Answers to Homework 10

Final answers in bold and red.

Here is the data for the combined small samples of m&m’s in both classes on Monday. Round all answers in this part to the nearest tenth of a percent. The formula for the margin of error for 95% is 1.96*sqrt(p-hat*q-hat/n) and the endpoints of the 95% confidence interval are p-hat-MoE95% and p-hat+MoE95% .

n = 580 f(red) = 72 f(blue) = 134

p-hat(red) = 72/580 = 12.4% p-hat(blue) = 134/580 = 23.1%

MoE95%(red) = 1.96*sqrt(.124*.876/580) = .0268... = 2.7%

MoE95%(blue) = 1.96*sqrt(.231*.769/580) = .0343... = 3.4%

We are 95% confident the true proportion of blue milk chocolate m&m’s currently being

produced in New Jersey is between (23.1+3.4)% = 26.5% and (23.1-3.4)% = 19.7%.

The formula for finding the minimum n for a particular margin of error given a 95% confidence interval is n >= 1.96^2*p-hat*q-hat/MoE95%^2 , where n should be rounded up to the next highest integer if the formula produces a non integer answer.

If we want a MoE95% of 2.5% and we can assume p-hat = .5 , what is the minimum n?

Answer: n >= 1.96^2*.5*.5/.025^2 = 1536.64, so n >= 1537

If we want a MoE95% of 2.5% and we can assume p-hat = .55 , what is the minimum n?

Answer: n >= 1.96^2*.55*.45/.025^2 = 1521.2736, so n >= 1522

In the last poll before the 2008 election, Obama lead McCain 51% to 47% in a sample of 600 voters in Virginia. Since 51% + 47% = 98%, we can use the confidence of victory method. Round the proportions and standard deviation to the nearest tenth of a percent. Round the confidence of victory number to the nearest percent. The z-score formula is (p-hat(leader) - .5)/sp-hat.

f(Obama) = .51*600 = 306 f(McCain) = .47*600 = 282 new n = 306 + 282 = 588


new p-hat(Obama) = 306/588 = .5204 = 52.0%

new z(Obama) = .520/sqrt(.520*.480/588) = .97

Finish this sentence about confidence of victory.

Given this sample, if the election were held when the poll was taken, we are 83% confident the underlying population would favor Obama and he would win the election in Virginia.