Question

25 men from Pinellas County were randomly drawn from a population of 100,000 men and weighed. The average weight of a man from the sample was found to be 150 pounds with a standard deviation of 54 pounds, Assuming the survey follows a normal distribution, find the 95% confidence interval for the true mean weight of men.

Group of answer choices

A) (127.25, 172.75)

B) (128.832, 171.168)

C) (127.757, 172.243)

D) (127.71, 172.29)

Answer 1

Solution:
Given in the question
Sample mean = 150 pounds
Sample standard deviation = 54 pounds
No. of sample = 25 men
Here also given that survey follows a normal distribution so 95% confidence interval for the true mean weight of men can be calculated as
Mean +/- (Zalpha/2)*S/sqrt(n)
from Z test
alpha = 0.05, alpha/2 = 0.05/2 = 0.025
From Z table we found Zalpha/2 = 1.96
So 95% confidence interval is
150 +/- 1.96*54/sqrt(25)
150 +/- 21.168
So 95% confidence interval is 128.832 to 171.168
So its answer is B i.e. (128.832, 171.168)