_1+n_2)\). Putting everything together, if we assume \(p_1=p_2\), then the sampling distribution of \(\hat{p}_1-\hat{p}_2\) will be approximately normal with mean 0 and standard error of \(\sqrt{p^*(1-p^*)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\), under certain conditions. \(z^*=\dfrac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}\) ...will follow a standard normal distribution. Finally, we can develop our hypothesis test for \(p_1-p_2\). Hypothesis Testing for Two-Sample Proportions Conditions : \(n_1\hat{p}_1\), \(n_1(1-\hat{p}_1)\), \(n_2\hat{p}_2\), and \(n_2(1-\hat{p}_2)\) are all greater than five Test Statistic: \(z^*=\dfrac{\hat{p}_1-\hat{p}_2-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}\) ...where \(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}\). The critical values, p-values, and decisions will all follow the same steps as those from a hypothesis test for a one-sample proportion. Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Hypothesis testing with two independent samples test statistics I'm learning a bit about hypothesis testing with two independent samples (continuous outcome), and I'm just curious about where some of the equations for the test statistics are derived and/or what they're measuring. Suppose our first sample is of size $n_1$, has mean $\bar{X_1}$, and standard deviation $\sigma_1$, and analogously, our second sample is of size $n_2$ with mean $\bar{X_2}$, and standard deviation $\sigma_2$. We want to test the null hypothesis $H_0: \bar{X_1} = \bar{X_2}$ and assuming our sample sizes are large enough, we can use the test statistic: $$z = \dfrac{\bar{X_1}-\bar{X_2}}{S_p\,\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$ where $S_p$ is the pooled estimate of the common standard deviation: $$S_p = \sqrt{\dfrac{(n_1-1)\,\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2}}$$ For this example, let's suppose the alternative hypothesis is the first mean is larger than the second i.e. $H_1:\bar{X_1}>\bar{X_2}$. I'm trying to get an intuitive feel for what the above test statistic $z$ is measuring and where the equation comes from and what the meaning of "pooled estimate of the common standard deviation" means... hypothesis-testing mathematical-statistics $\begingroup$ I'm puzzled by "assuming the sample sizes are large enough". If your sample sizes are large enough to compute an $S_p$ (the smallest possible being samples of 1 and 2 in either order), then you'll have a perfectly valid $t$-statstic, which will have a $t$-distribution under the null (and given the assumptions of the test) $\endgroup$ –  Glen_b Commented Oct 5, 2017 at 22:55 $\begingroup$ Ah I mentioned that because I thought $n_1$ and $n_2$ need to be large enough to use a z score because if $n_1$ and $n_2$ are small then you use a $t$ test because then your data will have a $t$ distribution. $\endgroup$ –  clueless_undergrad37 Commented Oct 5, 2017 at 23:09 $\begingroup$ But the statistic you gave is a t-statistic! Calling the t-statistic "z" doesn't do anything. [... Edit:] Oh, wait, I think I get it -- I guess you might want to invoke Slutsky's theorem for the ratio (assuming the sample sizes are large enough that you could treat the estimate in the denominator as essentially having no error) and then apply the CLT to the numerator. Well, okay but you could need quite large sample sizes before that's going to work well. Doesn't alter the motivation for the form of the statistic, apart from the claims about "best estimates" of various quantities. $\endgroup$ –  Glen_b Commented Oct 5, 2017 at 23:54 The numerator of the test statistic is the difference in sample means (representing an estimated difference in population means) But in order to judge whether that's more than would reasonably be seen if the null were true, we need some idea of scale. So we want to divide by the standard deviation of the difference in means. So the denominator is an estimate of the standard deviation of distribution of the difference in means. If the assumptions of the t-test hold or hold nearly enough, it will be a good estimate of that standard deviation. So the test statistic is a standardized difference in means; it's a kind of "internally-standardized z-score" for the difference, which - because we had to estimate the variance term - will have a t-distribution. The test assumes equal variance in the two groups. $S_p^2$ is (in a particular sense) our best estimate of that common variance, $\sigma^2$. We have two estimates of the same variance, $\sigma^2$. These are $s_1^2$ and $s_2^2$. We'd like to "average" them in some way to get a good estimate of $\sigma^2$. But an estimate derived from a larger sample is more precise -- it should get more weight in the average; the right weight to use (the one that minimizes the variance of the estimate of $\sigma^2$) weights by the degrees of freedom $\text{df}_i=n_i-1$ (they each lose a degree of freedom in estimating the sample mean). So we estimate $\sigma^2$ by a weighted average, $S_p^2 = w_1 s_1^2 + w_2 s_2^2$, where $w_1 = \frac{\text{df}_1}{\text{df}_1+\text{df}_2}$ and $w_2=\frac{\text{df}_2}{\text{df}_1+\text{df}_2}$. Note that the weights add to $1$. This gives the formula for $S_p^2$: $$S_p^2 = \frac{\text{df}_1}{\text{df}_1+\text{df}_2} s_1^2 + \frac{\text{df}_2}{\text{df}_1+\text{df}_2} s_2^2 = \frac{n_1-1}{n_1-1+n_2-1} s_1^2 + \frac{n_2-1}{n_1-1+n_2-1} s_2^2\\ = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2 }{n_1-1+n_2-1}= \frac{(n_1-1)s_1^2+(n_2-1)s_2^2 }{n_1+n_2-2}$$ If we knew $\sigma$ the variance of $\bar{X_1}$ would be $\sigma^2/n_1$ and the variance of $\bar{X_2}$ would be $\sigma^2/n_2$. Because we assume the observations in the two samples are independent of each other, the variance of the difference $\bar{X_1}-\bar{X_2}$ is the sum of their variances . So the variance of $\bar{X_1}-\bar{X_2}$ is $\frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}=\sigma^2(\frac{1}{n_1} + \frac{1}{n_2})$. Consequently its standard deviation (the standard error of the difference) is $\sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$. Once we replace $\sigma$ by our estimate for it, $S_p$, you have the denominator of the t-statistic. This is pretty much how t-statistics work in general -- they're estimates of some (raw) effect in the numerator, standardized by an estimate of the standard deviation of that effect in the denominator. If a number of conditions hold (generally a consequence of the assumptions for the test and the null being true) then the test statistic will have a t-distribution, with degrees of freedom coming from the d.f. of the estimate in the denominator. Your Answer Sign up or log in, post as a guest. Required, but never shown By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy . Not the answer you're looking for? 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Rekindled people are very different Stargate "instructional" videos Garage door not closing when sunlight is bright 224 IMAGES Hypothesis Testing with Two Samples Hypothesis Testing Example Two Sample t-Test Two Sample Z Hypothesis Test Ch8: Hypothesis Testing (2 Samples) Hypothesis Testing with Two Samples PPT COMMENTS Chapter 10 Test procedure: we state our hypotheses, set up a decision rule, insert the sample statistics, and make a decision. Two sample tests are used to compare sample results taken from two different ______. populations. If the population variances are unknown and not assumed equal, to calculate the test statistic we replace a2 and a2^2 with the Two Sample t-test: Definition, Formula, and Example If the p-value that corresponds to the test statistic t with (n 1 +n 2-1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis. Two Sample t-test: Assumptions. For the results of a two sample t-test to be valid, the following assumptions should be met: How t-Tests Work: 1-sample, 2-sample, and Paired t-Tests 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. 10: Hypothesis Testing with Two Samples 10.E: Hypothesis Testing with Two Samples (Exercises) These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax. You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. Two-sample hypothesis testing In statistical hypothesis testing, a two-sample test is a test performed on the data of two random samples, each independently obtained from a different given population. The purpose of the test is to determine whether the difference between these two populations is statistically significant . There are a large number of statistical tests that ... Two-Sample t-Test The results for the two-sample t-test that assumes equal variances are the same as our calculations earlier. The test statistic is 2.79996. The software shows results for a two-sided test and for one-sided tests. The two-sided test is what we want (Prob > |t|). Our null hypothesis is that the mean body fat for men and women is equal. Hypothesis Testing: Two Samples The Population Mean: This image shows a series of histograms for a large number of sample means taken from a population.Recall that as more sample means are taken, the closer the mean of these means will be to the population mean. In this section, we explore hypothesis testing of two independent population means (and proportions) and also tests for paired samples of population means. PDF Confidence Intervals and Hypothesis Tests: Two Samples 9.4 t-Test to Compare Two Population Means: Independent Samples (Equal Variances) Small-Sample Hypothesis Test of: HD 0 1 2 0: PP , 0 1 2 0: d:, or HD 0 1 2 0 PP t We will use the same seven steps as always; however, we will need a new t-test statistic. If we assume equal variances, the test statistic for these problems will be: PDF Chapter 10 One- and Two-sample Tests of Hypotheses CHAPTER 10ONE- AND TWO-SAMPLE TESTS OF HYPOTHESESCon dence intervals represent the rst of t. o kinds of inference that we study in this course. Hy-pothesis testing, or test of significance is the s. cond common type of formal statistical inference. . It has a di erent goal than con dence intervals.The big picture is that the test of hypothesis ... Two Sample t-test Calculator A two sample t-test is used to test whether or not the means of two populations are equal. This type of test assumes that the two samples have equal variances. ... Hi, I noticed that in using the calculators for hypothesis testing, the default level of significance is 0.05 and you cannot change it. ... The Four Assumptions Made in a T-Test A two sample t-test is used to test whether or not the means of two populations are equal.. This type of test makes the following assumptions about the data: 1. Independence: The observations in one sample are independent of the observations in the other sample. 2. Normality: Both samples are approximately normally distributed. 3. Homogeneity of Variances: Both samples have approximately the ... Two-Sample Hypothesis Tests Two-Sample Hypothesis Tests. A two-sample test compares samples with each other rather than comparing with a benchmark, as in a one-sample test. For two proportions, the samples may be pooled if the population proportions are assumed equal, and the test statistic is the difference of proportions divided by the standard error, the square root of ... 8: Hypothesis Testing with Two Samples 8.5: Matched or Paired Samples. When using a hypothesis test for matched or paired samples, the following characteristics should be present: Simple random sampling is used. Sample sizes are often small. Two measurements (samples) are drawn from the same pair of individuals or objects. Differences are calculated from the matched or paired samples. Chapter 15 Hypothesis Testing: Two Sample Tests 15.1.2 Two Sample t test approach. For this we can use the two-sample t-test to compare the means of these two distinct populations. Here the alternative hypothesis is that the lottery players score more points H A: μL > μN L H A: μ L > μ N L thus the null hypothesis is H 0: μL ≤ μN L. H 0: μ L ≤ μ N L. We can now perform the test ... Hypotheses for a two-sample t test (video) On the other hand, a two-sample T test is where you're thinking about two different populations. For example, you could be thinking about a population of men, and you could be thinking about the population of women. And you wanna compare the means between these two, say, the mean salary. So, you have the mean salary for men and you have the ... Two-sample t test for difference of means And to do this two sample T test now, we assume the null hypothesis. We assume our null hypothesis, and remember we're assuming that all of our conditions for inference are met. And then we wanna calculate a T statistic based on this sample data that we have. 5.5 5.5 - Hypothesis Testing for Two-Sample Proportions. We are now going to develop the hypothesis test for the difference of two proportions for independent samples. The hypothesis test follows the same steps as one group. These notes are going to go into a little bit of math and formulas to help demonstrate the logic behind hypothesis testing ... 11: Hypothesis Testing with Two Samples A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions. 11.3: Matched or Paired Samples When using a hypothesis test for matched or paired samples, the following characteristics should be present: Simple random sampling is used. Sample sizes are often small. Two ... Chapter 11: Two Sample Hypothesis Tests Flashcards µ₁=µ₂. What is the null hypothesis for the difference between two means? Independent. One assumption for testing a hypothesis between two means is the samples are ______ of each other. 2-SampTTest. What function should you use when testing a hypothesis about the difference between two means on a TI-84? p₁=p₂. Two Sample Z-Test: Definition, Formula, and Example A two sample z-test is used to test whether two population means are equal. This test assumes that the standard deviation of each population is known. This tutorial explains the following: The formula to perform a two sample z-test. The assumptions of a two sample z-test. An example of how to perform a two sample z-test. Let's jump in! Research Methods 2 Quiz 7 Flashcards Two-sample hypothesis testing assumes that the samples are _____. independent of each other Suppose you are interested in testing whether the mean earning of men in the General Social Survey is representative of the earning of the entire U.S. male population. Hypothesis testing with two independent samples test statistics I'm learning a bit about hypothesis testing with two independent samples (continuous outcome), and I'm just curious about where some of the equations for the test statistics are derived and/or what they're measuring. Two sample hypothesis testing assumes that the samples are In two sample hypothesis testing, there are few assumption considered before we can assure that the right test is applied to the given data. In a two sample hypothesis testing, the two sample are assumed: to be independent to each other. The two sample hypothesis testing aims to test and compare samples that are not related. essay homework paper phd project resume review speech study thesis