Answer:
The sample size must be 20 so that sample mean will not differ from the population mean by more than 2 units.
Step-by-step explanation:
We are given the following in the question:
Variance = 14
[tex]\sigma^2 = 14\\\sigma = \sqrt{14}[/tex]
We need to form a 9% confidence interval such that sample mean will not differ from the population mean by more than 2 units.
Thus, margin of error for the confidence interval is 2.
Formula for margin of error:
[tex]z_{critical}\times \dfrac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 2.33[/tex]
[tex]2 = 2.33\times \dfrac{\sqrt{14}}{\sqrt{n}}\\\\\sqrt{n} = \dfrac{2.33\times \sqrt{14}}{2}\\\\\sqrt{n}=4.359\\\Rightarrow n = 19.00115\approx 20[/tex]
Thus, the sample size must be 20 so that sample mean will not differ from the population mean by more than 2 units.
