Question

A random sample of 15 employees was selected. The average age in the sample was 31 years with a variance of 49 years. Assuming ages are normally distributed, the 98% confidence interval for the population average age is _____. a. 26.26 to 35.74 b. 11.54 to 18.46 c. 25.62 to 36.38 d. 27.82 to 34.18

Answer 1

Option a : 26.26 to 35.74 .

Explanation:

We are provided a random sample of 15 employees of which average age in the sample, xbar = 31 years and Standard Deviation, s =
`√(49)` = 7 years.

We know that
`(xbar - \mu)/((s)/(√(n) ) )` follows
`t_n_-_1`

So, 98% confidence interval is given by ;

P(-2.624 <
`t_1_4` < 2.624) = 0.98 {because at 14 degree of freedom t table

gives critical value of 2.624 at 1% level}

P(-2.624 <
`(xbar - \mu)/((s)/(√(n) ) )` < 2.624) = 0.98

P(-2.624*
`(s)/(√(n) )` <
`xbar - \mu` < 2.624*
`(s)/(√(n) )` ) = 0.98

P(xbar - 2.624*
`(s)/(√(n) )` <
`\mu` < xbar + 2.624*
`(s)/(√(n) )` ) = 0.98

98% Confidence Interval for
`\mu` = [xbar - 2.624*
`(s)/(√(n) )` , xbar + 2.624*
`(s)/(√(n) )` ]

= [
`31 - 2.624*(7)/(√(15) )` ,
`31 + 2.624*(7)/(√(15) )` ]

= [26.26 , 35.74]

Therefore, 98% confidence interval for the population average age is 26.26 to 35.74 .