Problem

Overweight Men For a random sample of 70 overweight men, the mean of the number of pounds that they were overwelght was 32 . The standard deviation of the population is 4.3 pounds.
Part 1 of 4
(a) The best point estimate of the mean is 32 pounds.
Part 2 of 4
(b) Find the $90 \%$ confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answers to one decimal place.
\[
31.2< \mu< 32.8
\]

Part 3 of 4 :
(c) Find the $95 \%$ confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answers to one decimal place.
\[
31< \mu< 33
\]

Part: $3 / 4$

Part 4 of 4
(d) Which interval is larger? Why?

The $\square \%$ confidence interval is larger. An interval with a Interval will be more likely to contain the true population

Answer

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Answer

Final Answer: The 90% confidence interval for the mean weight of the overweight men is approximately \(\boxed{(31.2, 32.8)}\) pounds and the 95% confidence interval is approximately \(\boxed{(31, 33)}\) pounds. The \(\boxed{95\%}\) confidence interval is larger.

Steps

Step 1 :The problem provides us with a sample of 70 overweight men, with a mean weight over the normal limit of 32 pounds and a standard deviation of 4.3 pounds.

Step 2 :We are asked to find the 90% and 95% confidence intervals for the mean weight of these men. The formula for a confidence interval is \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z\) is the Z-score (which depends on the confidence level), \(\sigma\) is the standard deviation of the population, and \(n\) is the sample size.

Step 3 :For a 90% confidence level, the Z-score is approximately 1.645. Substituting the given values into the formula, we get the 90% confidence interval as approximately \((31.2, 32.8)\) pounds.

Step 4 :For a 95% confidence level, the Z-score is approximately 1.96. Substituting the given values into the formula, we get the 95% confidence interval as approximately \((31, 33)\) pounds.

Step 5 :The problem also asks us to compare the two intervals. The larger the confidence level, the larger the interval. This is because a larger confidence level means that we are more certain that the true population mean lies within the interval, so we need a larger interval to be more certain. Therefore, the 95% confidence interval is larger than the 90% confidence interval.

Step 6 :Final Answer: The 90% confidence interval for the mean weight of the overweight men is approximately \(\boxed{(31.2, 32.8)}\) pounds and the 95% confidence interval is approximately \(\boxed{(31, 33)}\) pounds. The \(\boxed{95\%}\) confidence interval is larger.

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