Answer:
The narrowest confidence interval is obtained with a sample size of 2000 and a 95% confidence interval
Step-by-step explanation:
The confidence interval is the interval
[tex]\bf [\bar x -z\frac{s}{\sqrt{n}},\bar x +z\frac{s}{\sqrt{n}}][/tex]
where
[tex]\bf \bar x[/tex]= 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 [tex]\bf z\frac{s}{\sqrt{n}}[/tex], so we only need to see which value of [tex]\bf \frac{z}{\sqrt{n}}[/tex] 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
[tex]\bf \frac{z}{\sqrt{n}}=\frac{1.96}{31.6228}=0.061[/tex]
Sample size of 1000 and a 99% confidence interval
[tex]\bf \frac{z}{\sqrt{n}}=\frac{2.576}{31.6228}=0.0815[/tex]
Sample size of 2000 and a 95% confidence interval
[tex]\bf \frac{z}{\sqrt{n}}=\frac{1.96}{44.7214}=0.0438[/tex]
Sample size of 2000 and a 99% confidence interval
[tex]\bf \frac{z}{\sqrt{n}}=\frac{2.576}{44.7214}=0.0576[/tex]
And we see that the narrowest confidence interval is obtained with a sample size of 2000 and a 95% confidence interval
