Question

In a random sample of 625 messages, 20% were spam.

1. use this data to construct a 95% confidence interval for the true proportion os spam in all e-mail.

2. report the margin error associated with a 95% confidence interval for the treu proportion of spam in all email.

3. if you chose a 99% instead if 95% would you expect the margin of error to increase or decrease? why?

show all work.

Answer 1

Solution :

Given that,

n = 625

= 20% = 0.20

1 - = 1 - 0 20. = 0.80

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (((0.20 * 0.80) / 625 ) =0.03

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.20 - 0.03 < p < 0.20 + 0.03

0.17 < p < 0.23

The 99% confidence interval for the population proportion p is : ( 0.17 , 0.23)

3 ) 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 * (((0.20 * 0.80) / 625 ) =0.04

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.20 - 0.04 < p < 0.20 + 0.04

0.16 < p < 0.24

The 99% confidence interval for the population proportion p is : ( 0.16 , 0.24)

The confidence ineval leval is increase the margin of error is increase