Question

Suppose the U.S. president wants an estimate of the proportion of the population who support his current policy toward revisions in the health care system. The president wants the estimate to be within 0.03 of the true proportion. Assume a 98% level of confidence. The president's political advisors estimated the proportion supporting the current policy to be 0.58. (Use z Distribution Table.)

a.	How large of a sample is required? (Round the z-values to 2 decimal places. Round up your answer to the next whole number.)

Sample

b.	How large of a sample would be necessary if no estimate were available for the proportion supporting current policy? (Round the z-values to 2 decimal places. Round up your answer to the next whole number.)

Sample

Answer 1

Solution :

Given that,

margin of error = E = 0.03

At 98% confidence level the z is ,

$\alpha$ = 1 - 98% = 1 - 0.98 = 0.02

$\alpha$ / 2 = 0.02 / 2 = 0.01

Z_$\alpha$/2 = Z_0.01 = 2.33

(a)

$\hat p$ = 0.58

1 - $\hat p$ = 1 - 0.58 = 0.42

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (2.33 / 0.03)² * 0.58 * 0.42

= 1469.4 = 1470

sample size = 1470

(b)

$\hat p$ = 0.5

1 - $\hat p$ = 1 - 0.5 = 0.5

sample size = n = (Z_{$\alpha$ / 2} / E )² * $\hat p$ * (1 - $\hat p$ )

= (2.33 / 0.03)² * 0.5 * 0.5

= 1508.02 = 1509

sample size = 1509