Question

The random sample shown below was selected from a normal distribution. 3​, 7​, 4​, 10​, 7​, 5 Complete parts a and b. a. Construct a 95​% confidence interval for the population mean mu. left parenthesis nothing comma nothing right parenthesis ​(Round to two decimal places as​ needed.) b. Assume that sample mean x overbar and sample standard deviation s remain exactly the same as those you just calculated but that are based on a sample of nequals25 observations. Repeat part a. What is the effect of increasing the sample size on the width of the confidence​ intervals? The confidence interval is left parenthesis nothing comma nothing right parenthesis . ​(Round to two decimal places as​ needed.) What is the effect of the sample size on the width of the confidence​ interval? A. As the sample size​ increases, the width increases. B. As the sample size​ increases, the width stays the same. C. As the sample size​ increases, the width decreases.

Answer 1

a)

sample mean, xbar = 6
sample standard deviation, s = 2.5298
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.571

ME = tc * s/sqrt(n)
ME = 2.571 * 2.5298/sqrt(6)
ME = 2.655

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (6 - 2.571 * 2.5298/sqrt(6) , 6 + 2.571 * 2.5298/sqrt(6))
CI = (3.34 , 8.66)

b)

sample mean, xbar = 6
sample standard deviation, s = 2.5298
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.064

ME = tc * s/sqrt(n)
ME = 2.064 * 2.5298/sqrt(25)
ME = 1.044

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (6 - 2.064 * 2.5298/sqrt(25) , 6 + 2.064 * 2.5298/sqrt(25))
CI = (4.96 , 7.04)

C. As the sample size​ increases, the width decreases.