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?
Answer
A
62 plus-or-minus 1.65 StartRoot StartFraction 62 (1 minus 62) Over 100 EndFraction EndRoot
B
62 plus-or-minus 1.96 StartRoot StartFraction 62 (1 minus 62) Over 100 EndFraction EndRoot
C
0.62 plus-or-minus 1.65 StartRoot StartFraction 0.62 (1 minus 0.62) Over 100 EndFraction EndRoot
D
62 plus-or-minus 1.96 StartRoot StartFraction 0.62 (1 minus 0.62) Over 100 EndFraction EndRoot