
Christina, a restaurant manager, would like to make the claim that the average number of customers that her restaurant serves per hour is more than 57. Christina samples 29 hours over the course of a week and records the number of customers served and obtains a sample mean of 62 customers per hour.
At the 2.5% significance level, should Christina reject or fail to reject the null hypothesis given the sample data below?
- H0:μ=57 customers; Ha:μ>57 customers
- α=0.025 (significance level)
- test statistic=2.61
Use the graph below to select the type of test (left-, right-, or two-tailed).
Then set the α and the test statistic to determine the p-value. Use the results to determine whether to reject or fail to reject the null hypothesis.
Alternatively, use the table below to find the p-value:
z | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
2.6 | 0.995 | 0.995 | 0.996 | 0.996 | 0.996 | 0.996 | 0.996 | 0.996 | 0.996 | 0.996 |
2.7 | 0.997 | 0.997 | 0.997 | 0.997 | 0.997 | 0.997 | 0.997 | 0.997 | 0.997 | 0.997 |
2.8 | 0.997 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 |
2.9 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 | 0.998 | 0.999 | 0.999 | 0.999 |
Select the correct answer below:
1) Do not reject the null hypothesis because the value of z is positive.
2) Reject the null hypothesis because 2.61>0.025.
4) Do not reject the null hypothesis because the p-value 0.0045 is less than the significance level α=0.025.


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