High School

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Suppose a random sample of 40 American men is drawn and the weights of the men measured. If the sample mean is 182 lbs. and sample standard deviation is 18.4 lbs., construct a 90% confidence interval for the mean weight of the population of American men. Round each of the endpoints of your interval to one decimal place.

Option A (177.2, 186.8)
Option B none of the others
Option C (175.2, 189.2)
Option D (176.2, 184.6)
Option E (172.4, 188.6)

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.

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