Question

In a simple random sample of 1300 voters, 19% said they would prefer to vote online.

Which percents are in the 99% confidence interval for the percent of all voters who would prefer to vote online?

Choose all answers that are correct.

19.8%
17.4%
15.5%
14.7%

Answer 1

Formula for calculation of 99% confidence interval is given by:

=
`P \pm (\text{Z value} )\sqrt\frac {P*(1-P)}{N}`

P= Sample Proportion

N =Sample Size

Also,Number of voters who vote online= 19 % of 1300=19 × 13=247

Sample Population=
`=(247)/(1300)`=0.19

Z value for 99% confidence interval=2.58

The number of voters who vote online, who are in the 99% confidence interval for the percent of all voters who would prefer to vote online=
`0.19 \pm (2.58)\sqrt(0.19 * (1-0.19))/(1300)\\\\ =0.19 + (2.58)\sqrt(0.19 * (1-0.19))/(1300) {\text{or}} 0.19 - (2.58)\sqrt(0.19 * (1-0.19))/(1300)\\\\ 0.2180 {\text{or}} 0.1619`

So, Percent of voters , who are in the 99% confidence interval for the percent of all voters who would prefer to vote online= 100 × 0.2180 or 100 × 0.1619=21.80 % or 16.19 %

That is between, 16.19 % and 21.80 %.

→→Option (A) 19.8 % and Option (B)17.4 % are in 99 % confidence intervals.