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
state the null hypothesis.
h₀: customers are distributed evenly throughout the week.
hₐ: customers are not distributed evenly throughout the week.
h₀: customers are not distributed evenly throughout the week.
hₐ: customers are distributed evenly throughout the week.

Explanation:

In hypothesis testing, the null hypothesis ($H_0$) is a statement of no difference or no effect. Here, the assumption is that customers are distributed evenly throughout the week. The alternative hypothesis ($H_a$) is the statement we are trying to find evidence for. If the distribution is not even (the opposite of the null hypothesis), that is the alternative.

Answer:

$H_0$: Customers are distributed evenly throughout the week. $H_a$: Customers are not distributed evenly throughout the week.