Question

A sample of the reading scores of 35 fifth-graders has a mean of 82. The standard deviation of the population is 15. Find the 99% confidence interval of the mean reading scores of all fifth-graders.

Please round to the nearest whole number 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 99% confidence interval for the mean reading scores of all fifth-graders, we can use the formula for the confidence interval:

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

Where:
- \(\bar{x}\) = sample mean (82 in this case)
- \(z\) = z-score corresponding to the desired confidence level (for 99%, the z-score is approximately 2.576)
- \(\sigma\) = population standard deviation (15 in this case)
- \(n\) = sample size (35 in this case)

1. Calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{35}} \approx \frac{15}{5.916} \approx 2.535
\]

2. Find the margin of error (ME):

\[
ME = z \times SE \approx 2.576 \times 2.535 \approx 6.526
\]

3. Calculate the confidence interval:

\[
CI = 82 \pm 6.526
\]

This results in:

\[
CI = (82 - 6.526, 82 + 6.526) = (75.474, 88.526)
\]

4. Rounding to the nearest whole number:

\[
CI \approx (75, 89)
\]

Thus, the 99% confidence interval of the mean reading scores of all fifth-graders is:

\[
(75,89)
\]