Question

A simple random sample of size n is drawn. The sample​ mean, x​, is found to be 18.5, and the sample standard​ deviation, s, is found to be 4.6.

​(a) Construct a 95​% confidence interval about μ if the sample​ size, n, is 34.

Lower​ bound: ___

Upper​ bound: ___

​(Use ascending order. Round to two decimal places as​ needed.)

​(b) Construct a 95% confidence interval about μ if the sample​ size, n, is 61.

Lower​ bound: ___

Upper​ bound: ___

​(Use ascending order. Round to two decimal places as​ needed.)

How does increasing the sample size affect the margin of​ error, E?

A. The margin of error decreases.

B. The margin of error does not change.

C.The margin of error increases.

​(c) Construct a 99​% confidence interval about μ if the sample​ size, n, is 34.

Lower​ bound: ___

Upper​ bound: ___

​(Use ascending order. Round to two decimal places as​ needed.)

Compare the results to those obtained in part​ (a). How does increasing the level of confidence affect the size of the margin of​ error, E?

A. The margin of error increases.

B.The margin of error decreases.

C. The margin of error does not change.

​(d) If the sample size is 14​, what conditions must be satisfied to compute the confidence​ interval?

A. The sample data must come from a population that is normally distributed with no outliers.

B. The sample must come from a population that is normally distributed and the sample size must be large.

C. The sample size must be large and the sample should not have any outliers.

Answer 1

a)

sample mean, xbar = 18.5
sample standard deviation, s = 4.6
sample size, n = 34
degrees of freedom, df = n - 1 = 33

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.035

ME = tc * s/sqrt(n)
ME = 2.035 * 4.6/sqrt(34)
ME = 1.605

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (18.5 - 2.035 * 4.6/sqrt(34) , 18.5 + 2.035 * 4.6/sqrt(34))
CI = (16.89 , 20.11)

Lower bound = 16.89

Upper bound = 20.11

b)

sample mean, xbar = 18.5
sample standard deviation, s = 4.6
sample size, n = 61
degrees of freedom, df = n - 1 = 60

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

ME = tc * s/sqrt(n)
ME = 2 * 4.6/sqrt(61)
ME = 1.178

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (18.5 - 2 * 4.6/sqrt(61) , 18.5 + 2 * 4.6/sqrt(61))
CI = (17.32 , 19.68)

Lower bound = 17.32
Upper bound = 19.68

C.The margin of error increases.

c)
sample mean, xbar = 18.5
sample standard deviation, s = 4.6
sample size, n = 34
degrees of freedom, df = n - 1 = 33

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.733

ME = tc * s/sqrt(n)
ME = 2.733 * 4.6/sqrt(34)
ME = 2.156

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (18.5 - 2.733 * 4.6/sqrt(34) , 18.5 + 2.733 * 4.6/sqrt(34))
CI = (16.34 , 20.66)

Lower bound = 16.34
Upper bound = 20.66

A. The margin of error increases.

d)

B. The sample must come from a population that is normally distributed and the sample size must be large.

c)