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.