Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean,

x overbarx​,

is found to be

109109​,

and the sample standard​ deviation, s, is found to be

1010.

​(a) Construct

aa

9090​%

confidence interval about

muμ

if the sample​ size, n, is

1515.

​(b) Construct

aa

9090​%

confidence interval about

muμ

if the sample​ size, n, is

2626.

​(c) Construct

aa

9898​%

confidence interval about

muμ

if the sample​ size, n, is

1515.

​(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 = = 109

sample standard deviation = s = 10

sample size = n = 15

Degrees of freedom = df = n - 1 = 14

a)

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,14 = 1.761

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

= 1.761 * ( 10/ 15)

= 4.547

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

- E < < + E

109 - 4.547 < < 109 + 4.547

104.453 < < 113.547

(104.453 , 113.547)

c)

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,14 = 2.624

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

= 2.624* ( 10/ 15)

= 6.775

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

- E < < + E

109 - 6.775 < < 109 + 6.775

102.225 < < 115.775

(102.225, 115.775)

d)

No