Question

1. A poll was taken of 1010 U.S. employees sampled at random. The employees sampled were asked whether they “play hooky”; that is, call in sick at least once a year when they simply need time to relax; 202 responded “yes.” Calculate a 90% confidence interval for the true proportion of U.S. employees who play hooky. Then find A-E.

A. In the United Kingdom, approximately 24% of employees report that they call in sick at least once a year when they need time to relax. Is there enough evidence to say that US employees do this less often than employees in the UK? Use a significance level of .05. State the hypotheses that you will test.

B. Calculate the test statistic.

C.What is the P-value?

D. State your conclusion.

Answer 1

Q 1)

The proportion of U.S. employee who play hooky is p = X/n = 202/1010 = 0.2

Thr 90% CI for the true proportion of U.S. employees who play hooky is

90% CI is = (0.1793, 0.2207)

A)

Null Hypothesis

Altrenative Hypothesis

B) Under H0, the test statistic is

C) The P-Value is 0.00145

D) Since p value is less than significance level, Reject H0. Hence at 5% significance level, we have enough evidence to say that US employees do this less often than employees in the UK.