Question

In a poll of 175 randomly selected U.S. adults, 94 said they favored a new proposition. Based on this poll, compute a 95% confidence interval for the proportion of all U.S. adults in favor of the proposition (at the time of the poll). Then complete the table below Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.) What is the lower limit of the 95% confidence interval? What is the upper limit of the 95% confidence interval?

Answer 1

Solution :

Given that,

n = 175

x = 94

Point estimate = sample proportion = $\hat p$ = x / n = 94 / 175 = 0.537

1 - $\hat p$ = 1 - 0.537 = 0.463

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 = 1.96

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

= 1.96 * ($\sqrt$((0.537 * 0.463) / 175)

= 0.074

A 95% confidence interval for population proportion p is ,

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

0.537 - 0.074 < p < 0.537 + 0.074

0.463 < p < 0.611

(0.463 , 0.611)

Lower limit = 0.463

Upper limit = 0.611