Q.44 - The Practice of Statistics for AP

Question

School vouchers A national opinion poll found that $44$% of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. The result was based on a small sample.

(a)  How large an SRS is required to obtain a margin of error of $0.03$ (that is, $\pm 3\%$) in a $99$% confidence interval? Answer this question using the previous poll’s result as the guessed value for $\stackrel{\wedge }{p}$.

(b)  Answer the question in part (a) again, but this time use the conservative guess $\stackrel{\wedge }{p}=0.5$. By how much do the two sample sizes differ? 

Step-by-Step Solution

Answer

From the given information,

a) The required sample size is $1816$

b) The sample size is increased by $105$ when the guessed value $\stackrel{\wedge }{p}$ is changes to $0.5$.

part (a) Step 1: Given Information

It is given in the question that, the margin of error $E=0.03$

sample proportion $\stackrel{\wedge }{p }=44\%$

confidence interval =$90\%$

How large an SRS is required to obtain a margin of error of $0.03$ (that is, $\pm 3$%) in a $99$% confidence interval?

part (a) Step 2:Explanation

The confidence interval is $99\%$

convert  $99\%$ into decimal.

$\frac{99}{100}=0.99$

For confidence interval $0.99,$use tableA.

$z_{\alpha /2}=2.575$

Sample proportion $\stackrel{\wedge }{p}is 44\%$

Convert $44\%$ into decimal.

$\frac{44}{100}=0.44$

Now, find the sample size. Use the formula $n=\frac{{\left[z_{a/2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$.

$n=\frac{{\left[z_{a/2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$n=\frac{{2.575}^2\times 0.44\times (1-0.44)}{{0.03}^2}$

$=1816$

Hence, the required sample size is $1816$

part (b) Step1: Given Information

It is given in the question that, the margin of error

sample proportion $\stackrel{\wedge }{p}=0.5$

confidence interval =

By how much do the two sample sizes differ? 

part (b) Step 2: Explanation

The confidence interval is

convert   into decimal.

For confidence interval use tableA.

$z_{\alpha /2}=2.575$

From part (a) sample size $n=1816$

Calculate the new sample size. Use the formula $n=\frac{{\left[z_{a/2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$.

$n\text{'}=\frac{{\left[z_{a/2}\right]}^2\hat{p}(1-\hat{p})}{E^2}$

$n^{\mathrm{\prime }}=\frac{{2.575}^2\times 0.5\times (1-0.5)}{{0.03}^2}$

    $=1842$

Thus, the new sample size is $1842$.

For the difference between the sample sizes, subtract $n$ from $n\text{'}$

 $n^{\mathrm{\prime }}-n=1842-1816$

             =$26$