Question

A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds. Unfortunately, the company has a problem with the variability of the weight. In a sample of 10 of the bowling balls the sample standard deviation was found to be 0.94 pounds. Construct a 95% confidence interval for the variance of the bowling ball weight. Assume normality.

Answer 1

Answer:

95% confidence interval for the variance of the bowling ball weight is (0.418 , 2.945).

Step-by-step explanation:

We are given that the company has a problem with the variability of the weight. For this, a sample of 10 of the bowling balls, the sample standard deviation was found to be 0.94 pounds.

So, firstly the pivotal quantity for 95  % confidence interval for the population standard deviation is given by;

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

where, $s^{2}$ = sample variance = $0.94^{2}$

           $\sigma^{2}$ = population variance

            n = sample of bawling balls = 10

So, 95% confidence interval for population standard deviation, is;

P(2.7 < $\chi^{2} __1_0_-_1$ < 19.02) = 0.95 {As the table of $\chi^{2}$ at 9 degree of freedom

                                          gives critical values of 2.7 & 19.02}

P(2.7 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 19.02) = 0.95

P( $\frac{2.7}{(n-1)s^{2} }$ < $\frac{1}{\sigma^{2} }$ < $\frac{19.02}{(n-1)s^{2} }$ ) = 0.95

P($\frac{ (n-1)s^{2}}{19.02}$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{2.7}$ ) = 0.95

95% confidence interval for  = ( $\frac{ (n-1)s^{2}}{19.02}$ , $\frac{ (n-1)s^{2}}{2.7}$ )

                                                  = ( $\frac{ (10-1) \times 0.94^{2}}{19.02}$ , $\frac{ (10-1) \times 0.94^{2}}{2.7}$ )

                                                  = (0.418 , 2.945)

Therefore, 95% confidence interval for the variance of the bowling ball weight is (0.418 , 2.945).

Answer 2

Answer:

95% confidence interval for the variance of the bowling ball weight is between a lower limit of 13.328 pounds and an upper limit of 14.672 pounds.

Step-by-step explanation:

Confidence interval is given as weight +/- margin of error (E)

weight = 14 pounds

sample sd = 0.94 pound

n = 10

degree of freedom = n - 1 = 10 - 1 = 9

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value (t) corresponding to 9 degrees of freedom and 5% significance level is 2.262

E = t × sample sd/√n = 2.262×0.94/√10 = 0.672 pounds

Lower limit of weight = weight - E = 14 - 0.672 = 13.328 pounds

Upper limit of weight = weight + E = 14 + 0.672 = 14.672 pounds.

95% confidence interval is (13.328, 14.672)