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Answer :
To construct a 90% confidence interval for the mean weight of the population of American men based on the given sample data, we'll use the formula for a confidence interval for a population mean when the sample size is large enough (n ≥ 30), using the sample mean, sample standard deviation, and the standard normal distribution (Z-distribution).
Here are the step-by-step calculations:
Identify the Sample Statistics:
- Sample mean ([tex]\bar{x}[/tex]): 182 lbs
- Sample standard deviation (s): 18.4 lbs
- Sample size (n): 40
Determine the Confidence Level and Z-Score:
- We're constructing a 90% confidence interval.
- For a 90% confidence level, the Z-score is approximately 1.645 (from standard Z-tables).
Calculate the Standard Error (SE):
[tex]SE = \frac{s}{\sqrt{n}} = \frac{18.4}{\sqrt{40}} \approx 2.91[/tex]Compute the Margin of Error (ME):
[tex]ME = Z \times SE = 1.645 \times 2.91 \approx 4.79[/tex]Construct the Confidence Interval:
- Lower bound: [tex]\bar{x} - ME = 182 - 4.79 \approx 177.2[/tex]
- Upper bound: [tex]\bar{x} + ME = 182 + 4.79 \approx 186.8[/tex]
Therefore, the 90% confidence interval for the mean weight of the population of American men is approximately [tex](177.2, 186.8)[/tex].
The correct multiple-choice option is Option D (177.2, 186.8).
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