Question

A random sample of 100 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 the 90% confidence interval for the population mean, we can use the formula:

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

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 \(n = 100\)
- For a 90% confidence level, the z-score (from z-tables or standard normal distribution) is approximately 1.645.

Now, we can calculate:

1. Calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3
\]

2. Calculate the margin of error (ME):

\[
ME = z \cdot SE = 1.645 \cdot 0.3 \approx 0.4935
\]

3. Calculate the confidence interval:

\[
\text{Lower limit} = \bar{x} - ME = 68 - 0.4935 \approx 67.5065
\]
\[
\text{Upper limit} = \bar{x} + ME = 68 + 0.4935 \approx 68.4935
\]

Rounding to the nearest hundredths, we have:

\[
\text{Lower limit} \approx 67.51
\]
\[
\text{Upper limit} \approx 68.49
\]

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

\[
(67.51, 68.49)
\]

Final answer:

(67.51,68.49)