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This Quiz: 22 pts possit Use the standard normal distribution or the t-distribution to construct a 99% confidence interval fo

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Answer #1

solution:

sample size =n= 40

sample mean = \bar X = 26.9

sample standard deviation = S = 6.03

As the sample is randomly selected and standard deviation of the population is unknown so we use the t-distribution.

correct answer

B ) using a t distribution because the sample is random , the population is normal and the \sigma is unknown

step 2

significance level =\alpha= 1 - 0.99 = 0.01

degree of freedom = df = n - 1 = 40-1 = 39

critical value of t by using the t table = t_{\alpha/2,df}=t_{0.01/2,39}=2.708

margin of error(E) = t_{\alpha/2,df}*\frac{S}{\sqrt{n}}=2.708*\frac{6.03}{\sqrt{40}}=2.58

confidance interval = \bar X \pm E

lower bound of interval = 26.9 - 2.58 = 24.32

upper bound of interval = 26.9 + 2.58 = 29.48

correct choice is A) The 99% confidence interval is (24.32, 29.48)

part c)

interpretation of confidance interval

B) with 99% confidance , it can be said that the population mean BMI is between the bounds of confidence interval.

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