A random sample of size 9 produced a sample mean equal to 13.5 and a standard deviation of 3.2. The margin of error associated with a 95% confidence interval estimate for the population mean is approximately 2.46.

Question

Quizplus · Accepted Answer

The formula for margin of error is (z-score)*(standard deviation of sample/ sqrt(sample size)).

Here, we are given the following information:
Sample size (n) = 9
Sample mean (x̄) = 13.5
Standard deviation (s) = 3.2
Confidence level = 95%
Since the sample size is less than 30 and the population standard deviation is unknown, we need to use the t-distribution to find the margin of error.

The critical t-value for a 95% confidence interval with 8 degrees of freedom (9-1) is approximately 2.306. Therefore, the margin of error is:
2.306 * (3.2 / √9) = 2.4584 (rounded to two decimal places)

This is approximately 2.46, as given in the question. Therefore, the statement is true, and the answer is A.

A Random Sample of Size 9 Produced a Sample Mean