To determine the sample size needed to provide a 95% confidence interval with a margin of error of 6, given that the population standard deviation is 30, follow these steps:
1. Identify key values:
- Confidence level ([tex]\( \alpha \)[/tex]) = 0.95
- Margin of error (E) = 6
- Population standard deviation ([tex]\( \sigma \)[/tex]) = 30
2. Determine the critical value for a 95% confidence level:
For a 95% confidence level, the critical value (z*) is the z-score that corresponds to the middle 95% of the standard normal distribution. This leaves 2.5% in each tail of the distribution.
Using standard z-tables or normal distribution tables, the z-value corresponding to a cumulative probability of 0.975 is approximately 1.96.
3. Use the formula for the sample size:
The sample size [tex]\( n \)[/tex] needed for a given margin of error can be found using the formula:
[tex]\[
n = \left( \frac{z^* \cdot \sigma}{E} \right)^2
\][/tex]
Plug in the values:
[tex]\[
n = \left( \frac{1.96 \cdot 30}{6} \right)^2
\][/tex]
4. Calculate the sample size:
[tex]\[
n = \left( \frac{58.8}{6} \right)^2
\][/tex]
Simplify the fraction:
[tex]\[
n = \left( 9.8 \right)^2
\][/tex]
Finally, square the result:
[tex]\[
n = 96.04
\][/tex]
5. Round to the next whole number:
Since sample size cannot be a fraction, round 96.04 to the next whole number.
[tex]\[
n = 97
\][/tex]
Therefore, to achieve a 95% confidence interval with a margin of error of 6, given that the population standard deviation is 30, you should select a sample size of 97.
