Question

A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found to be 17.6​, and the sample standard​ deviation, s, is found to be 4.1. LOADING... Click the icon to view the table of areas under the​ t-distribution. ​(a) Construct a 95​% confidence interval about mu if the sample​ size, n, is 35. Lower​ bound: nothing​; Upper​ bound: nothing ​(Use ascending order. Round to two decimal places as​ needed.) ​(b) Construct a 95​% confidence interval about mu if the sample​ size, n, is 51. Lower​ bound: nothing​; Upper​ bound: nothing ​(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 does not change. B. The margin of error increases. C. The margin of error decreases. ​(c) Construct a 99​% confidence interval about mu if the sample​ size, n, is 35. Lower​ bound: nothing​; Upper​ bound: nothing ​(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 does not change. B. The margin of error decreases. C. The margin of error increases. ​(d) If the sample size is 16​, 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 size must be large and the sample should not have any outliers. C. The sample must come from a population that is normally distributed and the sample size must be large. Click to select your answer(s)

Answer 1

a)

sample mean 'x̄=	17.600
sample size   n=	35.00
sample std deviation s=	4.100

for 95% CI; and 34 df, value of t=		2.0320
margin of error E=t*std error    =		1.408
lower bound=sample mean-E =		16.19
Upper bound=sample mean+E =		19.01

b)

for 95% CI; and 50 df, value of t=		2.0090
margin of error E=t*std error    =		1.153
lower bound=sample mean-E =		16.45
Upper bound=sample mean+E =		18.75

. C. The margin of error decreases

c)

for 99% CI; and 34 df, value of t=		2.7280
margin of error E=t*std error    =		1.891
lower bound=sample mean-E =		15.71
Upper bound=sample mean+E =		19.49

  C. The margin of error increases.

d)

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