Question

A random sample of 25 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 90% 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 90% confidence interval for the population mean, we can use the formula for the confidence interval when the population standard deviation is known:

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

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

Given:
- Sample mean, \(\bar{x} = 68\)
- Population standard deviation, \(\sigma = 3\)
- Sample size, \(n = 25\)

First, find the z-score for a 90% confidence level. The critical z-value that corresponds to 90% confidence level (which has 5% in each tail of the standard normal distribution) is approximately 1.645.

Now, we compute the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{25}} = \frac{3}{5} = 0.6
\]

Now we can calculate the margin of error (ME):

\[
ME = z \cdot SE = 1.645 \cdot 0.6 \approx 0.987
\]

Now we can find the confidence interval:

\[
CI = \bar{x} \pm ME = 68 \pm 0.987
\]

Calculating the lower and upper bounds:

\[
\text{Lower bound} = 68 - 0.987 \approx 67.013
\]
\[
\text{Upper bound} = 68 + 0.987 \approx 68.987
\]

Finally, rounding to the nearest hundredths, we get:

\[
CI \approx (67.01, 68.99)
\]

Thus, the 90% confidence interval estimate for the population mean of exam scores is:

\[
(67.01, 68.99)
\]