Question

A national chain of coffee shops claimed that 60% their servers are notified of changes in their work schedules in advance. A random sample of 200 servers showed that 100 were notified of changes in their schedule in advance. Perform a hypothesis test at 95% confidence to determine if the claim could be true. To get full credit state your null and alternate hypothesis, show test statistic and p-value and state your conclusion: could the company’s claim be true? YES or NO

Answer 1

Solution:

Given:

Claim: A national chain of coffee shops claimed that 60% their servers are notified of changes in their work schedules in advance.

Sample size = n = 200

x = Number of servers showed that they were notified of changes in their schedule in advance = 100

Confidence level = c = 95% = 0.95

then level of significance =

Step 1) State H0 and H1:

Vs

This is two tailed test.

Step 2) Test statistic:

where

thus

Step 3) Find z critical values:

Since this is two tailed, we find : Area =

Look in z table for area = 0.0250 or its closest area and find z value

Area 0.0250 corresponds to -1.9 and 0.06

thus z critical value = -1.96

Since this is two tailed test, we have two z critical values: ( -1.96 , 1.96)

Step 4) Decision Rule:
Reject null hypothesis ,if z  test statistic value < z critical value=-1.96 or z  test statistic value > z critical value=1.96 , otherwise we fail to reject H0.

Since z  test statistic value = < z critical value=-1.96 , we reject null hypothesis H0.

Step 5) Conclusion:

At 0.05 level of significance , we have sufficient evidence to reject the claim of a national chain of coffee shops that "60% their servers are notified of changes in their work schedules in advance."

Could the company’s claim be true?

No.

that is: the population proportion of servers were notified of changes in their schedule in advance is different from 60%.