Question

A) If n=570 and ˆ p (p-hat) =0.21, find the margin of error at a 99% confidence level Give your answer to three decimals

B) If n=15, ¯xx¯(x-bar)=30, and s=20, construct a confidence interval at a 95% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

C) You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 38.

Find the critical value that corresponds to a confidence level of 85%.
(Report answer accurate to three decimal places with appropriate rounding.)

Answer 1

Solution :

a ) Given that

n = 570

= 0.210

1 - = 1 - 0.210 = 0.790

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 * (((0.210 * 0.790) / 570)

= 0.044

Margin of error =0.044

b ) Given that,

= 30

s = 20

n = 15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,14 =2.145

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

= 2.145 * (20 / 15)

= 11.07

Margin of error = 11.07

The 95% confidence interval estimate of the population mean is,

- E < < + E

30 - 11.07 < < 30 + 11.07

18.92 < < 341.07

(18.92, 41.07 )

c ) Given that

n = 38

Degrees of freedom = df = n - 1 = 38 - 1 = 37

At 85% confidence level the t is ,

  = 1 - 85% = 1 - 0.85 = 0.15

/ 2 = 0.15 / 2 = 0.075

t _/2,df = t_0.075,37=1.470

The critical value =1.470