One study reports that 33% of newly hired MBAs are confronted with unethical bus
ID: 3325022 • Letter: O
Question
One study reports that 33% of newly hired MBAs are confronted with unethical business practices during their first year of employment. One business school dean wondered if her MBA graduates had similar experiences. She surveyed recent graduates from her school's MBA program to find that 31% of the 125 graduates from the previous year claim to have encountered unethical business practices in the workplace. Can she conclude that her graduates' experiences are different? What are the null and alternative hypotheses? A. Ho:p=0.33 vs. HA:p0.31 Ho: p 0.33 vs. Ha: p . 0.33 Ho: p = 0.33 vs. HA: p > 0.33 A. The test statistic is | | (Round to two decimal places as needed.) B. The assumptions and conditions are not met, so the test cannot proceed What is the P-value of the test statistic? O A. P-value-(Round to three decimal places as needed.) O B. The assumptions and conditions are not met, so the test cannot proceed. Can she conclude that her graduates' experiences are different? Assume = 0.05. OA. Fail to reject the null hypothesis. There is sufficient evidence that the rate of unethical business practice exposure is different from that reported in the study. ( B. Reject the null hypothesis. There is insufficient evidence that the rate of unethical business practice exposure is different from that reported in the study. OC. Fail to reject the null hypothesis. There is insufficient evidence that the rate of unethical business practice exposure is different from that reported in the study. O D Reject the null hypothesis. There is sufficient evidence that the rate of unethical business practice exposure is different from that reported in the study. ( E. No conclusion can be drawn because the conditions and assumptions were not met for the hypothesis test.Explanation / Answer
A) Option-C)
B) The test statistic Z = (p - P)/sqrt(P * (1 - P)/n)
= (0.31 - 0.33)/sqrt(0.33 * 0.67/125)
= -0.30
The p-value = 2 * P(Z < -0.30)
= 2 * 0.3821 = 0.7642
As the p-value is greater than the significance level (0.7642 > 0.05), so the null hypothesis is not rejected.
Option-C is the correct answer.
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