Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar x​, is found to be 107​, and the sample standard​ deviation, s, is found to be 10. (a) Construct a 98​% confidence interval about mu μ if the sample​ size, n, is 18.    (b) Construct a 98​% confidence interval about mu μnif the sample​ size, n, is 12. c) Construct a 96​% confidence interval about mu μ if the sample​ size, n, is 8. (d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed. state the upper bound and lower bound for each

Answer 1

solution:-

given that x̅ = 107 , standard deviation s = 10

(a) 98% confidence interval

here given n = 18

degree of freedom df = n - 1 = 18 - 1 = 17

we look into t table with df and with 98% confidence interval

critical value t = 2.567

confidence interval formula

=> x̅ +/- t * s/sqrt(n)

=> 107 +/- 2.567*10/sqrt(18)

=> (100.9 , 113.1)

lower bound = 100.9 , upper bound = 113.1

b) 98% confidence interval

here given n = 12

degree of freedom df = n - 1 = 12 - 1 = 11

we look into t table with df and with 98% confidence interval

critical value t = 2.718

confidence interval formula

=> x̅ +/- t * s/sqrt(n)

=> 107 +/- 2.718*10/sqrt(12)

=> (99.2 , 114.8)

lower bound = 99.2 , upper bound = 114.8

c) 96% confidence interval

here given n = 8

degree of freedom df = n - 1 = 8 - 1 = 7

we look into t table with df and with 96% confidence interval

critical value t = 2.517

confidence interval formula

=> x̅ +/- t * s/sqrt(n)

=> 107 +/- 2.517*10/sqrt(8)

=> (98.1 , 115.9)

lower bound = 98.1 , upper bound = 115.9

d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed.

=> No , If the population is not normally distribution, then we cannot compute the confidence interval, because we require that the sampling distribution of the sample mean is approximately norma