Question

Friedman and Rosenman (1974) have classified people into two categories: Type A personalitites and Type B personalities. Type A's are hard-driving, competitive, and ambitious. Type B's are more relaxed, easy-going people. One factor that dfferentiates these two groups is the chronically high level of frustration experienced by Type A's. To demonstrate this phenomenon, separate samples of Type A's and Type B's were obtained, with n = 8 in each sample. The individual subjects were all given a frustration inventory measuring level of frustration. The average score for the Type A's was M1 = 85.50 with SS1 = 718.33 and the Type B's average was M2 = 70.25 with SS2 = 642.50. Do these data indicate a significant difference between the two groups? Use a two-tailed test with α = .01.

What is the correct statistical analysis for this study?

a. t-test for independent samples

b. t-test for related samples (repeated measures)

What is the null hypothesis (H0) for this study?

a. μ1 - μ2 = 0

b. μD = 0

c. μ1 - μ2 ≠ 0

d. μD ≠ 0

Answer 1

What is the correct statistical analysis for this study?

a. t-test for independent samples

What is the null hypothesis (H0) for this study?

a. μ1 - μ2 = 0

Now performing the actual test:

(Assuming equal variances)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: ​

Ha: ​

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.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df=14. 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​=2.977, for α=0.01 and df=14.

The rejection region for this two-tailed test is R={t:∣t∣>2.977}

(3) Test Statistics

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

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=1.176≤tc​=2.977, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.2591, and since p=0.2591≥0.01, 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 0.01 significance level.

Graphically

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