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

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 = $\sqrt{49}$ = 7 years.

We know that  $\frac{xbar - \mu}{\frac{s}{\sqrt{n} } }$ 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 < $\frac{xbar - \mu}{\frac{s}{\sqrt{n} } }$ < 2.624) = 0.98

  P(-2.624*$\frac{s}{\sqrt{n} }$ < $xbar - \mu$ < 2.624*$\frac{s}{\sqrt{n} }$ ) = 0.98

  P(xbar - 2.624*$\frac{s}{\sqrt{n} }$ < $\mu$ < xbar + 2.624*$\frac{s}{\sqrt{n} }$ ) = 0.98

98% Confidence Interval for $\mu$ = [xbar - 2.624*$\frac{s}{\sqrt{n} }$ , xbar + 2.624*$\frac{s}{\sqrt{n} }$ ]

                                                  = [ $31 - 2.624*\frac{7}{\sqrt{15} }$ , $31 + 2.624*\frac{7}{\sqrt{15} }$ ]

                                                  = [26.26 , 35.74]

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