(2) A survey of households in a small town showed that in 500 of 2,000 sampled...

(2) A survey of households in a small town showed that in 500 of 2,000 sampled households, at least one member attended a town meeting during the year. What is the 95% confidence interval for the proportion of households represented at a town meeting?

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Expert Solution

Solution :

Given that,

n = 2000

x = 500

= x / n = 500 /2000 = 0.25

1 - = 1 - 0.25= 0.75

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} * $\sqrt$ (( * (1 - )) / n)

= 1.96 * ( $\sqrt$ ((0.25* 0.75) / 2000) = 0.0190

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.25 - 0.0190 < p < 0.25+ 0.0190

0.2310 < p < 0.2690

The 90% confidence interval for the population proportion p is : ( 0.2310 , 0.2690 )

orchestra answered 1 year ago

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