Chapter 10 The t Test for Two Independent Samples

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The t statistic is1.95 .

1.95 .

You conduct an independent-measures t test. Given your null and alternative hypotheses, this is atwo-tailed test. To use the Distributions tool to find the rejection region, you first need to set the degrees of freedom. The degrees of freedom is46 .

two-tailed 46 .

two-tailed 55 .

two-tailed test. 68 .

The critical t-scores that form the boundaries of the rejection region for α = 0.05 are ±1.995 .

±1.995 .

The critical t-scores that form the boundaries of the rejection region for α = 0.01 are

±2.668 .

Sample Size Degrees of Freedom Sample Mean Standard Deviation Sums of Squares Sample 1n₁ = 21 df₁ = 20 M₁ = 27.6 s₁ = 4.5 SS₁ = 405

Sample 2 n₂ = 11 df₂ = 10 M₂ = 21.5 s₂ = 5.5 SS₂ = 302.5

Given the results of the F-max test, the researcher's conclusion is that

the data do not provide evidence that the homogeneity of variance assumption has been violated

Given the results of the F-max test, the researcher's conclusion is that

the data suggest the homogeneity of variance assumption has been violated .

The t statisticdoes not lie in the critical region for a one-tailed hypothesis test. Therefore, the null hypothesis isnot rejected . The psychologistcannot conclude that African American couples have less relationship satisfaction than Latino couples.

- does not - not rejected - cannot

In calculating theestimated standard error of M₁ - M₂ , you typically first need to calculate thepooled variance . Theestimated standard error of M₁ - M₂ is the value used in the denominator of the t statistic for the independent-measures t test.

- estimated standard error of M₁ - M₂ - pooled variance - estimated standard error of M₁ - M₂

Calculate the t statistic. The t statistic is-1.70 .

-1.70 .

The t statistic is-1.70 .

-1.70 .

With α = 0.01, the critical t-score (the value for a t-score that separates the tail from the main body of the distribution, forming the critical region) is

-2.423

With α = 0.01, the critical t-score (the value for a t-score that separates the tail from the main body of the distribution, forming the critical region) is-2.423 .

-2.423 .

The researcher decides to use a Hartley's F-max test. The value of F-max is1.604 .

1.604 .

The estimated standard error of the difference in sample means for your study is

1.807 .

The estimated standard error of M₁ - M₂ for your study is2.083 .

2.083 .

Calculate the t statistic. The t statistic is2.45 . (Hint: For the most precise results, retain four decimal places from your previous calculation to calculate the t statistic.)

2.45 .

The t statistic is2.55 .

2.55 .

The critical t-scores that form the boundaries of the rejection region for α = 0.01 are ±2.687

2.687 .

The pooled variance for your study is23.583 . (Note: You are being asked for this value to three decimal places, because you will need to use it in succeeding calculations. For the most accurate results, retain these three decimal places throughout the calculations.)

23.583

The t statistic for your independent-measures t test, when the null hypothesis is that the two population means are the same, is3.38 .

3.38 .

The degrees of freedom for this t statistic is30 .

30 .

Using the partial Critical Values for the F-max Statistic table below, the critical value for the researcher's F-max with α = 0.01 is

4.07 .

The researcher decides to use a Hartley's F-max test. The value of F-max is

4.592 .

The pooled variance for your study is60.240 . (Note: You are being asked for this value to three decimal places, because you will need to use it in succeeding calculations. For the most accurate results, retain these three decimal places throughout the calculations.)

60.240

Using the partial Critical Values for the F-max Statistic table below, the critical value for the researcher's F-max with α = 0.05 is

7.15

To calculate the t statistic, you first need to calculate the estimated standard error of the difference in means. To calculate this estimated standard error, you first need to calculate the pooled variance. The pooled variance is

72.1304 . The estimated standard error of the difference in means is2.4538

In order to calculate the t statistic, you first need to calculate the standard error under the assumption that the null hypothesis is true. In order to calculate the standard error, you first need to calculate the pooled variance. The pooled variance is s2pp2 =72.8696 . The standard error is s(M1 - M2)(M1 - M2) =2.4664 .

72.8696 2.4664 .

91.0182 2.5308 .

For the independent-measures t test, which of the following describes the pooled variance (whose symbol iss2pp2 )?

A weighted average of the two sample variances (weighted by the sample sizes)

For the independent-measures t test, which of the following describes the estimated standard error of M₁ - M₂ (whose symbol iss(M1 - M2)(M1 - M2) )?

An estimate of the standard distance between the difference in sample means (M₁ - M₂) and the difference in the corresponding population means (μ₁ - μ₂)

For the independent-measures t test, which of the following describes the estimated standard error of the difference in sample means (whose symbol iss(M1 - M2)(M1 - M2) )?

An estimate of the standard distance between the difference in sample means (M₁ - M₂) and the difference in the corresponding population means (μ₁ - μ₂)

Suppose a clinical psychologist sets out to look at the role of intactness of family of origin in relationship longevity. He decides to measure marital satisfaction in a group of couples from divorced families and a group of couples from intact families. He chooses the Marital Satisfaction Inventory, because it refers to "partner" and "relationship" rather than "spouse" and "marriage," which makes it useful for research with both traditional and nontraditional couples. Higher scores on the Marital Satisfaction Inventory indicate greater satisfaction. There is one score per couple. Assume that these scores are normally distributed and that the variances of the scores are the same among couples from divorced families as among couples from intact families.

H₀: μcouples from divorced familiescouples from divorced families≥ μcouples from intact familiescouples from intact families Points: 0 / 1 H₁: μcouples from divorced familiescouples from divorced families< μcouples from intact familiescouples from intact families Points: 0 / 1 This is aone- tailed test.

The statistician selects a random sample of size n₁ = 35 nights the sample mean M₁ = 9.09 and the sample variance s21s12= 28. Likewise, she selects a random sample of size n₂ = 32 nights that the northside restaurant is open. For each night in the sample, she collects data on the number of filet mignons sold at the northside location and computes the sample mean M₂ = 7.49 and the sample variance s22s22= 27. The statistician checks and concludes that the data satisfy the required assumptions for the independent-measures t test. Then she computes the 95% confidence interval for estimating the difference between the mean number of filet mignons sold per night at the southside restaurant and the mean number of filet mignons sold per night at the northside restaurant. This 95% confidence interval is 1.60 ± 2.5624 filet mignons.

If she were to formulate null and alternative hypotheses as H₀: µ₁ - µ₂ = 0, H₁: µ₁ - µ₂ ≠ 0 and conduct a hypothesis test with α = 0.05, the null hypothesis is not rejected based on the result that a difference of zero is in the computed interval. Hence, she would conclude that thereis not a significant difference between the mean nightly sales of filet mignons between the two restaurants. answer: is not is is not

Assumptions underlying the independent-measures t test A professor believes that students at her large university who exercise daily perform better in statistics classes. Since all students at the university are required to take Introduction to Statistics, she randomly selects 17 students who exercise daily and 22 students who exercise at most once per week. She obtains their scores in the final exam in Introduction to Statistics and finds that the students who did not exercise daily primarily produced scores in the 90s, with some scores in the 80s and a very few scores in the 70s and 60s. The students who did exercise daily also had a large number of scores in the 90s and an almost equal number in the 60s, with very few scores in between. Would it be valid for the professor to use the independent-measures t test to test whether students at her large university who exercise daily perform better in statistics classes?

No, because the two populations from which the samples are selected do not appear to be normally distributed.

Suppose you conduct a study using an independent-measures research design, and you intend to use the independent-measures t test to test whether the means of the two independent populations are the same. The following is a table of the information you gather. Fill in any missing values.

Sample Size Degrees of Freedom Sample Mean Standard Deviation Sums of Squares Sample 1n₁ = 41df₁ = 40 M₁ = 14.3s₁ = 8.2SS₁ = 2,689.6 Sample 2n₂ = 21df₂ = 20 M₂ = 13.6s₂ = 6.8 SS₂ = 924.8 The degrees of freedom for each sample, df₁ and df₂, are simply the respective sample sizes minus one. Therefore, df₁ = 41 - 1 = 40, and df₂ = 21 - 1 = 20. Being able to switch between the standard deviation, s, and the sum of squares, SS, is important, because sometimes you may have the sample standard deviation and sometimes you may have the sample sum of squares. You will need to know when to use which. Recall that s² = SS/df. So, if you have the standard deviation (s) and want to find the sum of squares (SS), you first need to solve the equation for SS (by multiplying each side of the equation by df): SSSS = = (df)(s²)(df)(s²) You know the standard deviation of Sample 1: s₁ = 8.2. So to get SS₁, you have to square the standard deviation to get the variance, and then multiply the variance (s²) by the degrees of freedom: SS₁ =(df₁) (s2112) = (40)(8.2)² = 2,689.6. For Sample 2, you are given SS₂ and need to calculate s₂. Since SS₂ = (df₂)(s2222), s₂ = √(SS₂ / df₂) = √(924.8/20) = 6.8.

The psychologist collects the data. A group of 25 African American couples scored an average of 26.3 with a sample standard deviation of 8 on the Marital Satisfaction Inventory. A group of 23 Latino couples scored an average of 20.3 with a sample standard deviation of 9. Use the t distribution table. To use the table, you will first need to calculate the degrees of freedom.

The degrees of freedom are46 .

The psychologist collects the data. A group of 30 couples from divorced families scored an average of 21.5 with a sample standard deviation of 10 on the Marital Satisfaction Inventory. A group of 27 couples from intact families scored an average of 25.8 with a sample standard deviation of 9. Use the t distribution table. To use the table, you will first need to calculate the degrees of freedom.

The degrees of freedom are55 .

The t statistic for your independent-measures t test, when the null hypothesis is that the two population means are the same, is0.34 .

The degrees of freedom for this t statistic is60 .

Suppose a social psychologist sets out to look at the role of race in relationship longevity. She decides to measure marital satisfaction in a group of African American couples and a group of Latino couples. She chooses the Marital Satisfaction Inventory, because it refers to "partner" and "relationship" rather than "spouse" and "marriage," which makes it useful for research with both traditional and nontraditional couples. Higher scores on the Marital Satisfaction Inventory indicate greater satisfaction. There is one score per couple. Assume that these scores are normally distributed and that the variances of the scores are the same among African American couples as among Latino couples.

The psychologist thinks that African American couples will have less relationship satisfaction than Latino couples. She identifies the null and alternative hypotheses as: H₀: μAfrican American couplesAfrican American couples≥ μLatino couplesLatino couples H₁: μAfrican American couplesAfrican American couples< μLatino couplesLatino couples This is a one- tailed test.

The researcher's null hypothesis is that the rats given the plant compound consume the same number of calories as the rats given a placebo. Since the rats in the treatment group are independent of the rats in the control group, the researcher realizes that the design of his study is an independent-measures design. The researcher knows that the total calories consumed by a rat over the course of a week are normally distributed. However, the researcher is not sure that the variance (σ²) of the total calories consumed by a rat over the course of a week is the same regardless of whether the rat is given the plant compound or a placebo. Suppose the mean number of weekly calories consumed by the rats given the plant compound (the treatment group) is 839 with a standard deviation of 42, and the mean number of calories consumed by the rats in the control group is 881 with a standard deviation of 90.

The researcher is unsure of which of the following required assumptions for the independent-measures t test? answer: Is there homogeneity of variance?

A researcher wants to investigate claims that a new plant compound is effective in curbing appetite. She randomly assigns 12 rats to a treatment or control group (6 rats in each group). Rats in the treatment group are given the plant compound daily for a week, while rats in the control group are given a placebo. Over the course of the week, the researcher monitors the total number of calories consumed by all 12 rats. All the rats are kept separate from one another, but in their typical environments, for the duration of the week. The researcher's null hypothesis is that the rats given the plant compound consume the same number of calories as the rats given a placebo. Since the rats in the treatment group are independent of the rats in the control group, the researcher realizes that the design of her study is an independent-measures design.

The researcher knows that the total calories consumed by a rat over the course of a week are normally distributed. However, the researcher is not sure that the standard deviation (σ) of the total calories consumed by a rat over the course of a week is the same regardless of whether the rat is given the plant compound or a placebo. Suppose the mean number of weekly calories consumed by the rats given the plant compound (the treatment group) is 881 with a standard deviation of 76, and the mean number of calories consumed by the rats in the control group is 948 with a standard deviation of 60. The researcher is unsure of which of the following required assumptions for the independent-measures t test? Answer: Is there homogeneity of variance?

According to the theory of stereotype threat, situational pressures can lead to decreased performance on tests of cognitive abilities. Joshua Aronson tested how Caucasian engineering students performed on a math test when placed under a form of situational pressure by randomly assigning 100 of these students to either the control or experimental groups. The control group was told that they were taking a test of general math ability. Members of the experimental group were presented with several news articles discussing the increasing difference in math scores between Asian and Caucasian students and were told that the purpose of the test was to explore these differences. Would it be valid for Dr. Aronson to use the independent-measures t test to test whether drawing attention to stereotypes about racial groups and math ability affects math scores?

Yes, because the sample size is large and there is no reason to believe the assumptions of the independent-measures t test are violated.

Can you still use the independent-measures t test if the homogeneity of variance assumption is violated?

Yes, you don't use a pooled variance and you recompute the degrees of freedom so that the resulting df is smaller.

You use Professor Wakefield's research to design an experiment with two groups of high school students from Israel. You show the tobacco company's youth smoking prevention ads to one of the groups of students every day for a week at the beginning of their math class. After the week, you assess the groups on the degree to which the students prefer to date/not date smokers, using a 5-point Likert scale. You do not have a prediction about the effect of the ads, as they are supposed to prevent smoking, but Professor Wakefield's findings suggested the ads might have the opposite effect if they have any effect at all. There are 49 students in each group. The first group, who saw the ads, scored an average of 3.4 with a sample standard deviation of 0.8 on the 5-point Likert scale. The second group, who did not see the ads, scored an average of 2.4 with a sample standard deviation of 0.9 on the 5-point Likert scale.

You find that the estimated standard error of the difference in means is 0.1720, the estimated Cohen's d is1.174 , the t statistic is 5.81, and the r² is0.26

You use Professor Wakefield's research to design an experiment with two groups of middle school students from Norway. You show the tobacco company's youth smoking prevention ads to one of the groups of students every day for a week at the beginning of their math class. After the week, you assess the groups on the degree to which the students perceive smoking as harmless/harmful, using a 5-point Likert scale. You do not have a prediction about the effect of the ads, as they are supposed to prevent smoking, but Professor Wakefield's findings suggested the ads might have the opposite effect if they have any effect at all. There are 64 students in each group. The first group, who saw the ads, scored an average of 2.9 with a sample standard deviation of 1.4 on the 5-point Likert scale. The second group, who did not see the ads, scored an average of 2.0 with a sample standard deviation of 1.5 on the 5-point Likert

You find that the estimated standard error of the difference in means is 0.2565, the estimated Cohen's d is0.620 , the t statistic is 3.51, and the r² is0.09 .

Which of the following experiments uses independent samples?

You want to compare the mean SAT score of students who take a new SAT preparation course to that of students who don't take an SAT preparation course. So you compare the mean SAT score of a randomly selected group of students who took a new SAT preparation course to the mean SAT score of a randomly selected group of students who didn't take an SAT preparation course.

Using Cohen's d and Cohen's criteria for evaluating Cohen's d, this is amedium to large effect size. Using r² and Cohen's criteria for evaluating r², this is amedium effect size.

d = 0.2Small effectd = 0.5Medium effectd = 0.8Large effect r² = 0.01Small effectr² = 0.09Medium effectr² = 0.25Large effect

Using Cohen's d and Cohen's criteria for evaluating Cohen's d, this is alarge effect size. Using r² and Cohen's criteria for evaluating r², this is alarge effect size.

d = 0.2Small effectd = 0.5Medium effectd = 0.8Large effect r² = 0.01Small effectr² = 0.09Medium effectr² = 0.25Large effect

The t statisticdoes not lie in the critical region for a one-tailed hypothesis test. Therefore, the null hypothesis isnot rejected . The psychologistcannot conclude that couples from divorced families have less relationship satisfaction than couples from intact families.

does not not rejected cannot

The t statisticdoes not lie in the rejection region. Therefore, the null hypothesis isnot rejected . Youcannot conclude that bullies have a different mean anxiety score than bystanders. Thus, it can be said that these two means are not significantly different from one another.

does not lie not rejected cannot not significantly

he t statisticdoes not lie in the rejection region. Therefore, the null hypothesis isnot rejected . Youcannot conclude that bully-victims have a different mean depression score than bystanders. Thus, it can be said that these two means are not significantly different from one another.

does not lie not rejected cannot not significantly

The t statisticlies in the rejection region. Therefore, the null hypothesis isrejected . Youcan conclude that victims have a different mean anxiety score than bully-victims. Thus, it can be said that these two means are significantly different from one another.

lies rejected can significantly

In calculating s(M1 - M2)(M1 - M2) , you typically first need to calculate s2pp2 . s(M1 - M2)(M1 - M2) is the value used in the denominator of the t statistic for the independent-measures t test.

s(M1 - M2)(M1 - M2) s2pp2 s(M1 - M2)(M1 - M2)

In order to calculate the t statistic, you first need to calculate the standard error under the assumption that the null hypothesis is true. In order to calculate the standard error, you first need to calculate the pooled variance. The pooled variance is s2pp2 =108.7941 . The standard error is s(M1 - M2)(M1 - M2) =2.5098 .

s2pp2 =108.7941 s(M1 - M2)(M1 - M2) =2.5098 .

In order to calculate the t statistic, you first need to calculate the standard error under the assumption that the null hypothesis is true. In order to calculate the standard error, you first need to calculate the pooled variance. The pooled variance is s2pp2 =91.0182 . The standard error is s(M1 - M2)(M1 - M2) =2.5308 .

Chapter 10 The t Test for Two Independent Samples

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