## Question 864:

1

No answer provided yet.1 . A 99% confidence interval will be wider. As the level of confidence increase, the interval gets wider when the sample size stays the same.

2. The confidence interval will be wider. As the sample sizes gets smaller the width of the confidence interval increases.

3.
The margin of error is found as the standard error times the critical value. The standard error is SQRT(pq/n) = SQRT(23*.77/1000) = .0133 and the critical value is 1.96 getting a margin as .0261. Adding and subtracting this from the mean gets a 95% confidence interval between 20.4% and 25.6% will likely vote for an incumbent. The question asks for the margin of error which is 2.6%

4. We need to work backwards from the confidence interval. The formula is (z^2*s^2) /d^2 where z is the critical value of 1.96 s is the standard deviation = SQRT(.23*.77) = .421 and d = the desired margin of error 2.  The formula result is [1.96^2 * .421^2 ]/ 2^2 = 1700.8. Rounding up to the nearest person gets us a result of needing 1701.  The question just asks for additional number needed over the 1000 from the last question, so 701.