Question

# A simple random sample of size n is drawn. The sample​ mean, x overbar​, is found...

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)

a)

 sample mean 'x̄= 17.6 sample size   n= 35 sample std deviation s= 4.1
 for 95% CI; and 34 df, value of t= 2.032 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= 2.009 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.

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