Question

In a study of academic procrastination, the authors of a paper reported that for a sample of 451 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.74 hours and the standard deviation of study times was 3.10 hours. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of students taking introductory psychology at this university.

(a) Construct a 95% confidence interval to estimate μ, the mean time spent studying for the final exam for students taking introductory psychology at this university. (Round your answers to three decimal places.) TWO VALUES
(b) The paper also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the exam.
n = 451 x = 43.78 s = 21.96
Construct a 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam. (Round your answers to three decimal places.)

Answer 1

a) The 95% confidence interval to estimate μ is (7.435, 8.045) hours.

b) The 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam is (41.079%, 45.481%).

Explanation:

Question a:

We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.

That is z with a pvalue of
, so Z = 1.96.

Now, find the margin of error M as such

In which
is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 7.74 - 0.305 = 7.435 hours

The upper end of the interval is the sample mean added to M. So it is 7.74 + 0.305 = 8.045 hours.

The 95% confidence interval to estimate μ is (7.435, 8.045) hours.

Question b:

90% confidence interval means that

The margin of error is:

43.78 - 1.701 = 41.079%

43.78 + 1.701 = 45.481%

The 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam is (41.079%, 45.481%).