A survey of 61,646 people included several questions about office relationships.

Question

A survey of 61,646 people included several questions about office relationships. Of the respondents, 26.3% reported that bosses scream at employees. Use a 0.01 significance level to test the claim that more than 1/4 of people say that bosses scream at employees. How is the conclusion affected after learning that the survey is an online survey in which Internet users chose whether to respond? 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

What is the test statistics ?

What is the P-Value?

Explanation / Answer

We have to test the null hypothesis H0: p=1/4=0.25 against the alternative Ha: p>0.25

Sample proportion phat=0.263

Sample size n=61646

We are given to use normal approximation.

So, Test statistic Z= (phat-0.13)/sqrt(p*(1-p)/n) =(0.263-0.25)/sqrt(0.25*(1-0.25)/61646) = 7.45

p-value=P(Z>7.45)<0.01 as this is more than 3 standard deviations.

As calculated p<0.01, we reject the null hypothesis and conclude that we mmore than ¼ of the people say that bosses scream at employees.

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