Question

A.

If

Upper X overbar

=61​,

Upper S=27​,

and

n=49​,

and assuming that the population is normally​ distributed, construct a

90 %

confidence interval estimate of the population​ mean,

B. If

n=200

and

X=40​,

construct a

99%

confidence interval estimate of the population proportion.

C.If

n=400

and

X=140​,

construct a

99%

confidence interval estimate of the population proportion.

Answer 1

Solution :

Given that,

A ) = 61

s = 27​

n = 49

Degrees of freedom = df = n - 1 = 49 - 1 = 48

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,48 =1.677

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

= 1.677 * (27​ / 49)

= 6.468

Margin of error = 6.468

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

- E < < + E

61- 6.468 < < 61 + 6.468

54.532 < < 67.468

(54.532 , 67.468 )

B ) Given that,

n = 200

x = 40

= x / n = 40 / 200 = 0.200

1 - = 1 - 0.200 = 0.800

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.200 * 0.800) / 200)

= 0.073

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.200 - 0.073 < p < 0.200 + 0.073

0.127 < p < 0.273

The 99% confidence interval for the population proportion p is : ( 0.127 , 0.200)

C ) Given that,

n = 400

x = 140

= x / n = 140 / 400 = 0.350

1 - = 1 - 0.350 = 0.650

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.350 * 0.650) / 400)

= 0.061

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.350 - 0.061 < p < 0.350 + 0.061

0.289 < p < 0.411

The 99% confidence interval for the population proportion p is : ( 0.289,  0.411)