Question

A survey of the age of people denied promotion was conducted. In a random sample of 23 people the average age was 47.0 with a sample standard deviation of 7.2 Assume this sample comes from a population that is normally distributed. a. Calculate the EBM then construct a 95% confidence interval for the average of the age of people denied (round the values of the interval to 1 decimal place). b. If a manager claimed that the average age of people denied promotion was 51, would you agree or disagree based on your confidence interval? Explain your reasoning.

Answer 1

(a) 95% confidence interval for the average of the age of people denied is [43.9 , 50.1].

(b) Based on our confidence interval, we would disagree with the manager's claim.

Explanation:

We are given that a survey of the age of people denied promotion was conducted. In a random sample of 23 people the average age was 47.0 with a sample standard deviation of 7.2.

Assume this sample comes from a population that is normally distributed.

Firstly, the Pivotal quantity for 95% confidence interval for the population mean is given by;

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

where,
`\bar X` = sample average age = 47.0

`\sigma` = sample standard deviation = 7.2

n = sample of people = 23

`\mu` = true average of the age of people denied

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 population mean,
`\mu` is ;

P(-2.074 <
`t_1_0_0` < 2.074) = 0.95 {As the critical value of t at 22 degree

of freedom are -2.074 & 2.074 with P = 2.5%}

P(-2.074 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 2.074) = 0.95

P(
`-2.074 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`2.074 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.074 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+2.074 * {(s)/(√(n) ) }` ) = 0.95

(a) 95% confidence interval for
`\mu` = [
`\bar X-2.074 * {(s)/(√(n) ) }` ,
`\bar X+2.074 * {(s)/(√(n) ) }` ]

= [
`47-2.074 * {(7.2)/(√(23) ) }` ,
`47+2.074 * {(7.2)/(√(23) ) }` ]

= [43.9 , 50.1]

Therefore, 95% confidence interval for the average of the age of people denied is [43.9 , 50.1].

(b) Based on our confidence interval, we would disagree with the manager's claim that the average age of people denied promotion was 51 because the interval doesn't contain the value 51.