In a study of academic procrastination, the authors of a paper reported that for a sample...

In a study of academic procrastination, the authors of a paper reported that for a sample of 431 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 7.24 hours and the standard deviation of study times was 3.50 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.)

(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 = 431 x = 43.18 s = 21.26

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.)
( , ) ( , )

Solutions

Expert Solution

a) Given that, sample size (n) = 431, sample mean = 7.24 hours and standard deviation (sd) = 3.50 hours

Since, sample size is very large we used standard normal distribution to find the confidence interval.

A 95% confidence level has significance level of 0.05 and critical value is,

The 95% confidence interval for the population mean is,

Therefore, a 95% confidence interval to estimate μ, the mean time spent studying for the final exam for students taking introductory psychology at this university is (6.910, 7.570).

b) Given that, sample size (n) = 431, sample mean = 43.18 hours and standard deviation (sd) = 21.26 hours

Since, sample size is very large we used standard normal distribution to find the confidence interval.

A 90% confidence level has significance level of 0.10 and critical value is,

The 90% confidence interval for the population mean is,

Therefore, a 90% confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam is (41.495, 44.485).

orchestra answered 1 year ago

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