## Question 860:

1## Answer:

No answer provided yet.Problem #1Since we are told the population standard deviation we can use the normal distribution instead of the t-distribution to compute the 95% confidence interval.

The margin of error is the standard error times the z(normal) critical value for a 95% confidence level which is 1.96.

The standard error is found as the standard deviation divided by the square root of the sample size. The standard deviation of this sample is 4.5, making the standard error = 4.5/SQRT(24) = .919. The margin of error is then .919*4.5 = 1.8, which is what the question is looking for (1.8kg). "Give a 95% confidence interval for , the mean of the population from which the sample is drawn. It is x bar +/ margin of error = 61.79 +/- 1.8kg."

Problem #2

We're given the standard error, so we don't need to compute it. All we need to do is find the critical value from the normal distribution for a 90% confidence level. It is 1.644. So the margin of error is the standard error * the critical value = 5.1*1.644 = 8.38.

The answer is the 96.0 +/- **8.38 days.**

Problem #3

A paired t-test was conducted on the pre and post-test scores and we are given all pertinent information. The last entry which says "Confidence Level(90.0%)" refers to the margin of error around the average difference between pre and post test. This is what we're looking for as it generates a confidence interval between .211 and 2.68.

Give a 90% confidence interval for the mean increase in listening score due to the intensive training. It is 1.45 +/- 1.238506