A company produces a women’s bowling ball that is supposed to weigh exactly 14 pounds. Unfortunately, the company has a problem with the var

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 11 of the bowling balls the sample standard deviation was found to be 0.71 pounds. Construct a 95% confidence interval for the variance of the bowling ball weight. Assume normality.

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Aubrey 2 weeks 2021-11-22T02:21:19+00:00 1 Answer 0

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    2021-11-22T02:22:41+00:00

    Answer: 0.2461<\sigma^2<1.5511

    Step-by-step explanation:

    Given : A company produces a women’s bowling ball that is supposed to weigh exactly 14 pounds.

    Sample size : n=11

    Degree of freedom =n-1=10

    Sample standard deviation : s= 0.71 pounds

    Significance level for 95% confidence interval :\alpha=1-0.95=0.05

    We assume that the bowling ball weight is normally distributed.

    Using chi-square distribution table, the required critical values are :-

    \chi^2_{df, \alpha/2}=20.48

    \chi^2_{df, 1-\alpha/2}=3.25

    Then, the 95% confidence interval for the variance of the bowling ball weight will be :

    \dfrac{s^2(n-1)}{\chi_{\alpha/2}}<\sigma^2<\dfrac{s^2(n-1)}{\chi_{1-\alpha/2}}

    =\dfrac{(0.71)^2(10)}{20.48}<\sigma^2<\dfrac{(0.71)^2(10)}{3.25}\\\\=0.2461<\sigma^2<1.5511

    ∴ The 95% confidence interval for the variance of the bowling ball weight will be : 0.2461<\sigma^2<1.5511

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