The times it took for 35 loggerhead sea turtle eggs to hatch in a simple random sample are normally distributed, with a mean of 50 days and a standard deviation of 2 days. Assuming a [tex]95 \%[/tex] confidence level (z score of 1.96), what is the margin of error for the population mean?

Remember, the margin of error, [tex]ME[/tex], can be determined using the formula [tex]ME = \frac{z \cdot S}{\sqrt{n}}[/tex].

A. 0.06
B. 0.11
C. 0.34
D. 0.66



Answer :

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].