**Table: Statistical Analysis of Male Waist Size**- **Mean:** 41.74- **Standard Error:** 1.406- **Median:** 39- **Mode:** 39- **Standard Deviation:** 6.127- **Sample Variance:** 37.54- **Kurtosis:** -0.86- **Skewness:** 0.582- **Range:** 20- **Minimum:** 32- **Maximum:** 52- **Sum:** 793- **Count:** 19- **Confidence Level (99.0%):** 4.046**Questions:**1. **What is the sample statistic?** \( \bar{x} = \) [Your Answer Here]2. **What is the point estimate for the average waist size?** [Your Answer Here]3. **What is the level of confidence as a decimal?** Ex: 0.554. **What is the margin of error?** Ex: 1.2345. **What is the confidence interval?** \([ \, \, , \, \, ]\)This detailed table and accompanying questions facilitate the understanding of key statistical measures and concepts, enhancing learning about sample statistics, confidence levels, and intervals.

Suppose that µ is the population average waist size of men.

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The gas mileages (in miles per gallon) of 23 randomly selected sports cars are listed in the accompanying table. Assume the mileages are not normally distributed. Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. Click the icon to view the sports car gas mileages. gas mileages :2131172120241824252020292625221621252320222119 Let σ be the population standard deviation and let n be the sample size. Which distribution should be used to construct the confidence interval? ▼ be used to construct the confidence interval, since
Suppose you have two samples with the following values: Sample 1: 2, 8, 6, 5, 4, 7, 5, 9, 6 Sample 2: 3, 5, 4, 8, 7, 5, 5, 7,6 For Sample 1, what are the values for: Mean Median Mode Standard Deviation Variance
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Low-fat or low-carb? Are low-fat diets or low-carb diets more effective for weight loss? A sample of 77 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.7 kilograms with a sample standard deviation of 7.16 kilograms. A second sample of 79 subjects went on a low-fat diet. Their sample mean weight loss was 2.6 kilograms with a standard deviation of 5.90 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let 4, denote the mean weight lost on the low-carb diet and u, denote the mean weight lost on the low-fat diet. Use the a = 0.01 level and the TI-84 Plus calculator. Compute the -value. Round the answer to four decimal places. Determine whether to reject H0 State a conclusion
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The state of Connecticut claims that the average time on death row is 15 years. A random survey of 75 death row inmates revealed that the average length of time on death row is 17.8 years with a standard deviation of 6.8 years. Conduct a hypothesis to test the state of Connecticut's claim. Use 5% significance level. What type of test should be run? t-test of a mean z-test of a proportion The alternative hypothesis indicates a right-tailed test left-tailed test two-tailed test Calculate the p-value. Round to 4 decimal places. What is the decision? We fail to reject the claim that the average time on death row is 15 years We reject the claim that the average time on death row is 15 years
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