Question

Consider taking a random sample from a population with \(p=0.40\) a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be larger for samples of size 100 or samples of size \(200 ?\) c. If the sample size were doubled from 100 to 200 , by what factor would the standard error of \(\hat{p}\) decrease?

Short answer

a. The standard error of \(\hat{p}\) for random samples of size 100 is 0.049 b. The standard error of \(\hat{p}\) would be smaller for samples of size 200. c. The standard error of \(\hat{p}\) would decrease by a factor of \(\sqrt{2}\) when the sample size is doubled from 100 to 200.

Step-by-step solution

Calculating the Standard Error

To calculate the standard error of \(\hat{p}\) for random samples of size 100, the following formula is applied:\[SE=\sqrt{\frac{p(1-p)}{n}}\]where \(p\) is the population proportion and \(n\) is the sample size. Substituting \(p=0.40\) and \(n=100\), we get \[SE=\sqrt{\frac{0.40(1-0.40)}{100}}\]

Comparing Standard Errors

The standard error of \(\hat{p}\) would be smaller for larger samples \(n\). This is because the standard error formula has \(n\) (sample size) in the denominator. Hence, an increase in the sample size will decrease the standard error, thus the standard error for a sample size of 200 would be smaller than for a sample size of 100.

Determining the Factor of Decrease

To find the decrease factor of SE when the sample size is doubled (from 100 to 200), we consider that the standard error is inversely proportional to the square root of the sample size. So, if the sample size doubles, the standard error will decrease by a factor of \(\sqrt{2}\).