Question

Which of the following would result in the narrowest​ (most informative) confidence interval for the population​ proportion, if everything else were held​ constant? Using a sample size of 1000 and creating a​ 95% confidence interval. Using a sample size of 1000 and creating a​ 99% confidence interval. Using a sample size of 2000 and creating a​ 95% confidence interval. Using a sample size of 2000 and creating a​ 99% confidence interval. We need to know the sample proportion to determine the standard error for the confidence interval and answer this question.

Answer 1

The narrowest confidence interval is obtained with a sample size of 2000 and a 95% confidence interval

Explanation:

The confidence interval is the interval

`\bf [\bar x -z(s)/(√(n)),\bar x +z(s)/(√(n))]`

where

`\bf \bar x`= the mean of the sample

s = the standard deviation of the sample

n = the size of the sample

z is the z-score corresponding to the level of confidence

In any case, the width of the interval is 2 times
`\bf z(s)/(√(n))`, so we only need to see which value of
`\bf (z)/(√(n))` is the smallest between the different options.

The values of z can be computed either with tables or a spreadsheet.

Sample size of 1000 and a 95% confidence interval

`\bf (z)/(√(n))=(1.96)/(31.6228)=0.061`

Sample size of 1000 and a 99% confidence interval

`\bf (z)/(√(n))=(2.576)/(31.6228)=0.0815`

Sample size of 2000 and a 95% confidence interval

`\bf (z)/(√(n))=(1.96)/(44.7214)=0.0438`

Sample size of 2000 and a 99% confidence interval

`\bf (z)/(√(n))=(2.576)/(44.7214)=0.0576`

And we see that the narrowest confidence interval is obtained with a sample size of 2000 and a 95% confidence interval