We are interested in seeing if there is sex discrimination in the job hiring pro

Question

We are interested in seeing if there is sex discrimination in the job hiring process. So, we give hiring managers two resumes with equal qualifications, experience, etc. The only difference, one is a man and one is a woman which is obvious by the name on the resume. From our random sample of 398 hiring managers, the male would be given the job 212 times.

Suppose that we would like to test to see if there was any sex discrimination. Formally test this hypothesis using a 5% significance level. Be sure to show all steps! (5 points)

Use StatCrunch to verify your test statistic and p-value. Do these numbers match yours? Insert output HERE! (0.5 point)

Additional information was actually collected on this data. There were two different types of jobs – business jobs and design jobs (see data: gender.xls). Create two pie charts using StatCrunch (one for business jobs and one for design jobs) that look at who would be given the job. Interpret the graphs. Insert them HERE!! (1.5 points)

Conduct the appropriate test to determine if there is a relationship between the two variables.   Show all steps and use a 1% significance level. Be sure to include appropriate StatCrunch output HERE!! (4 points)

Which type of error could you have committed in part (e)? Explain how you know. (0.5 point)

Explanation / Answer

Solution:-

1)

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

Null hypothesis: P = 0.50
Alternative hypothesis: P 0.50

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

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.02506
z = (p - P) /

z = 1.30

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.30 or greater than 1.30. Thus, the P-value = 0.1936

Interpret results. Since the P-value (0.1936) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that there is sex discrimination in the job hiring process.

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