Answer :
Final Answer:
The estimated mean number of farms per state with 99% confidence is 38.1. This estimation is based on a given population standard deviation of 31, a critical value of 2.576 for a 99% confidence interval, and the formula for confidence intervals.
So, the correct option is c.
Explanation:
To estimate the mean number of farms per state with 99% confidence, we can use the formula for the confidence interval:
[tex]\[ \text{Confidence Interval} = \text{Sample Mean} ± \left( \frac{Z \cdot \sigma}{\sqrt{n}} \right) \][/tex]
Where:
- Sample Mean is the average number of farms in the sample.
- Z is the critical value for a 99% confidence interval, which is approximately 2.576.
- σ (population standard deviation) is given as 31.
- n is the sample size.
We don't have the sample mean or sample size, but we can calculate the margin of error using the given values. The margin of error is [tex]\(2.576 \cdot \frac{31}{\sqrt{n}}\)[/tex].
To estimate the mean number of farms with 99% confidence, we need to find the value that, when added to and subtracted from the sample mean, gives us the range within which the true population mean lies. The critical value 2.576 corresponds to a 99% confidence level. By rearranging the formula, we can find the sample size required to achieve a specific margin of error. In this case, the margin of error is [tex]\(2.576 \cdot \frac{31}{\sqrt{n}}\)[/tex]. We want to be 99% confident in our estimate, so the critical value is 2.576.
Therefore, the correct option is c.
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Final Answer:
The estimated mean number of farms per state with 99% confidence, assuming a population standard deviation (σ) of 31, is 44.9 .
thus the correct option is D.
Explanation:
To calculate the estimated mean number of farms per state with 99% confidence, we can use the formula for the confidence interval:
[tex]\[ \text{Confidence Interval} = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \][/tex]
Where:
- [tex]\(\bar{x}\)[/tex] is the sample mean.
[tex]- \(Z\)[/tex]is the Z-score corresponding to the desired confidence level.
[tex]- \(\sigma\)[/tex] is the population standard deviation.
[tex]- \(n\)[/tex]is the sample size.
In this case, we want a 99% confidence interval, which corresponds to a Z-score of 2.576 (you can find this value in standard normal distribution tables). We are not given the sample mean, but since we want to estimate the mean, we can use the formula:
[tex]\[ \text{Confidence Interval} = \mu \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \][/tex]
Where:
[tex]- \(\mu\)[/tex] is the population mean (what we want to estimate).
[tex]Z[/tex] is the Z-score.
[tex]- \(\sigma\)[/tex] is the population standard deviation.
[tex]- \(n\)[/tex] is the sample size.
We don't know the sample size [tex](\(n\))[/tex] from the question. However, as the sample size increases, the margin of error decreases, and the estimate becomes more accurate.
Given the choices, the closest estimate to the population mean[tex]\(\mu\)[/tex] is 44.9.
thus the correct option is D.
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