To determine the confidence interval in which Jeremy can be 99.7% sure that the sample mean will lie, we follow these steps:
1. Identify the given values:
- Population mean ([tex]\(\mu\)[/tex]) = 112.5 ounces
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 37.5 ounces
- Sample size ([tex]\(n\)[/tex]) = 96
2. Calculate the standard error of the mean ([tex]\(SE\)[/tex]):
The standard error is calculated using the formula:
[tex]\[
SE = \frac{\sigma}{\sqrt{n}}
\][/tex]
Substituting the given values,
[tex]\[
SE = \frac{37.5}{\sqrt{96}} \approx 3.83 \text{ ounces}
\][/tex]
3. Determine the z-score for a 99.7% confidence interval:
A 99.7% confidence interval corresponds to roughly 3 standard deviations from the mean in a standard normal distribution. Therefore, the [tex]\(z\)[/tex]-score is 3.
4. Calculate the margin of error (ME):
The margin of error is calculated using the formula:
[tex]\[
ME = z \times SE
\][/tex]
Substituting the previously calculated SE and the z-score,
[tex]\[
ME = 3 \times 3.83 \approx 11.48 \text{ ounces}
\][/tex]
5. Calculate the confidence interval:
The confidence interval is given by:
[tex]\[
\text{Confidence Interval} = \mu \pm ME
\][/tex]
Substituting the population mean and the margin of error,
[tex]\[
\text{Lower bound} = 112.5 - 11.48 \approx 101.02 \text{ ounces}
\][/tex]
[tex]\[
\text{Upper bound} = 112.5 + 11.48 \approx 123.98 \text{ ounces}
\][/tex]
So, the interval in which Jeremy can be 99.7% sure that the sample mean will lie is approximately between 101.02 ounces and 123.98 ounces.
