Question

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 7.34 hours and the standard deviation of study times was 3.40 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 = 43.48      s = 21.36

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.)
(  ,  )

Interpret the interval.

We are confident that the mean percent of study time that occurs in the 24 hours prior to the exam for all students taking introductory psychology at this university is within this interval at least 90% of the time.

We are 90% confident that the mean percent of study time that occurs in the 24 hours prior to the exam for all students taking introductory psychology at this university is within this interval.  

  We are 90% confident that the mean percent of study time for all students taking introductory psychology is within this interval.We are confident that 90% of the mean percent of study time that occurs in the 24 hours prior to the exam for all students taking introductory psychology at this university is within this interval.

Answer 1

a)

sample mean, xbar = 7.34
sample standard deviation, s = 3.4
sample size, n = 431
degrees of freedom, df = n - 1 = 430

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.965

ME = tc * s/sqrt(n)
ME = 1.965 * 3.4/sqrt(431)
ME = 0.3

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (7.34 - 1.965 * 3.4/sqrt(431) , 7.34 + 1.965 * 3.4/sqrt(431))
CI = (7.018 , 7.662)

b)

sample mean, xbar = 43.48
sample standard deviation, s = 21.36
sample size, n = 431
degrees of freedom, df = n - 1 = 430

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.648

ME = tc * s/sqrt(n)
ME = 1.648 * 21.36/sqrt(431)
ME = 1.7

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (43.48 - 1.648 * 21.36/sqrt(431) , 43.48 + 1.648 * 21.36/sqrt(431))
CI = (41.784 , 45.176)

c)
We are 90% confident that the mean percent of study time that occurs in the 24 hours prior to the exam for all students taking introductory psychology at this university is within this interval.