Question

In Exercises 6.109 to 6.111 , we examine the effect of different inputs on determining the sample size needed. Find the sample size needed to give, with \(95 \%\) confidence, a margin of error within ±10 . Within ±5 . Within ±1 . Assume that we use \(\tilde{\sigma}=30\) as our estimate of the standard deviation in each case. Comment on the relationship between the sample size and the margin of error.

Short answer

The margin of error being ±10 results in a sample size of roughly 35. For a margin of error of ±5, the sample size is approximately 138. Lastly, for the smallest margin of error, ±1, the sample size goes up to about 3456. The smaller the margin of error, the larger the sample size needed thus indicating that for more precision a larger sample size is required.

Step-by-step solution

Setup the equation

Firstly, get the z-score, \(Z_{\frac{\alpha}{2}}\), for the desired 95% confidence level. For a two-tail test, the z-score is approximately 1.96. Secondly, solve the formula for the sample size \(n = \frac{(Z_{\frac{\alpha}{2}} \cdot \sigma)^{2}}{E^{2}}\) for each of the three different margin of errors.

Calculate sample size for margin of error ±10

Substitute the values into the equation, where \(Z_{\frac{\alpha}{2}}\) is 1.96, \(\sigma\) is 30, and \(E\) is 10. Thus, calculate the sample size, where \(n = \frac{(1.96 \cdot 30)^{2}}{10^{2}}\).

Calculate sample size for margin of error ±5

Repeat the previous process but this time use \(E=5\), thus \(n = \frac{(1.96 \cdot 30)^{2}}{5^{2}}\).

Calculate sample size for margin of error ±1

Again, substitute the values into the equation, this time with \(E=1\), and solve for \(n\), resulting in \(n = \frac{(1.96 \cdot 30)^{2}}{1^{2}}\).

Comment on relationship

After obtaining all results, it can be noticed that the smaller the margin of error, the larger the calculated sample size. This explains that precision comes with the need for a larger sample size as each incremental decrease in margin of error requires an increase in the sample size.