Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x overbar​, is found to be 110​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct a 98​% confidence interval about mu if the sample​ size, n, is 17. ​(b) Construct a 98​% confidence interval about mu if the sample​ size, n, is 28. ​(c) Construct a 99​% confidence interval about mu if the sample​ size, n, is 17. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

sample mean = 110

standard​ deviation = 10

a) n = 17 df = 16

98​% confidence interval t = 2.584

standard error = = = 2.425

= 110 2.584(2.425)

= 110 + 6.266 = 110-6.266

upper bound = 116.266 lower nound = 103.734

b)

sample size n= 28

df = 28-1 =27

98​% confidence interval t(alpha/2) = 2.584

standard error = = 10/sqrt(28) = 1.889

= 110 (1.889)(2.584)

= 110+4.881 = 110-4 .881

upper limit = 114.881 lower limit = 105.119

c)  

99​% confidence interval

n =17 df = 16

t(alpha/2) = 2.921

standard error =   = = 2.425

= 110 2.921*2.425

= 110+7.083 = 110-7.083

upper limit = 117.083 lower limit = 102.917

d)

NO, if the population had not been normally​ distributed we could not compute the confidence interval