Saturday, October 9, 2010

Practice problems for confidence of victory and confidence intervals

Links to earlier posts about confidence of victory.

Data from recent polls.

Boxer vs. Fiorina U.S. Senate (CA)
Date: 10/2
Boxer: 49%
Fiorina: 44%
n = 448

Brown vs. Whitman Governor (CA)
Brown: 50%
Whitman: 43%
n = 448

For both of these polls:
1) Find the 95% confidence interval for both candidates
2) Since the two top candidate poll over 90% total, do the confidence of victory, rounding to the nearest 5% if the value is under 90% and to the nearest 1% if the value of over 90%.

Answers in the comments.

1 comment:

Prof. Hubbard said...

1) 95% confidence interval

Boxer: 1.96*sqrt(.49*.51/448) = 4.6%
We are 95% confident Boxer's support at the time of the poll was between 44.4% and 53.6% of the likely voters.

Fiorina: 1.96*sqrt(.44*.56/448) = 4.6%
We are 95% confident Fiorina's support at the time of the poll was between 39.4% and 48.6% of the likely voters.

Brown: 1.96*sqrt(.50*.50/448) = 4.6%
We are 95% confident Brown's support at the time of the poll was between 45.4% and 54.6% of the likely voters.

Whitman: 1.96*sqrt(.43*.57/448) = 4.6%
We are 95% confident Whitman's support at the time of the poll was between 38.4% and 47.6% of the likely voters.

Confidence of victory:
f(Boxer) = .49*448 =~ 220
f(Fiorina) = .44*448 =~ 197

new n = 220+197 = 417

p-hat(Boxer) = 220/417 = 52.8%

s_p-hat = sqrt(.528*.472/417) = .0244

z(Boxer) = (.528 - .5)/.0244 = 1.15

z = 1.15 corresponds to .8749, so it barely rounds to 85% as the nearest 5%

If the election were held when the poll was taken, we are 85% confident the lead Boxer shows in the poll is due to her being favored by the underlying population of likely California voters and she would win.


f(Brown) = .50*448 = 224
f(Whitman) = .43*448 =~ 193

new n = 224+193 = 417

p-hat(Brown) = 224/417 = 53.7%

s_p-hat = sqrt(.537*.463/417) = .0244

z(Brown) = (.537 - .5)/.0244 = 1.52

z = 1.52 corresponds to .9357, so it rounds to 94%, since we round to the nearest percent when over 90%.

If the election were held when the poll was taken, we are 94% confident the lead Brown shows in the poll is due to him being favored by the underlying population of likely California voters and he would win.