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)
- Calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{35}} \approx \frac{15}{5.916} \approx 2.535 \]
- Find the margin of error (ME):
\[ ME = z \times SE \approx 2.576 \times 2.535 \approx 6.526 \]
- Calculate the confidence interval:
\[ CI = 82 \pm 6.526 \]
This results in:
\[ CI = (82 - 6.526, 82 + 6.526) = (75.474, 88.526) \]
- 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:
\[ (75,89) \]