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%

Solution:

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

=

P= Sample Proportion

N =Sample Size

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

Sample Population= =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=

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.

sorry i just need to anwers this to get points

step-by-step explanation: