Question

A department store manager claims that her store has twice as many customers on Fridays and Saturdays than on any other day of the week (the store is closed on Sundays). That is, the probability that a customer visits the store Friday is \(2 / 8\), the probability that a customer visits the store Saturday is \(2 / 8\), while the probability that a customer visits the store on each of the remaining weekdays is \(1 / 8\). During an average week, the following numbers of customers visited the store: $$ \begin{array}{lr} \hline \text { Day } & \text { Number of Customers } \\ \hline \text { Monday } & 95 \\ \text { Tuesday } & 110 \\ \text { Wednesday } & 125 \\ \text { Thursday } & 75 \\ \text { Friday } & 181 \\ \text { Saturday } & 214 \end{array} $$ Can the manager's claim be refuted at the \(\alpha=.05\) level of significance?

Short answer

Answer: No, we cannot refute the manager's claim at the α = 0.05 level of significance based on the chi-square goodness-of-fit test.

Step-by-step solution

Calculate Expected Frequencies

First, calculate the total number of customers \((n)\) for the week: \(n = 95 + 110 + 125 + 75 + 181 + 214 = 800\) Using the manager's claim, calculate the expected frequencies for each day: - Monday, Tuesday, Wednesday, Thursday: \(\frac{1}{8} * 800 = 100\) - Friday, Saturday: \(\frac{2}{8} * 800 = 200\)

Calculate Chi-Square Test Statistic

Now that we have the expected frequencies, we can calculate the chi-square test statistic using the formula: \(\chi^2 = \sum\frac{(O_i - E_i)^2}{E_i}\) where \(O_i\) are the observed frequencies, and \(E_i\) are the expected frequencies. \(\chi^2 = \frac{(95 - 100)^2}{100} + \frac{(110 - 100)^2}{100} + \frac{(125 - 100)^2}{100} + \frac{(75 - 100)^2}{100} + \frac{(181 - 200)^2}{200} + \frac{(214 - 200)^2}{200} = 9.1\)

Find Critical Value

Now we need to find the critical value for the chi-square distribution with a significance level (\(\alpha\)) of 0.05 and degrees of freedom: Degrees of freedom (DF) = number of categories - 1 = 6 - 1 = 5 Using a chi-square distribution table or calculator, we find the critical value for \(\alpha = .05\) (95% confidence level) and DF = 5 is 11.07.

Compare Test Statistic and Critical Value

Finally, compare the chi-square test statistic and the critical value to determine if the manager's claim can be refuted: - \(\chi^2 = 9.1\) - Critical value = 11.07 Since the test statistic is less than the critical value (9.1 < 11.07), we fail to reject the null hypothesis, meaning we cannot refute the manager's claim at the \(\alpha = .05\) level of significance.

Key concepts

Probability

Probability is a measure of the likelihood that an event will occur. In the context of the department store manager's claim, probability quantifies the chances that a customer visits the store on a specific day of the week. According to the manager, the probability for a customer to come on Friday or Saturday is \(2/8\) each, and for the other weekdays, it's \(1/8\). An important aspect of probability in this exercise is its role in determining expected frequencies, which are crucial for conducting a chi-square test.

Hypothesis Testing

Hypothesis testing is a statistical method that allows us to decide whether there is enough evidence to reject a hypothesis about a population parameter. In our case, the null hypothesis is the manager's claim about customer distributions across the week. We confront this hypothesis with sample data (the actual customer counts), using a chi-square test to assess whether the observed frequencies align with the expected ones under the claim. If the test statistic is sufficiently large, we might reject the null hypothesis, thus refuting the manager's claim.

Expected Frequencies

Expected frequencies are the counts we would expect in each category if the null hypothesis is true. They are based on the specified probabilities and the total sample size. For instance, if the manager's claim holds, with a total of 800 customers, we expect 100 customers for Monday through Thursday and 200 customers for Friday and Saturday. It's fundamental to accurately compute these values since they serve as a benchmark for the chi-square test, comparing them to the observed frequencies to see if any significant differences arise.

Degrees of Freedom

Degrees of freedom in a chi-square test refer to the number of independent comparisons that can be made between the observed and expected frequencies. It's a critical component as it affects the shape of the chi-square distribution, which we use to determine the critical values for our test. In our exercise, the degrees of freedom are calculated by subtracting one from the number of categories (\(6 - 1 = 5\)), because the total frequency is fixed; altering one category's frequency necessitates changing another's to maintain the total.