Respuesta :
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 = [tex]\sqrt{49}[/tex] = 7 years.
We know that [tex]\frac{xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] follows [tex]t_n_-_1[/tex]
So, 98% confidence interval is given by ;
P(-2.624 < [tex]t_1_4[/tex] < 2.624) = 0.98 {because at 14 degree of freedom t table
gives critical value of 2.624 at 1% level}
P(-2.624 < [tex]\frac{xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.624) = 0.98
P(-2.624*[tex]\frac{s}{\sqrt{n} }[/tex] < [tex]xbar - \mu[/tex] < 2.624*[tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.98
P(xbar - 2.624*[tex]\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < xbar + 2.624*[tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.98
98% Confidence Interval for [tex]\mu[/tex] = [xbar - 2.624*[tex]\frac{s}{\sqrt{n} }[/tex] , xbar + 2.624*[tex]\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]31 - 2.624*\frac{7}{\sqrt{15} }[/tex] , [tex]31 + 2.624*\frac{7}{\sqrt{15} }[/tex] ]
= [26.26 , 35.74]
Therefore, 98% confidence interval for the population average age is 26.26 to 35.74 .
