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- Body fat Do the data shown in the table below indicatean association between Waist size and %Body Fat?a) Test an appropriate hypothesis and state your conclusion. *b) Give a 95% confidence interval for the mean %BodyFat found in people with 40-inch Waists. Waist(in.)Weight(lb)Body Fat(%)Waist(in.)Weight(lb)Body Fat(%)32 175 6 33 188 1036 181 21 40 240 2038 200 15 36 175 2233 159 6 32 168 939 196 22 44 246 3840 192 31 33 160 1041 205 32 41 215 2735 173 21 34 159 1238 187 25 34 146 1038 188 30 44 219 28Independent random samples were selected from two quantitative populations, with sample sizes, means, and variances given below. Population Sample Size Sample Mean Sample Variance 1 49 10.3 9.64 2 58 8.5 15.42 A 90% confidence interval for ₁-₂ is 0.68 to 2.92 and a 99% confidence interval for ₁-₂ is 0.05 to 3.55. Use these confidence intervals to answer the questions. Can you conclude with 99% confidence that there is a difference in the means for the two populations? Ⓒ Yes. The value #₁ - ₂2=0 is not in the interval which suggests that there is likely a difference between ₁ and 1₂. ₂ = 0 is not in the interval which does not suggest that there is a difference between and H1 #₂. O No. The value μ₁ - O No. The value μ₁ - ₂ = 0 is in the interval which does not suggest that there is a difference between and M₂ 1₂" and H₂ H₂² Yes. The value and H₂ H₂" = 0 is in the interval which suggests that there is likely a difference between M₁ - H₂ Can you conclude with 99% confidence that there is…#3 Use the given confidence interval to find the margin of error and the sample proportion. (0.629,0.659) E= (Type an integer or a decimal.) p%3(Type an integer or a decimal.) Type here to search ■ 直 0 F4 F5 F7 2 4.
- A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week. Assume the population is normally distributed. (Give your answer to 3 decimal places if necessary.) 20 13 1 10 20 0 10 13 16 10 1 17 18 3 9 (?,?)Please let me know if I calculated relative risk correctly. Thank you for your assistance.Answer?
- USE 3 DECIMALS A researcher believes that reading habit of newspaper is decreasing every year. In order to see the true population proportion of people who are reading newspaper s/he collect a randomly selected people on the yearly basis. The following table gives the number of people who do not read newspaper with respect to the years. Construct a 90 % confidence interval for the true population proportion difference of people who are reading newspaper in 2005 and 2010. (P2005 - P2015) 2010 2005 In 1200 1000 Number of people who do not read newspaper 300 400 KOncekib Statistic TInterval Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. Complete parts (a) through (c) (13.046,22.15) X = 17.598 Sx = 16.01712719 n= 50 rses below. e Home a. Express the confidence interval in the format that uses the "less than" symbol. Given that the original listed data use one decimal place, round the confidence interval limits accordingly. uncements past due gnments Mbps < µ< Mbps see score (Round to two decimal places as needed.) dy Plan b. Identify the best point estimate of u and the margin of error. adebook 1 of 1 see score The point estimate of u is Mbps. tatCrunch (Round to two decimal places as needed.) The margin of error is E = Mbps. 1 of 1 see score eText (Round to two decimal places as needed.) Chapter Contents c. In constructing the confidence interval estimate of u, why is it not necessary to confirm that the sample data appear to be from a population with a normal 1 of oo see score distribution? Tools…Boys and girls: The National Health Statistics Reports reported that a sample of 310 one-year-old boys had a mean weight of 25,3 pounds with a standard deviation of 3.3 pounds. In addition, a sample of 286 one-year-old girls had a mean weight of 24.3 pounds with a standard deviation of 3.6 pounds. Part: 0/2 Part 1 of 2 Construct a 98% confidence interval for the difference between the mean weights. Let u, denote the mean weight of one-year-old boys. Use the TI-84 plus calculator and round the answers to one decimal place. A 98% confidence interval for the difference between the mean weights is
- 品2 learn.zybooks.com DZL 4-3 Project One Submission -.. DZL 6-2 Participation Activities: Con... Z 5.2 - MAT 240: Applied Statistics DZL Module Five MAT 240: Applied Statistics home > 6.1: Confidence intervals E zyBooks CHALLENGE 6.1.1: Confidence intervals for population mean. ACTIVITY 389226.2543846.qx3zqy7 Jump to level 1 A clothing retailer is interested in the average waist size of men. A sample is taken with the results given below. Column1 Mean 41.32 Standard Error What is the sample statistic? = 41.32 C 1.447 Median 40 What is the point estimate for the average waist size? 41.32 Mode Standard Deviation 6.307 Sample Variance 39.78 What is the level of confidence as a decimal? Ex0.55 Kurtosis -0.79 Skewness 0.571 What is the margin of error? 2.509 Range Minimum 32 What is the confidence interval? [ Maximum 52 Sum 785 Count Confidence Level(90.0%) 60 LEGO Check Next References (*1) The Media Insight Project. "How Millennials Get News: Inside the Habits of America's First Digital…At what age do infants speak their first word of English? Here are 20 data of children (age in month) 15 26 10 9 15 20 18 11 8 20 7 9 10 11 11 10 12 17 11 10 whAt is 99% confidence interval for the mean age at which children speak first words? Upper limit? lower limit?Summary statistics for Chol: (a) Calculate a 95% confidence interval (Chapter 17) for the mean Cholesterol of heart disease patients. Use the t-distribution table and the Cholesterol summary statistics.Group by: HD HD n Mean Variance Std. dev. Std. err. Median Range Min Max Q1 Q3 No 164 242.64024 2857.606 53.45658 4.1742576 234.5 438 126 564 208.5 267.5 Yes 139 251.47482 2448.9468 49.486835 4.1974185 249 278 131 409 217 284
