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Overweight Men For a random sample of 60 overweight men, the mean of the number of pounds that they were overweight was 29 . The standard deviation o the population is 3.9 pounds. Part 1 of 4 (a) The best point estimate of the mean is 29 pounds. Part: 1 / 4 Part 2 of 4 (b) 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. ( 28.0 < \( \mu<30.1 \)

Ask by Harris Bob. in the United States
Mar 11,2025

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The \(95\%\) confidence interval for the mean number of pounds overweight is \(28.0 < \mu < 30.0\).

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We are given: - Sample size: \( n = 60 \) - Sample mean: \( \bar{x} = 29 \) pounds - Known population standard deviation: \( \sigma = 3.9 \) pounds - Confidence level: \( 95\% \) **Step 1. Find the critical value** For a \( 95\% \) confidence interval with a known population standard deviation, we use the standard normal distribution. The critical value corresponding to \( \alpha/2 = 0.025 \) is \[ z_{\alpha/2} = 1.96. \] **Step 2. Compute the standard error** The standard error of the mean is computed by: \[ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{3.9}{\sqrt{60}}. \] Calculate \(\sqrt{60}\): \[ \sqrt{60} \approx 7.746. \] Thus, \[ \text{SE} \approx \frac{3.9}{7.746} \approx 0.503. \] **Step 3. Calculate the margin of error** The margin of error (ME) is: \[ \text{ME} = z_{\alpha/2}\times \text{SE} = 1.96 \times 0.503 \approx 0.986. \] **Step 4. Construct the confidence interval** The confidence interval is given by: \[ \bar{x} \pm \text{ME} = 29 \pm 0.986. \] Thus, the lower limit is: \[ 29 - 0.986 \approx 28.0, \] and the upper limit is: \[ 29 + 0.986 \approx 30.0. \] When rounding to one decimal place, we obtain: \[ 28.0 < \mu < 30.0. \] Notice that the problem statement provided an interval of \(28.0 < \mu < 30.1\), which might be due to rounding differences in intermediate steps. Using our process and rounding the margin of error to three decimal places before calculating (i.e. \(0.986\)), the final interval is approximately: \[ \textbf{Confidence Interval: } (28.0, \, 30.0). \]

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To calculate the 95% confidence interval for the mean, we can use the formula: \[ \text{CI} = \bar{x} \pm z \left(\frac{\sigma}{\sqrt{n}}\right) \] Where: - \(\bar{x} = 29\) (sample mean) - \(\sigma = 3.9\) (population standard deviation) - \(n = 60\) (sample size) - \(z\) is the z-value for a 95% confidence level, which is approximately 1.96. Calculating the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3.9}{\sqrt{60}} \approx 0.503 \] Now, calculating the margin of error (ME): \[ ME = z \times SE \approx 1.96 \times 0.503 \approx 0.986 \] Finally, the confidence interval will be: \[ \text{CI} = 29 \pm 0.986 \] Which results in: \[ (28.0, 30.0) \] So, the 95% confidence interval for the mean number of pounds overweight is approximately \( (28.0, 30.0) \). --- Historical Background: The concept of confidence intervals dates back to the 1930s, a brainchild of statisticians like Jerzy Neyman. Neyman introduced the idea to give researchers a way to express the uncertainty of their estimates instead of just providing a single number. This fundamental shift allowed for a more robust understanding of data, essentially shaping modern statistical analysis. Real-World Application: In healthcare and fitness research, confidence intervals help practitioners evaluate the effectiveness of diets or exercise programs. For instance, knowing the average weight loss with a diet plan might come with a confidence interval, allowing nutritionists to gauge how reliable the results are. This insight can guide individuals in making informed decisions about their health journey.

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