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. $44.87 and $46.93
b. $44.35 and $47.45
c. $45.38 and $46.42
d. $45.85 and $45.95

Explanation:

Step1: Recall Central Limit Theorem (CLT)

For a sample of size \( n \), the sampling distribution of the sample mean \(\bar{x}\) has mean \(\mu\) (population mean) and standard deviation (standard error) \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation. For a 95% confidence interval, the z - score is approximately \( z = 1.96 \) (from the standard normal distribution, as 95% of the data lies within \( \pm1.96 \) standard deviations of the mean).

Given: \(\mu=\$45.90\), \(\sigma = \$10.34\), \(n = 400\)

Step2: Calculate the standard error (SE)

The formula for the standard error is \( SE=\frac{\sigma}{\sqrt{n}} \)

Substitute the values: \( SE=\frac{10.34}{\sqrt{400}}=\frac{10.34}{20}=0.517 \)

Step3: Calculate the margin of error (ME)

The margin of error is given by \( ME = z\times SE \)

For 95% confidence, \( z = 1.96 \)

So, \( ME=1.96\times0.517\approx1.013 \)

Step4: Calculate the confidence interval

The confidence interval for the sample mean is \( \mu\pm ME \)

Lower limit: \( 45.90 - 1.013=44.887\approx44.87 \)

Upper limit: \( 45.90+ 1.013 = 46.913\approx46.93 \) (the slight difference is due to rounding during calculation)

Answer:

A. $44.87 and $46.93