Certainly! If we opt to perform a Z-test instead of a $t$-test, we'll assume that either the sample size is large enough (which it is, with $n = 35$), or we know the population standard deviation (which we do, given the previous month's standard deviation of 1.2 minutes).

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 Z-test is calculated as follows: $Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$ Where:

• $\bar{x}$ is the sample mean
• $\mu_0$ is the hypothesized population mean (4.5 minutes)
• $\sigma$ is the population standard deviation (1.2 minutes)
• $n$ is the sample size (35)

Decision Rule

For a two-tailed test at the $\alpha = 0.05$ significance level, the critical Z-values are $\pm 1.96$. If the calculated Z-value falls outside this range, we reject the null hypothesis.

Steps to Solve

1. Calculate the sample mean ($\bar{x}$) from the provided data.
2. Compute the Z-statistic using the formula above.
3. Compare the Z-statistic to the critical Z-value of $\pm 1.96$.
4. Make a decision based on whether the Z-statistic falls within or outside the critical region.

Please provide the sample data so we can calculate the sample mean ($\bar{x}$) and proceed with the Z-test.