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 110, and the sample standard​ deviation, s, is found to be 10.

​(a) Construct a 90​% confidence interval about μ if the sample​ size, n, is 25.

​(b) Construct a 90% confidence interval about μ if the sample​ size, n, is 16.

​(c) Construct an 80​% confidence interval about μ if the sample​ size, n, is 25.

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

****NEED THE LOWER AND UPPER BOUNDS****

Answer 1

Solution :-

Given that,

Point estimate = sample mean = = 110

sample standard deviation = s = 10

( a )

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 90% confidence level

= 1 - 90%

=1 - 0.90 = 0.10

/2 = 0.05

t/2,df = t 0.05 , 24 = 1.711

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

= 1.711 * ( 10 / 24 )

Margin of error = E = 3.49

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

- E < < + E

110 - 3.49 < < 110 + 3.49

106.51 < < 113.49

( 106.51 , 113.49 )

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

Lower limit = 106.51

Upper limit = 113.49

( b )

Given that,:

sample size = n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

At 90% confidence level

= 1 - 90%

=1 - 0.90 = 0.10

/2 = 0.05

t/2,df = t 0.05 , 15 = 1.753

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

= 1.753 * ( 10 / 16 )

Margin of error = E = 4.38

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

- E < < + E

110 - 4.38 < < 110 + 4.38

105.62 < < 114.38

( 105.62 , 114.38 )

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

Lower limit = 105.62

Upper limit = 114.38

( c )

Given that,

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 80% confidence level

= 1 - 80%

=1 - 0.80 = 0.2

/2 = 0.1

t/2,df = t 0.1 , 24 = 1.318

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

= 1.318 * ( 10 / 24 )

Margin of error = E = 2.69

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

- E < < + E

110 - 2.69 < < 110 + 2.69

107.31 < < 112.69

( 107.31 ; 112.69 )

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

Lower limit = 107.31

Upper limit = 112.69

( d ) we have computed the confidence intervals in parts​ (a)-(c) if the population is normally​ distributed?