Question

The costs associated with conducting interviews for a job opening have skyrocketed over the years. According to a Harris interactive survey, 211 of 502 senior human resources executives at U.S companies believe that their hiring managers are interviewing too many people to find qualified candidates for the job. Suppose the true proportion of human resource executives who believe that their hiring managers are interviewing too many people is given by p. At a 5% level of significance can you conclude that the true proportion p differs from 40%?

Answer 1

Solution :

Null and alternative hypotheses :

The null and alternative hypotheses would be as follows :

Test statisic :

To test the hypothesis the most appropriate test is one sample z-test for proportion. The test statistic is given as follows :

Where, p̂ is sample proportion, p is hypothesized value of population proportion under H_{​​​​​​0}, q = 1 - p and n is sample size.

Sample proportion of senior human resources executives at U.S companies who believe that their hiring managers are interviewing too many people is given by,

p = 0.40, q = 1 - 0.40 = 0.60 and n = 502

The value of the test statistic is 0.9293.

P-value :

Since, our test is two-tailed test, therefore we shall obtain two-tailed p-value for the test statistic. The two-tailed p-value is given as follows :

p-value = 2.P(Z > |z|)

We have, |z| = 0.9293

p-value = 2.P(Z > 0.9293)

p-value = 0.3527

The p-value is 0.3527

Decision :

Significance level = 5% = 0.05

p-value = 0.3527

(0.3527 > 0.05)

Since, p-value is greater than the significance level of 5%, therefore we shall be fail to reject the null hypothesis (H_{​​​​​​0}) at 5% significance level.

Conclusion :

At 5% significance level, there is not sufficient evidence to conclude that the true proportion p differs from 40%.

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