Question

In an Accountemps survey of 255 senior executives, 113 said that the most common job interview mistake is to have little or no knowledge of the company. Test the claim, at the 5% level that in the population of all senior executives, 40% say that the most common job interview mistake is to have little or no knowledge of the company. What important lesson is learned from this survey?

1. State the null and alternative hypothesis
2. What is the test statistics?
3. What is the critical value?
4. What is the confidence interval?
5. What is the P-value?
6. What can you conclude?
7. Can a hypothesis test be used to “prove” that 40% say that the most common job interview mistake is to have little or no knowledge of the company?

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: p = 0.4
Alternative Hypothesis: p > 0.4

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.4431 - 0.4)/sqrt(0.4*(1-0.4)/255)
z = 1.40

This is right tailed test, for α = 0.05
Critical value of z is 1.64.

0.05 is the level of significance

Confidence interval
sample proportion, = 0.4431
sample size, n = 255
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.4431 * (1 - 0.4431)/255) = 0.0311

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0311
ME = 0.061

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.4431 - 1.96 * 0.0311 , 0.4431 + 1.96 * 0.0311)
CI = (0.3821 , 0.5041)

P-value Approach
P-value = 0.0808

As P-value >= 0.05, fail to reject null hypothesis.

no