Question

A random sample of 36 exam scores is taken from the normally distributed population with unknown mean and a populations standard deviation know and equal to 3. The sample mean score is 68. Find a 95% confidence interval estimate for the population mean of exam scores.

Please round to the nearest hundredths and write your answer in the form ( , ); for example (31,34) or (25,46). Do not type in any extra spaces.

Answer 1

To calculate a 95% confidence interval for the population mean when the population standard deviation is known, we can use the formula:

\[
\text{CI} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right)
\]

where:
- \(\bar{x}\) is the sample mean,
- \(z\) is the z-score corresponding to the desired level of confidence,
- \(\sigma\) is the population standard deviation,
- \(n\) is the sample size.

Given:
- Sample mean \(\bar{x} = 68\)
- Population standard deviation \(\sigma = 3\)
- Sample size \(n = 36\)

1. First, find the z-score for a 95% confidence level. The z-score for 95% confidence is approximately \(z = 1.96\).

2. Calculate the standard error (SE):
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{3}{6} = 0.5
\]

3. Now calculate the margin of error (ME):
\[
ME = z \times SE = 1.96 \times 0.5 = 0.98
\]

4. Finally, construct the confidence interval:
\[
\text{Lower limit} = \bar{x} - ME = 68 - 0.98 = 67.02
\]
\[
\text{Upper limit} = \bar{x} + ME = 68 + 0.98 = 68.98
\]

Thus, the 95% confidence interval estimate for the population mean of exam scores is:

\[
(67.02, 68.98)
\]