Question

A simple random sample of size n is drawn. The sample mean, x, is found to be 19.4, and the sample standard deviation, s, is found to be 4.9. Click the icon to view the table of areas under the t-distribution. (a) Construct a 95% confidence interval about if the sample size, n, is 35. Lower bound: :Upper bound: (Use ascending order. Round to two decimal places as needed.) (b) Construct a 95% confidence interval about 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? O A. The margin of error increases O B. The margin of error does not change. OC. The margin of error decreases. (c) Construct a 99% confidence interval about if the sample size, n, is 35. 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?

Answer 1

SOLUTION:

From given data,

Sample Mean = = 19.4

Sample standard deviation = S = 4.9

(a) Construct a 95% confidence interval about if the sample size,n,is 35

95% confidence interval

c = 95% = 95/100 = 0.95

= 1-c = 1-0.95 = 0.05

/2 = 0.05/2 = 0.025

Degree of freedom = df = n-1 = 35-1 = 34

Critical value = t_/2,df = t_0.025,34 = 2.0322

95% CI

- t_/2,df * (s/sqrt(n)) < < +t_/2,df * (s/sqrt(n))

19.4 -2.0322 * (4.9/sqrt(35)) < < 19.4+2.0322 * (4.9/sqrt(35))

19.4 -1.6831720269 < < 19.4+1.6831720269

17.72 < < 21.08

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

95% confidence interval

c = 95% = 95/100 = 0.95

= 1-c = 1-0.95 = 0.05

/2 = 0.05/2 = 0.025

Degree of freedom = df = n-1 = 61-1 = 60

Critical value = t_/2,df = t_0.025,60 = 2.00029

95% CI

- t_/2,df * (s/sqrt(n)) < < +t_/2,df * (s/sqrt(n))

19.4 -2.00029 * (4.9/sqrt(35)) < < 19.4+2.00029* (4.9/sqrt(35))

19.4 -1.6567425321 < < 19.4+1.6567425321

17.74 < < 21.06

How does increasing the sample size affect the margin of error E

The margin of error increases

Option (A) is correct

(c) Construct a 99% confidence interval about if the sample size,n,is 35

99% confidence interval

c = 99% = 99/100 = 0.99

= 1-c = 1-0.99 = 0.01

/2 = 0.01/2 = 0.005

Degree of freedom = df = n-1 = 35-1 = 34

Critical value = t_/2,df = t_0.005,34 = 2.72839

99% CI

- t_/2,df * (s/sqrt(n)) < < +t_/2,df * (s/sqrt(n))

19.4 -2.72839 * (4.9/sqrt(35)) < < 19.4+2.72839 * (4.9/sqrt(35))

19.4 -2.259792208 < < 19.4+2.259792208

17.14 < < 21.66

For the remaining problems please post again friend as per HOMEWORKLIB RULES first question should be the answer when multiple questions are posted. Please thumbs-up / vote up this answer if it was helpful. In case of any problem, please comment below. I will surely help. Down-votes are permanent and not notified to us, so we can't help in that case.