Question

For each combination of sample size and sample proportion, find the approximate margin of error for the 95% confidence level. (Round the answers to three decimal places.)

(a) n = 100, p̂ = 0.55.


(b) n = 700, p̂ = 0.55.


(c) n = 700, p̂ = 0.10.


(d) n = 700, p̂ = 0.90.


(e) n = 1000, p̂ = 0.60.

Answer 1

Solution :

Given that

a ) n = 100

$\hat p$ = 0.550

1 -$\hat p$ = 1 - 0.550 = 0.450

At 95% confidence level the z is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$(($\hat p$ * (1 - $\hat p$)) / n)

= 1.960 * ($\sqrt$((0.550 * 0.450) / 100)

= 0.097

A 95 % confidence interval for population proportion p is ,

$\hat p$- E < P <$\hat p$ + E

0.550 - 0.097 < p < 0.550 + 0.097

0.452 < p < 0.647

b ) n = 700

$\hat p$ = 0.550

1 -$\hat p$ = 1 - 0.550 = 0.450

At 95% confidence level the z is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$(($\hat p$ * (1 - $\hat p$)) / n)

= 1.960 * ($\sqrt$((0.550 * 0.450) / 700)

= 0.037

A 95 % confidence interval for population proportion p is ,

$\hat p$- E < P <$\hat p$ + E

0.550 - 0.037 < p < 0.550 + 0.037

0.513 < p < 0.587

c ) n = 700

$\hat p$ = 0.100

1 -$\hat p$ = 1 - 0.100 = 0.900

At 95% confidence level the z is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$(($\hat p$ * (1 - $\hat p$)) / n)

= 1.960 * ($\sqrt$((0.100 * 0.900) / 700)

= 0.097

A 95 % confidence interval for population proportion p is ,

$\hat p$- E < P <$\hat p$ + E

0.100 - 0.022 < p < 0.100 + 0.022

0.078 < p < 0.122

d ) n = 700

$\hat p$ = 0.900

1 -$\hat p$ = 1 - 0.900 = 0.100

At 95% confidence level the z is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$(($\hat p$ * (1 - $\hat p$)) / n)

= 1.960 * ($\sqrt$((0.900 * 0.100) / 700)

= 0.022

A 95 % confidence interval for population proportion p is ,

$\hat p$- E < P <$\hat p$ + E

900 - 0.022 < p < 900 + 0.022

0.878 < p < 0.922

e ) n = 1000

$\hat p$ = 0.600

1 -$\hat p$ = 1 - 0.600 = 0.400

At 95% confidence level the z is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$(($\hat p$ * (1 - $\hat p$)) / n)

= 1.960 * ($\sqrt$((0.600 * 0.400) / 1000)

= 0.030

A 95 % confidence interval for population proportion p is ,

$\hat p$- E < P <$\hat p$ + E

0.600 - 0.030 < p < 0.600 + 0.030

0.570 < p < 0.630