Question

multiple choice 4 points
calculate the test statistic \\( x ^ { 2 } \\)
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, calculate the test statistic \\( x ^ { 2 } \\).
6.67
9.24
10.31
13.45
15.86

Explanation:

Step1: Calculate the expected value for each day

Since the assumption is that customers are served uniformly throughout the week (5 days), the expected value \( E_i=\frac{\text{Total}}{\text{Number of days}}=\frac{200}{5} = 40\)

Step2: Calculate the \(\chi^{2}\) statistic

The formula for the \(\chi^{2}\) statistic is \(\chi^{2}=\sum\frac{(O_i - E_i)^{2}}{E_i}\)

• For Monday: \(O_1 = 40\), \(E_1=40\), \(\frac{(40 - 40)^{2}}{40}=0\)
• For Tuesday: \(O_2 = 33\), \(E_2 = 40\), \(\frac{(33 - 40)^{2}}{40}=\frac{(- 7)^{2}}{40}=\frac{49}{40}=1.225\)
• For Wednesday: \(O_3 = 35\), \(E_3 = 40\), \(\frac{(35 - 40)^{2}}{40}=\frac{(-5)^{2}}{40}=\frac{25}{40}=0.625\)
• For Thursday: \(O_4 = 32\), \(E_4 = 40\), \(\frac{(32 - 40)^{2}}{40}=\frac{(-8)^{2}}{40}=\frac{64}{40}=1.6\)
• For Friday: \(O_5 = 60\), \(E_5 = 40\), \(\frac{(60 - 40)^{2}}{40}=\frac{(20)^{2}}{40}=\frac{400}{40}=10\)

Step3: Sum up the values

\(\chi^{2}=0 + 1.225+0.625 + 1.6+10=13.45\)

Answer:

13.45