Answer:
Option a : 26.26 to 35.74 .
Step-by-step explanation:
We are provided a random sample of 15 employees of which average age in the sample, xbar = 31 years and Standard Deviation, s = = 7 years.
<em>We know that </em><em> follows </em><em />
So, 98% confidence interval is given by ;
P(-2.624 < < 2.624) = 0.98 {because at 14 degree of freedom t table
gives critical value of 2.624 at 1% level}
P(-2.624 < < 2.624) = 0.98
P(-2.624* < < 2.624* ) = 0.98
P(xbar - 2.624* < < xbar + 2.624* ) = 0.98
98% Confidence Interval for = [xbar - 2.624* , xbar + 2.624* ]
= [ , ]
= [26.26 , 35.74]
Therefore, 98% confidence interval for the population average age is 26.26 to 35.74 .