Question

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

Answer 1

Solution:

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

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

P= Sample Proportion

N =Sample Size

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

Sample Population= $=\frac{247}{1300}$=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= $0.19 \pm (2.58)\sqrt\frac{0.19 \times (1-0.19)}{1300}\\\\ =0.19 + (2.58)\sqrt\frac{0.19 \times (1-0.19)}{1300} {\text{or}} 0.19 - (2.58)\sqrt\frac{0.19 \times (1-0.19)}{1300}\\\\ 0.2180 {\text{or}} 0.1619$

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.