Question

Suppose you administer a certain aptitude test to a random sample of 9 students in your high school, and the average score is 105. We want to determine the mean μ of the population of all students in the school. Assume a standard deviation of σ = 15 for the test. Round all answers to 2 decimals.

1.What is the margin of error for a 98% confidence interval?

2.What would be the interval for a 98% confidence interval?

3. Write a sentence explaining the 98% confidence interval in context of the question. (There are only a few ways to do this correctly, so be sure to read the book closely.)

4. Write a few sentences explaining what would happen to the size of the CI if you wanted a 80% confidence interval?

5. Write a few sentences explaining what would happen to the size of the CI if you changed the sample size to be 81 and kept a 98% confidence interval.

Answer 1

Q1) From standard normal tables, we have here:
P(Z < 2.326 ) = 0.99

Therefore, due to symmetry, we have here:
P( -2.326 < Z < 2.326) = 0.98

Therefore the margin of error here is computed as:

Q2) The 98% confidence interval for the population mean is obtained here as:

This is the required 98% confidence interval for population mean here.

Q3) The interpretation of the 98% confidence interval here is that we are 98% confidence that the true population mean lies in the above obtained confidence interval.

Q4) As the confidence interval here is 80%, which is less than 98% confidence interval, therefore the critical z value would be lower for 80% confidence level compared to 98% confidence level. Therefore the size of the confidence interval would reduce when the confidence interval reduces.

Q5) As the sample size increases that is from 9 to 81, the margin of error decreases as it is inversely proportional to the square root of the sample size. Therefore the size of the confidence interval would decrease in that case.