Question

A study of 35 golfers showed that their average score on a particular course was 92. The standard deviation of the population is 5. Find the 95% confidence interval of the mean score for all golfers.

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 95% confidence interval for the mean score of all golfers, we can use the formula for the confidence interval when the population standard deviation is known:

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

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

Given:
- \( \bar{x} = 92 \)
- \( \sigma = 5 \)
- \( n = 35 \)

Step 1: Calculate the standard error (SE):

\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{35}} \approx \frac{5}{5.916} \approx 0.845
\]

Step 2: Calculate the margin of error (ME):

\[
ME = z \times SE = 1.96 \times 0.845 \approx 1.655
\]

Step 3: Calculate the confidence interval:

\[
\text{Lower Limit} = \bar{x} - ME \approx 92 - 1.655 \approx 90.345
\]
\[
\text{Upper Limit} = \bar{x} + ME \approx 92 + 1.655 \approx 93.655
\]

Step 4: Round to the nearest whole numbers:

- Lower Limit: \( \approx 90.345 \) rounds to 90
- Upper Limit: \( \approx 93.655 \) rounds to 94

Thus, the 95% confidence interval for the mean score is:

\[
(90, 94)
\]

So the final answer is:

(90,94)