A random sample of 36 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.

Question

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 a 90% confidence interval for the population mean, we can use the z-distribution since the population standard deviation is known.

1. Determine the sample mean (M):

   • \( M = 68 \)

2. Determine the population standard deviation (σ):

   • \( σ = 3 \)

3. Determine the sample size (n):

   • \( n = 36 \)

4. Calculate the standard error (SE): \[ SE = \frac{σ}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{3}{6} = 0.5 \]

5. Find the z-score for a 90% confidence level: The z-score that corresponds to a 90% confidence level (with a two-tailed test) can be found using a standard normal distribution table, or using a z-score calculator. For 90% confidence, we look for the z-score which leaves 5% in each tail, which is approximately 1.645.

6. Calculate the margin of error (ME): \[ ME = z \times SE = 1.645 \times 0.5 = 0.8225 \]

7. Construct the confidence interval: \[ \text{Confidence Interval} = (M - ME, M + ME) = (68 - 0.8225, 68 + 0.8225) \] \[ = (67.1775, 68.8225) \]

8. Round to the nearest hundredths: \[ = (67.18, 68.82) \]

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

\[ (67.18,68.82) \]