A statistics professor wishes to estimate the average mark on a term test for a course that has multiple sections and many students. A survey of some of the students registered for the course reveals the following results: Determine a 95% confidence interval for the term test results.
A) [48.0, 93.6]
B) [66.8, 74.8]
C) [62.9, 78.71]
D) [64.2, 77.3]
E) [62.6, 79.0]

Question

A statistics professor wishes to estimate the average mark on a term test for a course that has multiple sections and many students. A survey of some of the students registered for the course reveals the following results:  Determine a 95% confidence interval for the term test results.
A) [48.0, 93.6] 
B) [66.8, 74.8] 
C) [62.9, 78.71] 
D) [64.2, 77.3] 
E) [62.6, 79.0]

Quizplus · Accepted Answer

Based on the sample data, the sample mean is 70.8 and the standard deviation is 5.71. Using a t-distribution with n-1 degrees of freedom (where n is the sample size of 25), a 95% confidence interval can be calculated as:

sample mean ± t*(standard error)

where t is the critical value for a 95% confidence interval and can be found using a t-table or calculator. For a sample size of 25 and a 95% confidence interval, the critical value is approximately 2.06.

The standard error is calculated as:

standard deviation / sqrt(sample size)

Plugging in the values, we get:

70.8 ± 2.06*(5.71 / sqrt(25))
= 70.8 ± 2.48
= [68.32, 73.28]

Therefore, the best choice is E: [62.6, 79.0].

A Statistics Professor Wishes to Estimate the Average Mark on a Term