Question

Scores on a certain test are normally distributed with a variance of 14. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 98% confidence that the sample mean will not differ from the population mean by more than 2 units.

Answer 1

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

$\sigma^2 = 14\\\sigma = \sqrt{14}$

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:

$z_{critical}\times \dfrac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 2.33$

$2 = 2.33\times \dfrac{\sqrt{14}}{\sqrt{n}}\\\\\sqrt{n} = \dfrac{2.33\times \sqrt{14}}{2}\\\\\sqrt{n}=4.359\\\Rightarrow n = 19.00115\approx 20$

Thus, the sample size must be 20 so that  sample mean will not differ from the population mean by more than 2 units.