Respuesta :
Answer:
95% confidence interval for the true mean weight of men is [127.71 pounds , 172.30 pounds].
Step-by-step 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. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = 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, [tex]\mu[/tex] is ;
P(-2.064 < [tex]t_2_4[/tex] < 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 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.064) = 0.95
P( [tex]-2.064 \times }{\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.064 \times }{\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.064 \times }{\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.064 \times }{\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.064 \times }{\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.064 \times }{\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]150-2.064 \times }{\frac{54}{\sqrt{25} } }[/tex] , [tex]150+2.064 \times }{\frac{54}{\sqrt{25} } }[/tex] ]
= [127.71 pounds , 172.30 pounds]
Therefore, 95% confidence interval for the true mean weight of men is [127.71 pounds , 172.30 pounds].
