Question

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 Two Exams

n=20 n=20

M=73 M=68

a. If the first sample variance is s2 = 38 and the second sample has s2 = 42, do the data indicate that testing frequency has a significant effect on performance? Use a two-tailed test at the .05 level of significance. (Note: because the two sample are the same size, the pooled variance is simply the average of the two sample variances.)

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 x = .05.

c. Describe how the size of the variance effects the outcome of the hypothesis test.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = 2.00
DF = 20 + 20 -2

D.F = 38
t = [ (x₁ - x₂) - d ] / SE

t = 2.50

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 38 degrees of freedom is more extreme than 2.50; that is, less than -2.50 or greater than 2.50.

Thus, the P-value = 0.017

Interpret results. Since the P-value (0.017) is less than the significance level (0.05), we have to reject null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that testing frequency has a significant effect on performance.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = 3.0
DF = 20 + 20 -2

D.F = 38
t = [ (x₁ - x₂) - d ] / SE

t = 1.667

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 38 degrees of freedom is more extreme than 1.667; that is, less than -1.667 or greater than 1.667.

Thus, the P-value = 0.104

Interpret results. Since the P-value (0.104) is greater than the significance level (0.05), hence we failed to reject null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that testing frequency has a significant effect on performance.

c) As the size of the variance increases the the chances of rejecting the null hypothesis decreases. We can also say that the chances of getting significant results decreases.