Question 9 of 14, Step 1 of 1 A research company desires to know the mean consumption of meat per week among people over age 24. They believe that the meat consumption has a mean of 4.4 pounds, and want to construct a $$ 90 \% $$ confidence interval with a maximum error of 0.08 pounds. Assuming a variance of 0.36 pounds, what is the minimum number of people over age 24 they must include in their sample? Round your answer un to the next integer.

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UpStudy AI · Accepted Answer

We know that the margin of error $$ E $$ for a confidence interval is given by
$$
E = z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}},
$$
where
- $$ z_{\frac{\alpha}{2}} $$ is the z-score corresponding to a $$ 90\% $$ confidence level,
- $$\sigma$$ is the standard deviation,
- $$ n $$ is the sample size.

Given:
- $$ E = 0.08 $$,
- $$\sigma^2 = 0.36 $$ so that $$\sigma = \sqrt{0.36} = 0.6$$,
- For a $$ 90\% $$ confidence level, $$\alpha = 0.10$$ and hence $$\frac{\alpha}{2} = 0.05$$. The critical value is $$ z_{0.05} \approx 1.645 $$.

Substitute the values into the formula:
$$
0.08 = 1.645 \frac{0.6}{\sqrt{n}}.
$$

Solve for $$ n $$:
$$
\sqrt{n} = 1.645 \frac{0.6}{0.08}.
$$
First, calculate the numerator:
$$
1.645 \times 0.6 = 0.987.
$$
Then,
$$
\sqrt{n} = \frac{0.987}{0.08} \approx 12.3375.
$$

Square both sides:
$$
n \approx (12.3375)^2 \approx 152.1539.
$$

Since we must round up to the next whole number, the minimum sample size is
$$
n = 153.
$$
The minimum number of people over age 24 that must be included in the sample is 153.

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