Question

Dr. Patton is a professor of English. Recently she counted the number of misspelled works in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 44 students, the mean number of misspelled words per essay was 6.05. Construct a 90% confidence interval for the mean number of misspelled words in the population of student essays.

Multiple Choice

A- 5.445 to 6.6551

B-5.161 to 6.939

C-5.482 to 6.618

D-5.374 to 6.726

E-5.102 to 6.998

Answer 1

Solution :

Given that,

Sample size = n = 44

Z/2 = 1.645

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

= 1.645 * (2.44 / 44)

Margin of error = E = 0.605

At 90% confidence interval estimate of the population mean is,

- E < < + E

6.05 - 0.605 < < 6.05 - 0.605

5.445 < < 6.655

(5.445,6.655) , Option A is correct