Answer to Question #250372 in Statistics and Probability for smilynne

Question

In each problem, provide the following:

a. State the hypotheses and identify the claim.

b. Find the critical value(s)

c. Find the test value

d. Make the decision

e. Summarize the result

2. An industrial psychologist obtains scores on a job-selection test from 41 men and 31

women, with the following results: men, M = 48.75 (SD = 9.0); women, M = 46.07 (SD

= 10.0). Test this difference for significance at both the .05 and .01 levels (two-tailed).

(Use 𝑷-value method)

Answer 1

$H_0:\mu_1=\mu_2$

$H_a:\mu_1\neq \mu_2$

$t=\frac{\overline{x}_1-\overline{x}_2}{\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}}=\frac{48.75-46.07}{\sqrt{9^2/41+10^2/31}}=1.175$

$df=\frac{(\sigma_1^2/n_1+\sigma_2^2/n_2)^2}{\frac{(\sigma_1^2/n_1)^2}{n_1-1}+\frac{(\sigma_2^2/n_2)^2}{n_2-1}}=\frac{(9^2/41+10^2/31)^2}{\frac{(9^2/41)^2}{40}+\frac{(10^2/31)^2}{30}}=60.87$

t value for $\alpha=0.05$ :

$t=2$

The test statistic is lower than the t value. We fail to reject the hypothesis of equal means.

Men and women have same  scores on a job-selection test.

value for $\alpha=0.01$ :

$t=2.66$

The test statistic is lower than the t value. We fail to reject the hypothesis of equal means.

Men and women have same  scores on a job-selection test.

Using p- value:

p-value for t=1.175, df=60.87:

p-value = 0.2446

for $\alpha=0.05$ :

p-value > $\alpha$ , so we accept the null hypothesis.

for $\alpha=0.01$ :

p-value > $\alpha$ , so we accept the null hypothesis.