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 90% level of confidence. The president's political advisors estimated the proportion supporting the current policy to be 0.06.



a.
How large of a sample is required? (Round up your answer to the next whole number.)




b.
How large of a sample would be necessary if no estimate were available for the proportion that support current policy? (Round up your answer to the next whole number.)

Answer 1

Answer:  a. 170        b.  752

Step-by-step explanation:

The formula to find the sample size is given by :-

$n= p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

, where p is the prior estimate of population proportion.

$z_{\alpha/2}$ = Two-tailed z-value for $\alpha$ (significance level).

E= Margin of error .

Given : Margin of error =  0.03

Confidence level = 90%=0.90

$\alpha=1-0.90=0.10$

By z-value table : $z_{\alpha/2}=z_{0.05}=1.645$

a) The president's political advisors estimated the proportion supporting the current policy to be : p= 0.06.

Required sample size  :

$n= 0.06(1-0.06)(\dfrac{1.645}{0.03})^2$

$n=0.0564(3006.6944)$

$n=169.57756416\approx170$

∴ Required sample size = 170

b) If no prior estimate of population proportion is given , then we assume

p= 0.5

Required sample size  :

$n= 0.5(1-0.5)(\dfrac{1.645}{0.03})^2$

$n=0.25(3006.6944)$

$n=751.6736\approx752$

∴ Required sample size = 752

Answer 2

The probability computed shows that the sample size required will be 170.

How to calculate the probability?

The following have been given from the information:

Margin of error = 0.03

Confidence level = 90% = 0.90.

Z value = = 1.645

The sample size will be:

n = 0.06(1 - 0.06)(1.645/0.03)²

n = 170

The sample size that would be necessary if no estimate were available for the proportion that support current policy will be:

n = 0.5(1 - 0.5)(1.645/0.03)²

= 752

Therefore, the sample size is 752.

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