Respuesta :
Answer:
95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].
Step-by-step explanation:
We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.
An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.
Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample average height = 175 cm
[tex]\sigma[/tex] = population standard deviation = 20 cm
n = sample of men = 40
Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.
So, 95% confidence interval for the true mean, [tex]\mu[/tex] is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%
level of significance are -1.96 & 1.96}
P(-1.96 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times }{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.96 \times }{\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }[/tex] ]
= [ [tex]175-1.96 \times }{\frac{20}{\sqrt{40} } }[/tex] , [tex]175+1.96 \times }{\frac{20}{\sqrt{40} } }[/tex] ]
= [168.8 cm , 181.2 cm]
Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].
The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.
