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.

Answer 1

95% confidence interval for the true mean weight of men is [127.71 pounds , 172.30 pounds].

Explanation:

We are given that the average weight of a man from the sample was found to be 150 pounds with a standard deviation of 54 pounds

25 men from Pinellas County were randomly drawn from a population of 100,000 men.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample average weight of a man = 150 pounds

s = sample standard deviation = 54 pounds

n = sample of men = 25

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the true mean,
`\mu` is ;

P(-2.064 <
`t_2_4` < 2.064) = 0.95 {As the critical value of t at 24 degree of

freedom are -2.064 & 2.064 with P = 2.5%}

P(-2.064 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.064) = 0.95

P(
`-2.064 * }{(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.064 * }{(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.064 * }{(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.064 * }{(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-2.064 * }{(s)/(√(n) ) }` ,
`\bar X+2.064 * }{(s)/(√(n) ) }` ]

= [
`150-2.064 * }{(54)/(√(25) ) }` ,
`150+2.064 * }{(54)/(√(25) ) }` ]

= [127.71 pounds , 172.30 pounds]

Therefore, 95% confidence interval for the true mean weight of men is [127.71 pounds , 172.30 pounds].