Answer:
(A) The minimum sample size required achieve the margin of error of 0.04 is 601.
(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.
Step-by-step explanation:
Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.
(A)
The margin of error, <em>MOE</em> = 0.04.
The formula for margin of error is:
![MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%20%2F2%7D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
The critical value of <em>z</em> for 95% confidence interval is: ![z_{\alpha/2}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3D1.96)
Compute the minimum sample size required as follows:
![MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%20%2F2%7D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%5C%5C0.04%3D1.96%5Ctimes%20%5Csqrt%7B%5Cfrac%7B0.50%281-0.50%29%7D%7Bn%7D%7D%5C%5C%28%5Cfrac%7B0.04%7D%7B1.96%7D%29%5E%7B2%7D%20%3D%5Cfrac%7B0.50%281-0.50%29%7D%7Bn%7D%5C%5Cn%3D600.25%5Capprox%20601)
Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.
(B)
The margin of error, <em>MOE</em> = 0.02.
The formula for margin of error is:
![MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%20%2F2%7D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
The critical value of <em>z</em> for 95% confidence interval is: ![z_{\alpha/2}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3D1.96)
Compute the minimum sample size required as follows:
![MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%20%2F2%7D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%5C%5C0.02%3D1.96%5Ctimes%20%5Csqrt%7B%5Cfrac%7B0.50%281-0.50%29%7D%7Bn%7D%7D%5C%5C%28%5Cfrac%7B0.02%7D%7B1.96%7D%29%5E%7B2%7D%20%3D%5Cfrac%7B0.50%281-0.50%29%7D%7Bn%7D%5C%5Cn%3D2401.00%5Capprox%202401)
Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.