To determine the margin of error for the population mean, we will follow these steps:
1. Identify the given information:
- Sample size, [tex]\( n = 35 \)[/tex]
- Population mean, [tex]\( \mu = 50 \)[/tex] days
- Population standard deviation, [tex]\( \sigma = 2 \)[/tex] days
- [tex]\(95\% \)[/tex] confidence level, which corresponds to a [tex]\( z \)[/tex] score of [tex]\( 1.96 \)[/tex]
2. Understand the formula for margin of error (ME):
[tex]\[
ME = \frac{z \cdot \sigma}{\sqrt{n}}
\][/tex]
where:
- [tex]\( z \)[/tex] is the z-score associated with the confidence level,
- [tex]\( \sigma \)[/tex] is the population standard deviation,
- [tex]\( n \)[/tex] is the sample size.
3. Substitute the given values into the formula:
[tex]\[
ME = \frac{1.96 \cdot 2}{\sqrt{35}}
\][/tex]
4. Calculate the margin of error:
- First, calculate the square root of [tex]\( n \)[/tex]:
[tex]\[
\sqrt{35} \approx 5.92
\][/tex]
- Then, divide the standard deviation by the square root of the sample size:
[tex]\[
\frac{2}{5.92} \approx 0.3385
\][/tex]
- Finally, multiply by the z-score:
[tex]\[
1.96 \cdot 0.3385 \approx 0.6626
\][/tex]
Thus, the margin of error (ME) for the population mean is approximately [tex]\( 0.6626 \)[/tex].
Given the answer choices:
- [tex]\( 0.06 \)[/tex]
- [tex]\( 0.11 \)[/tex]
- [tex]\( 0.34 \)[/tex]
- [tex]\( 0.66 \)[/tex]
The correct answer is [tex]\( 0.66 \)[/tex].
