To find the 95% confidence interval for the population mean (\( \mu \)), we can use the formula for the confidence interval for the mean when the population standard deviation is unknown and the sample size is small. We will use the sample mean (\( \bar{x} \)), the sample standard deviation (\( s \)), and the critical value from the t-distribution.
Given:
- Sample size (\( n \)) = 37
- Sample mean (\( \bar{x} \)) = 17.0 pounds
- Sample standard deviation (\( s \)) = 3.3 pounds
- Confidence level = 95%
Steps to calculate the confidence interval:
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Find the critical value (\( t^ \))*: Since \( n = 37 \), the degrees of freedom (\( df \)) = \( n - 1 = 36 \). Using a t-table or calculator, we find the critical value \( t^* \) for 95% confidence and 36 degrees of freedom.
The critical value \( t^* \) for 95% confidence and 36 degrees of freedom is approximately 2.030 (you can confirm this in a t-distribution table).
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Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{3.3}{\sqrt{37}} \approx 0.542 \]
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Calculate the margin of error (ME): \[ ME = t^* \cdot SE = 2.030 \cdot 0.542 \approx 1.099 \]
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Calculate the confidence interval: \[ \text{Lower limit} = \bar{x} - ME = 17.0 - 1.099 \approx 15.901 \] \[ \text{Upper limit} = \bar{x} + ME = 17.0 + 1.099 \approx 18.099 \]
Final Confidence Interval:
Rounding the lower and upper limits to the same number of decimal places as the sample mean (which has one decimal place):
\[ 15.9 \text{ lb} < \mu < 18.1 \text{ lb} \]
So, the correct answer is B) 15.9 lb < μ < 18.1 lb.
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