Question

In a survey of 155155 senior​ executives, 48.448.4​% said that the most common job interview mistake is to have little or no knowledge of the company. Use a 0.050.05 significance level to test the claim that in the population of all senior​ executives, 4040​% say that the most common job interview mistake is to have little or no knowledge of the company. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

Answer 1

We conclude that in the population of all senior​ executives, different from 40​% say that the most common job interview mistake is to have little or no knowledge of the company.

Explanation:

We are given that in a survey of 155 senior​ executives, 48.4% said that the most common job interview mistake is to have little or no knowledge of the company.

We have to use a 0.05 significance level to test the claim 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.

Let p = population proportion of senior​ executives who said that the most common job interview mistake is to have little or no knowledge of the company

So, Null Hypothesis,
`H_0` : p = 40% {means 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}

Alternate Hypothesis,
`H_A` : p
`\\eq` 40% {means that in the population of all senior​ executives, different from 40​% say that the most common job interview mistake is to have little or no knowledge of the company}

The test statistics that would be used here One-sample z test for proportions;

T.S. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of senior​ executives who said that the most common job interview mistake is to have little or no knowledge of the company = 48.4%

n = sample of senior​ executives = 155

So, the test statistics =
`\frac{0.484-0.40}{\sqrt{(0.484(1-0.484))/(155) } }`

= 2.09

The value of z test statistics is 2.09.

Also, P-value of the test statistics is given by;

P-value = P(Z > 2.09) = 1 - P(Z
`\leq` 2.09)

= 1 - 0.9817 = 0.0183

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that in the population of all senior​ executives, different from 40​% say that the most common job interview mistake is to have little or no knowledge of the company.