In a study of academic procrastination the authors of a pape

In a study of academic procrastination, the authors of a paper reported that for a sample of 411 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 7.54 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 , 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 = 411      x = 43.28      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.)
( , )

Solution

a)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    7.54          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    3.1          
n = sample size =    411          
              
Thus,              
Margin of Error E =    0.299701471          
Lower bound =    7.240298529          
Upper bound =    7.839701471          
              
Thus, the confidence interval is              
              
(   7.240298529   ,   7.839701471   ) [ANSWER]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    43.28          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    21.26          
n = sample size =    411          
              
Thus,              
Margin of Error E =    1.724922575          
Lower bound =    41.55507743          
Upper bound =    45.00492257          
              
Thus, the confidence interval is              
              
(   41.55507743   ,   45.00492257   ) [ANSWER]