Question

1. (AM#6c) Is there a difference in how much males and females use aggressive behavior to improve an angry mood? For the "Anger-Out" scores, compute a 99% confidence interval on the difference between gender means. (relevant section)
   
   
2. (AM#10) Calculate the 95% confidence interval for the difference between the mean Anger-In score for the athletes and non-athletes. What can you conclude? (relevant section)

25) (AM#10) Calculate the 95% confidence interval for the difference between the mean Anger-In score for the athletes and non-athletes. What can you conclude? (relevant section)

Learning Objectives

State the assumptions for computing a confidence interval on the difference between means

Compute a confidence interval on the difference between means

Format data for computer analysis

It is much more common for a researcher to be interested in the difference between means than in the specific values of the means themselves. We take as an example the data from the "Animal Research" case study. In this experiment, students rated (on a 7-point scale) whether they thought animal research is wrong. The sample sizes, means, and variances are shown separately for males and females in Table 1.

Table 1. Means and Variances in Animal Research study.

Condition n Mean Variance

Females 17 5.353 2.743

Males 17 3.882 2.985

As you can see, the females rated animal research as more wrong than did the males. This sample difference between the female mean of 5.35 and the male mean of 3.88 is 1.47. However, the gender difference in this particular sample is not very important. What is important is the difference in the population. The difference in sample means is used to estimate the difference in population means. The accuracy of the estimate is revealed by a confidence interval.

In order to construct a confidence interval, we are going to make three assumptions:

The two populations have the same variance. This assumption is called the assumption of homogeneity of variance.

The populations are normally distributed.

Each value is sampled independently from each other value.

The consequences of violating these assumptions are discussed in a later section. For now, suffice it to say that small-to-moderate violations of assumptions 1 and 2 do not make much difference.

A confidence interval on the difference between means is computed using the following formula:

Lower Limit = M1 - M2 -(tCL)()

Upper Limit = M1 - M2 +(tCL)()

where M1 - M2 is the difference between sample means, tCL is the t for the desired level of confidence, and is the estimated standard error of the difference between sample means. The meanings of these terms will be made clearer as the calculations are demonstrated.

We continue to use the data from the "Animal Research" case study and will compute a confidence interval on the difference between the mean score of the females and the mean score of the males. For this calculation, we will assume that the variances in each of the two populations are equal.

The first step is to compute the estimate of the standard error of the difference between means (). Recall from the relevant section in the chapter on sampling distributions that the formula for the standard error of the difference in means in the population is:

In order to estimate this quantity, we estimate σ2 and use that estimate in place of σ2. Since we are assuming the population variances are the same, we estimate this variance by averaging our two sample variances. Thus, our estimate of variance is computed using the following formula:

where MSE is our estimate of σ2. In this example,

MSE = (2.743 + 2.985)/2 = 2.864.

Note that MSE stands for "mean square error" and is the mean squared deviation of each score from its group's mean.

Since n (the number of scores in each condition) is 17,

== = 0.5805.

The next step is to find the t to use for the confidence interval (tCL). To calculate tCL, we need to know the degrees of freedom. The degrees of freedom is the number of independent estimates of variance on which MSE is based. This is equal to (n1 - 1) + (n2 - 1) where n1 is the sample size of the first group and n2 is the sample size of the second group. For this example, n1= n2 = 17. When n1= n2, it is conventional to use "n" to refer to the sample size of each group. Therefore, the degrees of freedom is 16 + 16 = 32.

Online: Calculator: Find t for confidence interval

From either the above calculator or a t table, you can find that the t for a 95% confidence interval for 32 df is 2.037.

We now have all the components needed to compute the confidence interval. First, we know the difference between means:

M1 - M2 = 5.353 - 3.882 = 1.471

We know the standard error of the difference between means is

= 0.5805

and that the t for the 95% confidence interval with 32 df is

tCL = 2.037

Therefore, the 95% confidence interval is

Lower Limit = 1.471 - (2.037)(0.5805) = 0.29

Upper Limit = 1.471 + (2.037)(0.5805) = 2.65

We can write the confidence interval as:

0.29 ≤ μf - μm ≤ 2.65

where μf is the population mean for females and μm is the population mean for males. This analysis provides evidence that the mean for females is higher than the mean for males, and that the difference between means in the population is likely to be between 0.29 and 2.65.

Formatting data for Computer Analysis

Most computer programs that compute t tests require your data to be in a specific form. Consider the data in Table 2.

Table 2. Example Data.

Group 1 Group 2

3 5

4 6

5 7

Here there are two groups, each with three observations. To format these data for a computer program, you normally have to use two variables: the first specifies the group the subject is in and the second is the score itself. For the data in Table 2, the reformatted data look as follows:

Table 3. Reformatted Data.

G Y

1 3

1 4

1 5

2 5

2 6

2 7

To use Analysis Lab to do the calculations, you would copy the data and then

Click the "Enter/Edit User Data" button. (You may be warned that for security reasons you must use the keyboard shortcut for pasting data.)

Paste your data.

Click "Accept Data."

Set the Dependent Variable to Y.

Set the Grouping Variable to G.

Click the t-test confidence interval button.

The 95% confidence interval on the difference between means extends from -4.267 to 0.267.

Computations for Unequal Sample Sizes (optional)

The calculations are somewhat more complicated when the sample sizes are not equal. One consideration is that MSE, the estimate of variance, counts the sample with the larger sample size more than the sample with the smaller sample size. Computationally this is done by computing the sum of squares error (SSE) as follows:

where M1 is the mean for group 1 and M2 is the mean for group 2. Consider the following small example:

Table 4. Example Data.

Group 1 Group 2

3 2

4 4

5

M1 = 4 and M2 = 3.

SSE = (3-4)2 + (4-4)2 + (5-4)2 + (2-3)2 + (4-3)2 = 4

Then, MSE is computed by: MSE = SSE/df

where the degrees of freedom (df) is computed as before:

df = (n1 -1) + (n2 -1) = (3-1) + (2-1) = 3.

MSE = SSE/df = 4/3 = 1.333.

The formula

=

is replaced by

=

where nh is the harmonic mean of the sample sizes and is computed as follows:

nh = = = 2.4

and

= = 1.054.

tCL for 3 df and the 0.05 level = 3.182.

Therefore the 95% confidence interval is

Lower Limit = 1 - (3.182)(1.054)= -2.35

Upper Limit = 1 + (3.182)(1.054)= 4.35

We can write the confidence interval as:

-2.35 ≤ μ1 - μ2 ≤ 4.35

24)(AM#6c) Is there a difference in how much males and females use aggressive behavior to improve an angry mood? For the "Anger-Out" scores, compute a 99% confidence interval on the difference between gender means.

Learning Objectives

State the assumptions for computing a confidence interval on the difference between means

Compute a confidence interval on the difference between means

Format data for computer analysis

It is much more common for a researcher to be interested in the difference between means than in the specific values of the means themselves. We take as an example the data from the "Animal Research" case study. In this experiment, students rated (on a 7-point scale) whether they thought animal research is wrong. The sample sizes, means, and variances are shown separately for males and females in Table 1.

Table 1. Means and Variances in Animal Research study.

Condition n Mean Variance

Females 17 5.353 2.743

Males 17 3.882 2.985

As you can see, the females rated animal research as more wrong than did the males. This sample difference between the female mean of 5.35 and the male mean of 3.88 is 1.47. However, the gender difference in this particular sample is not very important. What is important is the difference in the population. The difference in sample means is used to estimate the difference in population means. The accuracy of the estimate is revealed by a confidence interval.

In order to construct a confidence interval, we are going to make three assumptions:

The two populations have the same variance. This assumption is called the assumption of homogeneity of variance.

The populations are normally distributed.

Each value is sampled independently from each other value.

The consequences of violating these assumptions are discussed in a later section. For now, suffice it to say that small-to-moderate violations of assumptions 1 and 2 do not make much difference.

A confidence interval on the difference between means is computed using the following formula:

Lower Limit = M1 - M2 -(tCL)()

Upper Limit = M1 - M2 +(tCL)()

where M1 - M2 is the difference between sample means, tCL is the t for the desired level of confidence, and is the estimated standard error of the difference between sample means. The meanings of these terms will be made clearer as the calculations are demonstrated.

We continue to use the data from the "Animal Research" case study and will compute a confidence interval on the difference between the mean score of the females and the mean score of the males. For this calculation, we will assume that the variances in each of the two populations are equal.

The first step is to compute the estimate of the standard error of the difference between means (). Recall from the relevant section in the chapter on sampling distributions that the formula for the standard error of the difference in means in the population is:

In order to estimate this quantity, we estimate σ2 and use that estimate in place of σ2. Since we are assuming the population variances are the same, we estimate this variance by averaging our two sample variances. Thus, our estimate of variance is computed using the following formula:

where MSE is our estimate of σ2. In this example,

MSE = (2.743 + 2.985)/2 = 2.864.

Note that MSE stands for "mean square error" and is the mean squared deviation of each score from its group's mean.

Since n (the number of scores in each condition) is 17,

== = 0.5805.

The next step is to find the t to use for the confidence interval (tCL). To calculate tCL, we need to know the degrees of freedom. The degrees of freedom is the number of independent estimates of variance on which MSE is based. This is equal to (n1 - 1) + (n2 - 1) where n1 is the sample size of the first group and n2 is the sample size of the second group. For this example, n1= n2 = 17. When n1= n2, it is conventional to use "n" to refer to the sample size of each group. Therefore, the degrees of freedom is 16 + 16 = 32.

Online: Calculator: Find t for confidence interval

From either the above calculator or a t table, you can find that the t for a 95% confidence interval for 32 df is 2.037.

We now have all the components needed to compute the confidence interval. First, we know the difference between means:

M1 - M2 = 5.353 - 3.882 = 1.471

We know the standard error of the difference between means is

= 0.5805

and that the t for the 95% confidence interval with 32 df is

tCL = 2.037

Therefore, the 95% confidence interval is

Lower Limit = 1.471 - (2.037)(0.5805) = 0.29

Upper Limit = 1.471 + (2.037)(0.5805) = 2.65

We can write the confidence interval as:

0.29 ≤ μf - μm ≤ 2.65

where μf is the population mean for females and μm is the population mean for males. This analysis provides evidence that the mean for females is higher than the mean for males, and that the difference between means in the population is likely to be between 0.29 and 2.65.

Formatting data for Computer Analysis

Most computer programs that compute t tests require your data to be in a specific form. Consider the data in Table 2.

Table 2. Example Data.

Group 1 Group 2

3 5

4 6

5 7

Here there are two groups, each with three observations. To format these data for a computer program, you normally have to use two variables: the first specifies the group the subject is in and the second is the score itself. For the data in Table 2, the reformatted data look as follows:

Table 3. Reformatted Data.

G Y

1 3

1 4

1 5

2 5

2 6

2 7

To use Analysis Lab to do the calculations, you would copy the data and then

Click the "Enter/Edit User Data" button. (You may be warned that for security reasons you must use the keyboard shortcut for pasting data.)

Paste your data.

Click "Accept Data."

Set the Dependent Variable to Y.

Set the Grouping Variable to G.

Click the t-test confidence interval button.

The 95% confidence interval on the difference between means extends from -4.267 to 0.267.

Computations for Unequal Sample Sizes (optional)

The calculations are somewhat more complicated when the sample sizes are not equal. One consideration is that MSE, the estimate of variance, counts the sample with the larger sample size more than the sample with the smaller sample size. Computationally this is done by computing the sum of squares error (SSE) as follows:

where M1 is the mean for group 1 and M2 is the mean for group 2. Consider the following small example:

Table 4. Example Data.

Group 1 Group 2

3 2

4 4

5

M1 = 4 and M2 = 3.

SSE = (3-4)2 + (4-4)2 + (5-4)2 + (2-3)2 + (4-3)2 = 4

Then, MSE is computed by: MSE = SSE/df

where the degrees of freedom (df) is computed as before:

df = (n1 -1) + (n2 -1) = (3-1) + (2-1) = 3.

MSE = SSE/df = 4/3 = 1.333.

The formula

=

is replaced by

=

where nh is the harmonic mean of the sample sizes and is computed as follows:

nh = = = 2.4

and

= = 1.054.

tCL for 3 df and the 0.05 level = 3.182.

Therefore the 95% confidence interval is

Lower Limit = 1 - (3.182)(1.054)= -2.35

Upper Limit = 1 + (3.182)(1.054)= 4.35

We can write the confidence interval as:

-2.35 ≤ μ1 - μ2 ≤ 4.35

Answer 1

Following assumptions are required:

1. The two populations have the same variance.
2. The populations are normally distributed.
3. Each value is sampled independently from each other value.

Test and CI for Two Variances

Method

Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.05

Statistics

Sample N StDev Variance
1 17 1.656 2.743
2 17 1.728 2.985

Ratio of standard deviations = 0.959
Ratio of variances = 0.919

95% Confidence Intervals

CI for
Distribution CI for StDev Variance
of Data Ratio Ratio
Normal (0.577, 1.593) (0.333, 2.537)

Tests

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 16 16 0.92 0.868

Since p-value=0.868>0.05 so we can assume two population variances are same.

95% C.I. of difference i.e. (-2.35, 4.35) is also contain zero so we get same conclusion as above.