Question

a local retailer currently schedules employees based on the assumption that they serve customers uniformly throughout the week (the same number each day). management is starting to question this assumption and decides to collect data on the number of customers served each day of the week to perform a chi - square goodness - of - fit test at a 5% significance level. frequency table: monday = 40, tuesday = 33, wednesday = 35, thursday = 32, friday = 60, total = 200 provided the assumptions of the test are satisfied, state your conclusion to the test. note: the p - value = 0.01
cannot be determined.
there is insufficient evidence to conclude that customers are unevenly distributed throughout the week.
there is sufficient evidence to conclude that customers are unevenly distributed throughout the week.

Explanation:

Step1: Compare p - value and significance level

In hypothesis testing, if the p - value ($p = 0.01$) is less than the significance level ($\alpha=0.05$), we reject the null hypothesis.

Step2: State the conclusion

The null hypothesis for a chi - square goodness - of - fit test (where customers are uniformly distributed) is rejected. This means there is sufficient evidence to conclude that customers are unevenly distributed throughout the week.

Answer:

There is sufficient evidence to conclude that customers are unevenly distributed throughout the week.