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 115​, and the sample standard​ deviation, s, is found to be 10. ​(a) Construct an 80​% confidence interval about mu if the sample​ size, n, is 12. ​(b) Construct an 80​% confidence interval about mu if the sample​ size, n, is 27. ​(c) Construct a 98​% confidence interval about mu if the sample​ size, n, is 12. ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 115

sample standard deviation = s = 10

a)

sample size = n = 12

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

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.8 = 0.2

/ 2 = 0.2 / 2 = 0.1

t _/2,df = t_0.1,11 = 1.363

Margin of error = E = t_/2,df * (s /n)

= 1.363 * (10 / 12)

= 3.9

The 80% confidence interval estimate of the population mean is,

- E < < + E

115 - 3.9 < < 115 + 3.9

111.1 < < 118.9

(111.1, 118.9)

b)

sample size = n = 27

Degrees of freedom = df = n - 1 = 27 - 1 = 26

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.2

/ 2 = 0.2 / 2 = 0.1

t _/2,df = t_0.1,26 =1.315

Margin of error = E = t_/2,df * (s /n)

= 1.315 * (10 / 27)

= 2.5

The 80% confidence interval estimate of the population mean is,

- E < < + E

115 - 2.5 < < 115 + 2.5

112.5 < < 117.5

(112.5 , 117.5)

c)

sample size = n = 12

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

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,11 = 2.718

Margin of error = E = t_/2,df * (s /n)

= 2.718 * ( 10/ 12)

= 7.8

The 98% confidence interval estimate of the population mean is,

- E < < + E

115 - 7.8 < < 115 + 7.8

107.2 < < 122.8

(107.2 , 122.8)

d)

Yes