A simple random sample of size n is drawn. The sample mean, x bar, is found to b

Question

A simple random sample of size n is drawn. The sample mean, x bar, is found to be 18.2, and the sample standard deviation, s, is found to be 4.2. (a) Construct a 95% confidence interval about mu if the sample size, n, is 35. Lower bound: Upper bound: (b) Construct a 95% confidence interval about mu if the sample size, n, is 61. Lower bound:: Upper bound: How does increasing the sample size affect the margin of error, E? A. The margin of error increases. B. The margin of error does not change. C. The margin of error decreases. (c) Construct a 99% confidence interval about mu if the sample size, n, is 35. Lower bound:: Upper bound: 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 does not change. C. The margin of error decreases. (d) If the sample size is 17, 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.

Explanation / Answer

a)

here std error =std deviation/(n)1/2 =4.2/(35)1/2 =0.7099

for 95% CI and (n-1=34) degree of freedom ; t=2.0322

margin of error =t*std error =1.4428

hence lower bound =sample mean -margin of error =16.76

upper bound =sample mean +margin of error =19.64

b)

here std error =std deviation/(n)1/2 =4.2/(61)1/2 =0.5378

for 95% CI and (n-1=60) degree of freedom ; t=2.0003

margin of error =t*std error =1.0757

hence lower bound =sample mean -margin of error =17.12

upper bound =sample mean +margin of error =19.28

option C is correct; margin of error decreases

c)

here std error =std deviation/(n)1/2 =4.2/(35)1/2 =0.7099

for 99% CI and (n-1=34) degree of freedom ; t=2.7284

margin of error =t*std error =1.9370

hence lower bound =sample mean -margin of error =16.26

upper bound =sample mean +margin of error =20.14

option A is correct; margin of error increases.

d)

option A is correct

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