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 90​% confidence interval about mu if the sample​ size, n, is 23. ​(b) Construct a 90​% confidence interval about mu if the sample​ size, n, is 27. ​(c) Construct a 95​% confidence interval about mu if the sample​ size, n, is 23. ​(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 = = 110

sample standard deviation = s = 10

a) sample size = n = 23

Degrees of freedom = df = n - 1 = 23- 1 = 22

At 90% confidence level

= 1-0.90% =1-0.9 =0.10

/2 =0.10/ 2= 0.05

t/2,df = t_{0.05,22 = 1.72}

t _/2,df = 1.72

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

= 1.72 * (10 / 23)

Margin of error = E = 3.58

The 90% confidence interval estimate of the population mean is,

- E < <  + E

110+3.58 < < 110-3.58

106.42 < < 113.58

(106.42,113.58)

b)

sample size = n = 27

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

At 90% confidence level

= 1-0.90% =1-0.9 =0.10

/2 =0.10/ 2= 0.05

t/2,df = t_{0.05,22 = 1.72}

t _/2,df = 1.72

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

= 1.71 * (10 / 27)

Margin of error = E = 3.28

The 90% confidence interval estimate of the population mean is,

- E < <  + E

110+3.28 < < 110-3.28

106.72 < < 113.28

(106.72,113.28)

c)

sample size = n = 23

Degrees of freedom = df = n - 1 = 23- 1 = 22

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

t/2,df = t_{0.025,99 =2.06}

t _/2,df = 2.06

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

= 2.06 * (10 / 23)

Margin of error = E = 3.96

The 90% confidence interval estimate of the population mean is,

- E < <  + E

110+3.96 < < 110-3.96

106.04 < < 113.96

(106.04,113.96)

d) no.