Question

The Consumer Reports National Research Center conducted a telephone survey of 2000 adults to learn about the major economic concerns for the future.The survey showed that 1240 of the respondents think the future health of Social Security is a major economic concern. Construct a 96% confidence interval estimate for the population proportion of adults who think the future health of social security is a major economic concern. Please round z or t value to 2 decimal places

Answer 1

Solution :

n = 2000

x = 1240

$\hat p$ = x / n = 1240 / 2000 = 0.620

1 -$\hat p$ = 1 - 0.620 = 0.0.380

At 96% confidence level the z is ,

$\alpha$ = 1 - 96% = 1 - 0.96 = 0.04

$\alpha$ / 2 = 0.04 / 2 = 0.02

Z_$\alpha$/2 = Z_0.02 = 2.054

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

= 2.054 * ($\sqrt$((0.620 * 0.0.380) / 2000)

= 0.022

A 96 % confidence interval for population proportion p is ,

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

0.620 - 0.022 < p < 0.620 + 0.022

0.598 < p < 0.642