Question

For a large supermarket chain in a particular​ state, a​ women's group claimed that female employees...

For a large supermarket chain in a particular​ state, a​ women's group claimed that female employees were passed over for management training in favor of their male colleagues. The company denied this​ claim, saying it picked the employees from the eligible pool at random to receive this training.​ Statewide, the large pool of more than 1000 eligible employees who can be tapped for management training is 40​% female and 60​% male. Since this program​ began,30 of the 45 employees chosen for management training were male and 15 were female.

a. The company claims that it selected employees for training according to their proportion in the pool of eligible employees. Define a parameter of interest and state this claim as a hypothesis.

b. State the null and alternative hypotheses for a test to investigate the strength of evidence to support the​ women's claim of gender bias.​ (Hint: Gender bias means that either males or females are disproportionately​ selected.)

c. The table given in the problem shows results of using technology to do a​ large-sample analysis. Explain why the​ large-sample analysis is justified. Choose the correct answer below.

d. To what alternative hypothesis does the​ P-value in the table​ refer? Use it to find the​ P-value for the significance test you specified in part​ b, and interpret it. To what alternative hypothesis does the​ P-value in the table​ refer?

e. What decision would you make for a 0.05 significance​ level?

Homework Answers

Answer #1

(a) The parameter of interest is management training employees.

Claim: Either males or females are disproportionately​ selected

(b) The hypothesis being tested is:

H0: p1 = p2

Ha: p1 ≠ p2

(c) n*p, n*q 10

(d) The p-value is 0.0016.

p1 p2 pc
0.6667 0.3333 0.5 p (as decimal)
30/45 15/45 45/90 p (as fraction)
30. 15. 45. X
45 45 90 n
0.3333 difference
0. hypothesized difference
0.1054 std. error
3.16 z
.0016 p-value (two-tailed)

(e) Since the p-value (0.0016) is less than the significance level (0.05), we can reject the null hypothesis.

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