# Election and Voters

• April 13th 2013, 08:08 AM
Vinod
Election and Voters
In a certain constituency, the organizer for the ruling party believe that 70% of the local electorates will support the ruling party candidate at the forthcoming election. A random sample of 1000 voters is taken and it is ascertained that 333 voters support the opposition party candidate. Is this result consistent with the original estimate of the organizer for the ruling party?ANSWER With 99% level of confidence, this result is consistent with the estimate of the organizer for the ruling party. Z=-2.277What's your opinion?
• April 13th 2013, 08:32 AM
HallsofIvy
Re: Election and Voters
Quote:

Originally Posted by Vinod
In a certain constituency, the organizer for the ruling party believe that 70% of the local electorates will support the ruling party candidate at the forthcoming election. A random sample of 1000 voters is taken and it is ascertained that 333 voters support the opposition party candidate. Is this result consistent with the original estimate of the organizer for the ruling party?ANSWER With 99% level of confidence, this result is consistent with the estimate of the organizer for the ruling party. Z=-2.277What's your opinion?

More importantly, what's your opinion? Have you made no attempt to do this yourself? This is a binomial distribution in which it is believe that the 70% are for and 30% against. Since 1000 is a pretty large sample, a good approximation is the normal distribution with mean .7(1000)= 700 and standard deviation $\sqrt{1000(.7)(.3)}= 14.5$. Does 333 lie within .01 of the mean?
• April 14th 2013, 06:14 AM
Vinod
Re: Election and Voters
Standard error of proportion is square root of PQ/n where P is population proportion supporting ruling party candidate 0.7. So Q or (1-P) is 0.3. And n is 1000