Question

4.

A simple random sample of 800 elements generates a sample proportion j5 =

.70.

a.

Provide a 90% confidence interval for the population proportion.

b.

Provide a 95% confidence interval for the population proportion.

Answer 1

a)

Given that,

n = 800

= 0.70

1 - = 1 - 0.70= 0.30

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90= 0.1

/ 2 = 0.1 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * {[(0.70*0.30)] / 800}

= 0.0267

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.70 - 0.0267 < p < 0.70 + 0.0267

0.6733 < p < 0.7267

0.67 < p < 0.73

b)

n = 800

= 0.70

1 - = 1 - 0.70= 0.30

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 = 1.96

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

= 1.96* {[(0.70*0.30)] / 800}

= 0.0318

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.70 - 0.0318< p < 0.70 + 0.0318

0.6682< p < 0.7318

0.67< p < 0.73