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 7 of the bowling balls the sample standard deviation was found to be 0.64 pounds. Construct a 95% confidence interval for the variance of the bowling ball weight. Assume normality. a) What is the lower limit of the 95% interval? Give your answer to three decimal places. b) What is the upper limit of the 95% interval? Give your answer to three decimal places. c) Which of the following assumptions should be checked before constructing the above confidence interval? the data need to follow a normal distribution the data need to have small variance the data need to follow a chi-square distribution the data need to be right skewed

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Answer:

Step-by-step explanation:

Mean = 14

Std deviation of sample s = 0.64

n = sample size =7

Std error = [tex]\frac{s}{\sqrt{n} } =0.2419[/tex]

t critical for 95% two tailed = 2.02

Margin of error = 2.02*SE = 0.4886

a)Conf interval lower bound = 14-0.4886 = 13.5114

b)Upper bound = 14+0.4886 = 14.4886

c)Assumption

the data need to follow a normal distribution

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