Question

The manager of the service department of a local car dealership has noted that the service times of a sample of 30 new automobiles has a standard deviation of 6 minutes. A 95% confidence interval estimate for the standard deviation of the service times for all their new automobiles is

Answer 1

Answer:

95% confidence interval estimate for the standard deviation = [4.78 , 8.06]

Step-by-step explanation:

We are given that the manager of the service department of a local car dealership has noted that the service times of a sample of 30 new automobiles has a standard deviation of 6 minutes, i.e. n = 30 and s = 6.

The pivotal quantity for 95% confidence interval for the population variance is given by;

                   $\frac{(n-1)s^{2} }{\sigma^{2} }$  ~ $\chi^{2}__n_-_1$

95% confidence interval for the population variance is;

P(16.05 < $\chi^{2}__2_9$ < 45.72) = 0.95

P(16.05 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 45.72) = 0.95

P( $\frac{16.05}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{45.72}{(n-1)s^{2} }$ ) = 0.95

P( $\frac{(n-1)s^{2}}{ 45.72}$ < $\sigma^{2}$ <$\frac{(n-1)s^{2}}{ 16.05}$  ) = 0.95

95% Confidence interval for $\sigma^{2}$ = [ $\frac{(n-1)s^{2}}{ 45.72}$ , $\frac{(n-1)s^{2}}{ 16.05}$ ]

                                                = [ $\frac{(30-1)6^{2}}{ 45.72}$ , $\frac{(30-1)6^{2}}{ 16.05}$ ]

                                                = [ 22.835 , 65.047 ]

So, 95% Confidence interval for $\sigma$ = [ $\sqrt{ 22.835} ,\sqrt{ 65.047}$ ] = [ 4.78 , 8.06 ]

Therefore, 95% confidence interval for the standard deviation of the service times for all their new automobiles is [ 4.78 , 8.06 ] .