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 7.64 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 u, 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 = 21.56 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. o 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. o 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. O 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. O We are 90% confident that the mean percent of study time for all students taking introductory psychology is within this interval.

Answer 1

( a )

The provided sample mean is X = 7.64 and the sample standard deviation is 8 = 3.50. The size of the sample is n = 441 and the required confidence level is 95% The number of degrees of freedom are df = 441 - 1 = 440, and the significance level is a = 0.05 Based on the provided information, the critical t-value for a = 0.05 and df = 440 degrees of freedom is te = 1.965 The 95% confidence for the population mean y is computed using the following expression C1 = (x - * x + y ) Therefore, based on the information provided, the 95% confidence for the population mean is 441 (7.64 _ 1.965 x 3.50 64 1.965 x 3.50 441 = (7.64 -0.328,7.64 +0.328) = (7.312,7.968)

( b )

The provided sample mean is X = 43.78 and the sample standard deviation is 8 = 21.56. The size of the sample is n = 441 and the required confidence level is 90% The number of degrees of freedom are df = 441 - 1 = 440, and the significance level is a = 0.1 Based on the provided information, the critical t-value for a = 0.1 and df = 440 degrees of freedom is te = 1.648 The 90% confidence for the population mean y is computed using the following expression ci= (x - Foto 8 + ) Therefore, based on the information provided, the 90 % confidence for the population mean is 1.648 1.648 x 21.56 CI= 43.78 - 1.648 x 21.56 2.43.78 V441 441 = (43.78 - 1.692, 43.78 + 1.692) = (42.088, 45.472)

Interpretation :

Answer : Option ( 1 )