To determine whether the population mean waiting time at the billing counter has changed during the sales season compared to the previous month's average of 4.5 minutes, we can perform a hypothesis test. Specifically, we'll use a one-sample $t$-test, given that we have a sample mean and standard deviation but do not know the population standard deviation exactly.

Hypothesis Setup

• Null Hypothesis ($H_0$): The mean waiting time during the sales season is equal to the previous month's average waiting time. $H_0: \mu = 4.5 \text{ minutes}$

• Alternative Hypothesis ($H_1$): The mean waiting time during the sales season is different from the previous month's average waiting time. $H_1: \mu \neq 4.5 \text{ minutes}$

Significance Level

The significance level ($\alpha$) is 0.05.

Test Statistic

The test statistic for a one-sample $t$-test is calculated as follows: $t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$ Where:

• $\bar{x}$ is the sample mean
• $\mu_0$ is the hypothesized population mean (4.5 minutes)
• $s$ is the sample standard deviation
• $n$ is the sample size

Degrees of Freedom

The degrees of freedom ($df$) for the $t$-test is $n - 1$.

Decision Rule

Compare the calculated $t$-statistic to the critical $t$-value from the $t$-distribution table at $df = 34$ (since $n = 35$) for a two-tailed test at $\alpha = 0.05$. If $|t|$ is greater than the critical value, we reject the null hypothesis.

Steps to Solve

1. Calculate the sample mean ($\bar{x}$) and sample standard deviation ($s$) from the data provided.
2. Compute the $t$-statistic using the formula above.
3. Determine the critical $t$-value from the $t$-distribution table for $df = 34$ at $\alpha = 0.05$.
4. Compare the $t$-statistic to the critical value and make a decision.

You can provide the sample data for the next steps, and I'll help you calculate the necessary values and conclude the hypothesis test.