Question

me that the population has a normal distribution. 2) The principal randomly selected six students to take an aptifude test. Their scores were: 81.087077.185.870.275.1 Construct a 90% confidence interval for the mean score for all students. (a) Identify the best point estimate for μ, Round your answers to two decimal places. You can use technology without showing your work. (b) Find the critical values by first sketching the distribution curve and indicating the given confidenoe level on the graph. (c) Find the margin of error. Round your answers to two decimal places. (d) Construct the confidence interval using your results from Parts (b) and (c).

Answer 1

• a) The best point estimate for μ is the sample mean is 83.50.
• b) The critical values for a 90% confidence interval are -1.645 (for the lower tail) and 1.645 (for the upper tail).
• c)The margin of error is 1.24.
• d)The confidence interval is therefore 81.087077 - 1.24 to 81.087077 + 1.24, or (79.847, 82.327).

Step-by-step explanation:

To construct a 90% confidence interval for the mean score for all students, we need to determine the best point estimate for μ, find the critical values, calculate the margin of error, and construct the confidence interval.

(a) The best point estimate for μ is the sample mean, which is calculated by finding the average of the scores: (81 + 87 + 77 + 81 + 87 + 77) / 6 = 83.50.

(b) The critical values can be found by referring to the standard normal distribution table. At a 90% confidence level, we need to split the remaining 10% evenly between the upper and lower tails.

This means we need to find the z-score that leaves 5% in the upper tail, which corresponds to a z-score of 1.645..

(c) The margin of error is calculated by multiplying the critical value by the standard deviation of the sample divided by the square root of the sample size.

In this case, the margin of error is (1.645)(2.49) / √6 ≈ 1.24.

(d) Finally, we can construct the confidence interval by subtracting the margin of error from the sample mean and adding it to the sample mean.

The 90% confidence interval is therefore 83.50 - 1.24 to 83.50 + 1.24, or (82.26, 84.74).

Answer 2

The best point estimate for the mean score is 89.5. The critical values for a 90% confidence interval are -1.645 and 1.645. The margin of error is 75.974. The confidence interval is (13.526, 165.474).

Step-by-step explanation:

The best point estimate for μ (the mean score for all students) is the sample mean, which is found by summing the scores of the six students and dividing by the sample size. In this case, the sample mean is (81 + 0 + 87 + 77 + 185 + 87) / 6 = 89.5.

To find the critical values for a 90% confidence interval, we need to look up the z-value corresponding to an alpha of 0.10 divided by 2, because we have a two-tailed test. Using a standard normal distribution table, the critical values are approximately -1.645 and 1.645.

The margin of error can be calculated by multiplying the standard deviation of the sample (which can be approximated as the sample range divided by 4) by the critical value. The range of the sample is 185 - 0 = 185, so the standard deviation is 185 / 4 = 46.25. Therefore, the margin of error is 46.25 * 1.645 = 75.974.

Finally, we can construct the confidence interval by subtracting the margin of error from the sample mean to get the lower bound, and adding the margin of error to the sample mean to get the upper bound. In this case, the confidence interval is (89.5 - 75.974, 89.5 + 75.974), which simplifies to (13.526, 165.474).