Question

On the morning of November 9, 1994-the day after the electoral landslide that had returned Republicans to power in both branches of Congress-several key races were still in doubt. The most prominent was the Washington contest involving Democrat Tom Foley, the reigning speaker of the house. An Associated Press story showed how narrow the margin had become (120): With 99 percent of precincts reporting, Foley trailed Republican challenger George Nethercutt by just 2,174 votes, or 50.6 percent to 49.4 percent. About 14,000 absentee ballots remained uncounted, making the race too close to call. Let p = P(Absentee voter prefers Foley). How small could p have been and still have given Foley a 20% chance of overcoming Nethercutt's lead and winning the election?

Answer 1

p = 0.574197

Explanation:

There are 14000 ballots left.

The lead is 2174, so he needs to lead at least 2175 in these 14000 ballots.

He does so if he gets at least 8088 votes.

If P(x >= 8088) = 0.20, then the corresponding z score to this is, by table/technology,

z = 0.841621234

Now, as

z = (x - u) / s

and

u= n p

s = sqrt(n p(1-p))

Then

0.841621234 = (8088 - 14000*p) / sqrt(14000p(1-p))

Solving for p here,

p = 0.574197