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 95% 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 find the 95% confidence interval for the mean reading scores of all fifth-graders, we can use the formula for the confidence interval:

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

where:
- \(\bar{x}\) is the sample mean (82),
- \(z\) is the z-score corresponding to the desired confidence level (for 95%, \(z \approx 1.96\)),
- \(\sigma\) is the population standard deviation (15),
- \(n\) is the sample size (35).

Now we calculate the standard error (SE):

\[
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{35}} \approx \frac{15}{5.916} \approx 2.53
\]

Next, we calculate the margin of error (ME):

\[
\text{ME} = z \times \text{SE} \approx 1.96 \times 2.53 \approx 4.96
\]

Now, we can calculate the confidence interval:

\[
\text{CI} = 82 \pm 4.96
\]

This gives us:

\[
\text{Lower limit} = 82 - 4.96 \approx 77.04 \\
\text{Upper limit} = 82 + 4.96 \approx 86.96
\]

Rounding to the nearest whole number, we obtain:

\[
\text{Lower limit} \approx 77 \\
\text{Upper limit} \approx 87
\]

Thus, the 95% confidence interval for the mean reading scores of all fifth-graders is:

\[
\boxed{(77,87)}
\]