Question

In a study of academic procrastination, the authors of a paper reported that for a sample of 441 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 3.40 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 = 441      x = 43.78      s = 21.46

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

Solution :

A )Given that,

= 7.74

= 3.40

n = 441

A ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Margin of error = E = Z_/2* (/n)

= 1.960 * (3.40 / 441 )

= 0.317

Margin of error = E = 0.317

At 95% confidence interval estimate of the population mean is,

- E < < + E

7.74 - 0.317 < < 7.74 + 0.317

7.423 < < 8.057

(7.423 , 8.057)

B ) Given that,

= 43.78

s = 21.46

n = 441

Degrees of freedom = df = n - 1 = 441 - 1 = 440

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,440 = 1.965

Margin of error = E = t_/2,df * (s /n)

= 1.965 * (21.46 / 441 )

= 2.008

Margin of error = 2.008

The 95% confidence interval estimate of the population mean is,

- E < < + E

43.78 - 2.008 < < 43.78 + 2.008

41.772 < < 45.788

(41.772, 45.788 )