In the last lecture, we worked on this sample problem:
In a recent poll of 200 households, it was found that 152 households had at least one computer. Estimate the proportion of households in the population that have at least one computer. Construct a 95% confidence interval to estimate the population proportion.
We found that we were 95% confident that the proportion of households in the population with at least one computer was between .701 and .819.
This is a point estimate of 0.76 with a margin of error of 0.059.
What size sample would I need to change the margin of error from 0.059 to 0.030 in a 95% confidence interval?
With a prior estimate, we would need 779 people in our sample. Without a prior estimate, we would need 1068 people in our sample!