Question

1. Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

95% confidence; n = 391, x = 42

2. Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

90% confidence; n = 383, x = 184

Answer 1

Solution :

1)

Given that,

n = 391

x = 42

= x / n = 41/391 = 0.1074

1 - = 1 - 0.1074 = 0.8926

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 * [(0.1074 * 0.8926) / 391]

        E = 0.0157

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.1074 - 0.0157 < p < 0.1074 + 0.0157

0.0917 < p < 0.1231

(0.0917, 0.1231)

2)

Solution :

Given that,

n = 383

x = 184

= x / n = 184/383 = 0.4804

1 - = 1 - 0.4804 = 0.5196

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * [(0.4804 * 0.5196) / 383)

   E = 0.0420

A 90% confidence interval for population proportion p is ,

- E < P < + E

0.4804 - 0.0420 < p < 0.4804 + 0.0420

0.4384 < p < 0.5224

(0.4384, 0.5224)