Question

A poll is taken in which 310 out of 525 randomly selected voters indicated their preference for a certain candidate. (a) Find a 90 % confidence interval for p . ≤ p≤

(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.

Answer 1

Solution :

Given that,

n = 525

x = 310

(a) Point estimate = sample proportion = = x / n = 310 / 525 = 0.590

1 - = 1 - 0.590 = 0.410

(a) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 ((0.590 * (0.410) / 525)

= 0.035

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.590 - 0.035 < p < 0.590 + 0.035

0.555 < p < 0.625

The 95% confidence interval for p is : 0.555 < p < 0.625

(b) Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 ((0.590 * (0.410) / 525)

= 0.035

Margin of error = E = 0.035

(c) The margin of error is smaller for a 80% confident interval.