Question

2. Suppose that a national survey finds that 73 percent of restaurant employees say that work stress has a negative impact on their personal lives. A random sample of 200 employees of a large restaurant chain finds that 141 employees say that work stress has a negative impact on their personal lives. Conduct a hypothesis testing to test whether the percentage of work-stressed employees for the restaurant chain differs from the national percentage. Is this an one-sample or two-sample test? Is this test for population mean or proportion? Is this a large sample or small sample? State the null and alternative hypotheses: is this a left-, right-, or two-tailed test? Find the critical value(s) at 0.01 level of significance. Compute the test statistic, Decision: do you reject or fail to reject the null hypothesis? What is the P-value? State your conclusion to the original research question: Compute the corresponding 95% confidence interval: Planning a future survey, how many employees does the company need to sample in order to have 2% margin of error for computing a 99% confidence interval?

Answer 1

This is a one sample Z test

This test is for population proportion

This is a large sample n> 30

$H_a: p \neq 0.73$

This is a two tailed test as we have to test if the proportion of work stressed employee for restraunt differs from the national proportion.

Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a two- tailed test is % = 2.58. The rejection region for this two-tailed test is R = {z: 17| > 2.58} Test Statistics The Z-statistic is computed as follows: P - Po 0.705 -0.73 -0.796 V po(1 - po)/n 0.73(1 - 0.73)/200 Decision about the null hypothesis Since it is observed that = = 0.796 <%= 2.58, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p = 0.4258, and since p = 0.4258 > 0.01, it is concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is different than po, at the a = 0.01 significance level.

P II X N 141 200 0.705 The critical value for a = 0.05 is me = 21-a/2 computed as shown below: 1.96. The corresponding confidence interval is p(1-) CI(Proportion) p(1-P) - Ô + V P - V n n2 0.705 - 1.96 0.705(1 - 0.705) 0.705 +1.96 x 200 0.705(1 - 0.705) 200 0.642<p<0.768,

Margin of error = 0.02

for 99% confidence $Z_{\frac{\alpha }{2}}=2.576$

$ME=Z_{\frac{\alpha }{2}}\sqrt{\frac{p(1-p)}{n}}$

for no prior estimate of p we assume p = 0.5

$n=(\frac{2.576}{0.02})^2*0.5*0.5=4148$