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. A 96% confidence interval for desired retirement age of all college students is:
54.30 to 55.70
54.55 to 55.45
54.58 to 55.42
54 60 to 55.40

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].

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