Calculate the test statistic [tex]\chi^2[/tex].

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 [tex]5 \%[/tex] significance level.

\begin{tabular}{|l|c|c|c|c|c|c|}
\hline & Monday & Tuesday & Wednesday & Thursday & Friday & Total \\
\hline
\begin{tabular}{c}
Number \\
Served
\end{tabular} & 40 & 33 & 35 & 32 & 60 & 200 \\
\hline
\end{tabular}

Provided the assumptions of the test are satisfied, calculate the test statistic [tex]\chi^2[/tex].



Answer :

To determine whether the observed frequencies of customers served align with the expected uniform distribution, we need to perform a Chi-Square goodness-of-fit test. Let's go through the steps in detail:

1. Observed Frequencies:
- The number of customers served each day are as follows:
[tex]\[ \text{Monday: } 40, \quad \text{Tuesday: } 33, \quad \text{Wednesday: } 35, \quad \text{Thursday: } 32, \quad \text{Friday: } 60 \][/tex]
This gives us the observed frequencies:
[tex]\[ \mathbf{Observed} = [40, 33, 35, 32, 60] \][/tex]

2. Total Number of Observations:
- The total number of customers served in the week is:
[tex]\[ \text{Total} = 200 \][/tex]

3. Expected Frequencies:
- If the customers are served uniformly throughout the week, we expect the same number of customers each day. Given there are 5 days, the expected frequency for each day can be calculated as:
[tex]\[ \text{Expected Frequency} = \frac{\text{Total}}{5} = \frac{200}{5} = 40 \][/tex]
This gives us the expected frequencies:
[tex]\[ \mathbf{Expected} = [40, 40, 40, 40, 40] \][/tex]

4. Calculate the Chi-Square Test Statistic:
- The Chi-Square test statistic [tex]\(\chi^2\)[/tex] is calculated using the formula:
[tex]\[ \chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}} \][/tex]
Substituting the observed and expected frequencies into this formula, we get:
[tex]\[ \chi^2 = \frac{(40 - 40)^2}{40} + \frac{(33 - 40)^2}{40} + \frac{(35 - 40)^2}{40} + \frac{(32 - 40)^2}{40} + \frac{(60 - 40)^2}{40} \][/tex]
Simplifying each term, we get:
[tex]\[ \frac{(40 - 40)^2}{40} = \frac{0^2}{40} = 0 \][/tex]
[tex]\[ \frac{(33 - 40)^2}{40} = \frac{(-7)^2}{40} = \frac{49}{40} = 1.225 \][/tex]
[tex]\[ \frac{(35 - 40)^2}{40} = \frac{(-5)^2}{40} = \frac{25}{40} = 0.625 \][/tex]
[tex]\[ \frac{(32 - 40)^2}{40} = \frac{(-8)^2}{40} = \frac{64}{40} = 1.6 \][/tex]
[tex]\[ \frac{(60 - 40)^2}{40} = \frac{20^2}{40} = \frac{400}{40} = 10 \][/tex]
Summing these results:
[tex]\[ \chi^2 = 0 + 1.225 + 0.625 + 1.6 + 10 = 13.45 \][/tex]

Therefore, the test statistic [tex]\(\chi^2\)[/tex] is [tex]\(13.45\)[/tex].