Question

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 1

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

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

where,
`\bar X` = sample average desired retirement age = 55 years

`\sigma` = sample standard deviation = 3.4 years

n = sample of seniors = 101

`\mu` = 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,
`\mu` is ;

P(-2.114 <
`t_1_0_0` < 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 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.114) = 0.96

P(
`-2.114 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.114 * {(s)/(√(n) ) }` ) = 0.96

P(
`\bar X-2.114 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.114 * {(s)/(√(n) ) }` ) = 0.96

96% confidence interval for
`\mu` = [
`\bar X-2.114 * {(s)/(√(n) ) }` ,
`\bar X+2.114 * {(s)/(√(n) ) }` ]

= [
`55-2.114 * {(3.4)/(√(101) ) }` ,
`55+2.114 * {(3.4)/(√(101) ) }` ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].