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.

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)

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step-by-step explanation: