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Question 30
Assuming the population has an approximate normal distribution, if a sample size $n=18$ has a sample mean $\bar{x}=31$ with a sample standard deviation $s=9$, find the margin of error at a $90 \%$ confidence level. Round the answer to two decimal places.
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Answer
So, the margin of error at a 90% confidence level is approximately \(\boxed{3.48}\), rounded to two decimal places.
Step 1 ：First, calculate the denominator using the square root of the sample size, \(\sqrt{18} \approx 4.24\)
Step 2 ：Next, divide the sample standard deviation by the square root of the sample size, \(\frac{9}{4.24} \approx 2.12\)
Step 3 ：Finally, multiply this result by the Z-score to find the margin of error, \(E = 1.645 \times 2.12 \approx 3.48\)
Step 4 ：So, the margin of error at a 90% confidence level is approximately \(\boxed{3.48}\), rounded to two decimal places.
