Question

92.19-T Question Help A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 18.3, and the sample standard deviation s, is found to be 5.6. (a) Construct a 90% confidence interval about if the sample size, n, is 31. (b) Construct a 90% confidence interval about μ if the sample size, n' is 61 . How does increasing the sample size affect the margin of error, E? (c) Construct a 99% confidence interval about μ if the sample size, n' is 31 . How does increasing the level of confidence affect the size of the margin of error, E? (d) If the sample size is 18, what conditions must be satisfied to compute the confidence interval? (a) Construct a 90% confidence interval about if the sample size, n, is 31 Lower bound:Upper bound: (Round to two decimal places as needed.)

Answer 1

Solution :

Given that x-bar = 18.3 , s = 5.6

(a) sample size n = 31

=> df = n - 1 = 30

=> For 90% confidence interval , t = 1.697

=> The 90% confidence interval of the mean population is

=> x-bar +/- t*s/sqrt(n)

=> 18.3 +/- 1.697*5.6/sqrt(31)

=> (16.59 , 20.01)

=> Lower bound : 16.59 ; Upper bound : 20.01

=> Margin of error E = t*s/sqrt(n)

= 1.697*5.6/sqrt(31)

= 1.7068
  

(b) sample size n = 61

=> df = n - 1 = 60

=> For 90% confidence interval , t = 1.671

=> The 90% confidence interval of the mean population is

=> x-bar +/- t*s/sqrt(n)

=> 18.3 +/- 1.671*5.6/sqrt(61)

=> (17.10 , 19.50)

=> Lower bound : 17.10 ; Upper bound : 19.50

=> Margin of error E = t*s/sqrt(n)

= 1.671*5.6/sqrt(61)

= 1.1981

=> Therefore , when increase the sample size then the margin of error is decreases

(c) sample size n = 31

=> df = n - 1 = 30

=> For 99% confidence interval , t = 2.750

=> The 90% confidence interval of the mean population is

=> x-bar +/- t*s/sqrt(n)

=> 18.3 +/- 2.750*5.6/sqrt(31)

=> (15.53 , 21.07)

=> Lower bound : 15.53 ; Upper bound : 21.07

=> Margin of error E = t*s/sqrt(n)

= 2.750*5.6/sqrt(31)

= 2.7659

=> Therefore , the confidence level is increases then the margin of error is increases.

(d)

=> The sample data must come from a population that is normally distributed with no outliers