Question

1) An educational psychologist studies the effect of frequent testing on retention of class material. In one section of an introductory course, students are given quizzes each week. A second section of the same course receives only two tests during the semester. At the end of the semester, both sections receive the same final exam, and the scores are summarized below.

Frequent Quizzes n = 20

M = 73

SS = 722

Two Exams n = 20

M = 68

SS = 798

a. Do the data indicate that testing frequency has a significant effect on performance? Use a two-tailed test at the .05 level of significance. State the null hypothesis and tcritical. Do you need to be concerned about homogeneity of variance? Conclude with a suitable summary statement.

b. If the first sample variance is s2 = 84 and the second sample has s2 = 96, do the data indicate that testing frequency has a significant effect? Again, use a two-tailed test with α = .05. Conclude with a suitable summary statement

c. Describe how the size of the variance affects the outcome of the hypothesis test? Be specific in your description.

2) A researcher conducts an independent-measures study examining how the brain chemical serotonin is related to aggression. One sample of rats serves as a control group and receives a placebo that does not affect normal levels of serotonin. A second sample of rats receives a drug that lowers brain levels of serotonin. Then the researcher tests the animals by recording the number of aggressive responses each of the rats display. The data are as follows.

Control n = 10

M = 14

SS = 180.5

Low Serotonin n = 15

M = 19

SS = 130.0

a. Does the drug have a significant effect on aggression? Use an alpha level of .05, two-tailed. State the null hypothesis and tcritical. Do you need to be concerned about homogeneity of variance? Conclude with a suitable summary statement.

b. Compute Cohen’s d and r2 to evaluate the size of the treatment effect. Conclude with a suitable summary statement.

Answer 1

SOLUTION 1: Frequent Quizzes n = 20

M 1 = 73

SS1 = 722

VARIANCE 1 = 722/19= 38

SD1= SQRT(38)=6.16

Two Exams n = 20

M2 = 68

SS 2= 798

VARIANCE2= 798/19=42

SD2= SQRT(42)=6.48

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.396 and FU​=2.526, and since F=0.904, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region: Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=38. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=2.024, for α=0.05 and df=38.

The rejection region for this two-tailed test is R={t:∣t∣>2.024}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

X - X2 -1)(1+) (ni-1)si+n2-1) / 1 1 73-68 = 0.396 20-116.48 (20-16.16° +(20-16.482 20-20- 9 0 (1 )

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=0.396≤tc​=2.024, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.6946, and since p=0.6946≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

B] VARIANCE 1=84

SD 1= SQRT(84)=9.17

VARIANCE 2=96

SD 2= SQRT(96)= 9.80

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

9.172 9.802 = 0.876

The critical values are FL​=0.396 and FU​=2.526, and since F=0.876, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region: Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=38. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=2.024, for α=0.05 and df=38.

The rejection region for this two-tailed test is R={t:∣t∣>2.024}.

(3) Test Statistics:

Xi - X2 (n.1 – 1) si +(12-1) (3 + ) ni+n2-2 73-68 (20-1)9.17 +(20-1)9.802/ 1 80( 20-20-2 = = 0.176 +) 1 )

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=0.176≤tc​=2.024, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.8616, and since p=0.8616≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion:It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

c] As the size of variances increased the value of t statistic got decreased. For variance 1= 38 and variance 2= 42 t stat=0.396 while for New variances it is t=0.176 just half.

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