Answer:
96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].
Step-by-step explanation:
We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.
Firstly, the Pivotal quantity for 96% confidence interval for the population 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 desired retirement age = 55 years
[tex]\sigma[/tex] = sample standard deviation = 3.4 years
n = sample of seniors = 101
[tex]\mu[/tex] = true mean retirement age of all college students
Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 96% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.114 < [tex]t_1_0_0[/tex] < 2.114) = 0.96 {As the critical value of t at 100 degree
of freedom are -2.114 & 2.114 with P = 2%}
P(-2.114 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.114) = 0.96
P( [tex]-2.114 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.114 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.96
P( [tex]\bar X-2.114 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.114 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.96
96% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.114 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.114 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]55-2.114 \times {\frac{3.4}{\sqrt{101} } }[/tex] , [tex]55+2.114 \times {\frac{3.4}{\sqrt{101} } }[/tex] ]
= [54.30 , 55.70]
Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].
