Answer :
To construct a 90% confidence interval for the mean weight of the population of American men, we will use the formula for a confidence interval for a sample mean:
[tex]CI = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right)[/tex]
Where:
- [tex]\bar{x}[/tex] is the sample mean.
- [tex]z[/tex] is the z-score corresponding to the confidence level.
- [tex]s[/tex] is the sample standard deviation.
- [tex]n[/tex] is the sample size.
Given:
- Sample mean ([tex]\bar{x}[/tex]) = 182 lbs.
- Sample standard deviation ([tex]s[/tex]) = 18.4 lbs.
- Sample size ([tex]n[/tex]) = 40
For a 90% confidence level, the critical z-score ([tex]z[/tex]) is approximately 1.645. This value can be found using a standard normal distribution table or z-score calculator.
Now plug these values into the formula:
[tex]CI = 182 \pm 1.645 \left( \frac{18.4}{\sqrt{40}} \right)[/tex]
First, calculate the standard error:
[tex]\frac{18.4}{\sqrt{40}} = \frac{18.4}{6.3246} \approx 2.91[/tex]
Next, calculate the margin of error:
[tex]1.645 \times 2.91 \approx 4.786[/tex]
Finally, calculate the confidence interval:
[tex]CI = 182 \pm 4.786[/tex]
[tex]CI = (182 - 4.786, 182 + 4.786)[/tex]
[tex]CI = (177.214, 186.786)[/tex]
Rounding to one decimal place, the confidence interval is:
[tex]CI = (177.2, 186.8)[/tex]
Therefore, the correct answer is Option A (177.2, 186.8). This means we are 90% confident that the true mean weight of the population of American men falls within this interval.
