A local school board wants to determine the proportion of households in the district that would support a proposal to start the school year a week earlier. They ask a random sample of 100 households whether they would support the proposal, and 62 households stated that they would. Assuming that conditions have been met, what is the 90% confidence interval for the true proportion of households that would support the proposal?
62 ± 1.65 * sqrt(62(1 - 62) / 100)
62 ± 1.96 * sqrt(62(1 - 62) / 100)
0.62 ± 1.65 * sqrt(0.62(1 - 0.62) / 100)
0.62 ± 1.96 * sqrt(0.62(1 - 0.62) / 100)