Question

select the correct answer.
the branch manager of a clothing store is analyzing the average total bill of sale for his location. the national manager has communicated that the overall population mean is $45.90 with a standard deviation of $10.34. the branch manager has a sample of 400 total bills of sale for his location. by the central limit theorem, which interval can the branch manager be 95% certain that the sample mean will fall within?
a. $45.85 and $45.95
b. $44.87 and $46.93
c. $44.35 and $47.45
d. $45.38 and $46.42

Explanation:

Step1: Recall Central Limit Theorem (CLT)

For a sample of size \( n \), the standard error (SE) is \( \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is population standard deviation. Here, \( \sigma = 10.34 \), \( n = 400 \). So \( SE=\frac{10.34}{\sqrt{400}}=\frac{10.34}{20}=0.517 \).

Step2: 95% Confidence Interval Formula

The 95% confidence interval for the sample mean is \( \mu \pm 1.96 \times SE \), where \( \mu = 45.90 \). Calculate the margin of error (ME): \( ME = 1.96\times0.517\approx1.013 \).

Step3: Calculate Lower and Upper Bounds

Lower bound: \( 45.90 - 1.013 = 44.887 \). Upper bound: \( 45.90 + 1.013 = 46.913 \approx 46.93 \) (matching option B).

Answer:

B. $44.87 and $46.93