Question

(1 point) A poll is taken in which 399 out of 575 randomly selected voters indicated their preference for a certain candidate (a) Find a 90% confidence interval for p SPS (b) Find the margin of error for this 90% confidence interval for p (c) without doing any calculations, indicate whether the margin of error is larger or smaller or the same for an 80% confidence interval OA. larger B. smaller C. same

Answer 1

Solution :

Given that,

n = 575

x = 399

Point estimate = sample proportion = $\hat p$ = x / n = 399 / 575 = 0.694

1 - $\hat p$ = 1 - 0.594 = 0.406

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

(b)

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

= 1.645 * ($\sqrt$((0.694 * 0.406) / 575)

= 0.032

(b)

A 90% confidence interval for population proportion p is ,

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

0.694 - 0.032 < p < 0.694 + 0.032

0.662 < p < 0.726

(c)

smaller