Respuesta :
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, MOE = 0.04.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]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[/tex]
Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.
(B)
The margin of error, MOE = 0.02.
The formula for margin of error is:
[tex]MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}[/tex]
The critical value of z for 95% confidence interval is: [tex]z_{\alpha/2}=1.96[/tex]
Compute the minimum sample size required as follows:
[tex]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[/tex]
Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.
