Let's go through the hypothesis testing step by step:
1. Setting Up Hypotheses:
- Null Hypothesis \( H_0 \): The average customer receipt for the branch is equal to the chain's average, \( \mu = 72 \).
- Alternative Hypothesis \( H_a \): The average customer receipt for the branch is less than the chain's average, \( \mu < 72 \).
2. Significance Level:
- The significance level (\(\alpha\)) is 0.01 (1%).
3. Critical Value:
- From the provided table for a one-tailed test at the 1% significance level, the critical z-value is 2.58. Since we are conducting a lower-tailed test (looking for receipts less than the chain average), we consider the negative of this value, which is -2.58.
4. Sample Data:
- The mean of the branch's receipts (\(\bar{x}\)) is $67.00.
- The mean of the chain's receipts (\(\mu\)) is $72.00.
- Standard deviation (\(\sigma\)) is $11.00.
- Sample size (\(n\)) is 40.
5. Calculate Standard Error of the Mean (SEM):
- \( \text{SEM} = \frac{\sigma}{\sqrt{n}} \)
6. Calculate the z-score:
- \( z = \frac{\bar{x} - \mu}{\text{SEM}} \)
Using the given result:
- The standard error of the mean (SEM) is approximately 1.7393.
- The z-score is approximately -2.8748.
7. Decision Rule:
- If the computed z-score is less than the critical z-value (-2.58), we reject the null hypothesis \( H_0 \).
8. Conclusion:
- The computed z-score (-2.8748) is indeed less than the critical value (-2.58).
- Therefore, we reject the null hypothesis \( H_0 \).
Based on the hypothesis test, we reject \( H_0: \mu = 72 \) and accept \( H_a: \mu < 72 \). This means we have sufficient evidence to conclude that the average receipt for the branch is less than the chain's average, at the 1% significance level.
Hence, the correct choice is:
She should reject [tex]\( H_0: \mu = 72 \)[/tex] and accept [tex]\( H_a: \mu < 72 \)[/tex].
