4/8/2017 eBook with Connect Math Exercises 7-4 1. What distribution must be used when computing confidence intervals for variances and standard deviations? 2. What assumption must be made when computing confidence intervals for variances and standard deviations? 3. Using Table G, find the values for 2left and 2right. 1. = 0.05, n = 12 2. = 0.10, n = 20 3. = 0.05, n = 27 4. = 0.01, n = 6 5. = 0.10, n = 41 4. Lifetimes of Wristwatches Find the 90% confidence interval for the variance and standard deviation for the lifetimes of inexpensive wristwatches if a random sample of 24 watches has a standard deviation of 4.8 months. Assume the variable is normally distributed. Do you feel that the lifetimes are relatively consistent? 5. Carbohydrates in Yogurt The number of carbohydrates (in grams) per 8ounce serving of yogurt for each of a random selection of brands is listed below. Estimate the true population variance and standard deviation for the number of carbohydrates per 8ounce serving of yogurt with 95% confidence. Assume the variable is normally distributed. 174241203941351543 2538334223172534 6. Carbon Monoxide Deaths A study of generationrelated carbon monoxide deaths showed that a random sample of 6 recent years had a standard deviation of 4.1 deaths per year. Find the 99% confidence interval of the variance and standard deviation. Assume the variable is normally distributed. Source: Based on information from Consumer Protection Safety Commission. 7. Cost of Knee Replacement Surgery U.S. insurers' costs for knee replacement surgery range from $17,627 to $25,462. Estimate the population variance (standard deviation) in cost with 98% confidence based on a random sample of 10 persons who have had this surgery. The retail costs (for uninsured persons) for the same procedure range from $40,640 to $58,702. Estimate the population variance and standard deviation in cost with 98% confidence based on a sample of 10 persons, and compare your two intervals. Assume the variable is normally distributed. Source: Time Almanac. 8. Age of College Students Find the 90% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park College if a random sample of 24 students has a standard deviation of 2.3 years. Assume the variable is normally distributed. 9. NewCar Lease Fees A newcar dealer is leasing various brandnew randomly selected models for the monthly rates (in dollars) listed below. Estimate the true population variance (and standard deviation) in leasing rates with 90% confidence. Assume the variable is normally distributed. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4SqlWYTNf_6Sy8q?1oiPvmAdTNWxRtu... 8/15 4/8/2017 eBook with Connect Math Elementary Statistics: A StepByStep Approach, 9th Ed. Bluman O r der Pr inted Ver s ion jump to pg search ebook Chapter 7. Confidence Intervals and Sample Size Chapter 7 Sections Chapter 7. Introduction 7-1. Confidence Intervals for the Mean When Is Known 7-2. Confidence Intervals for the Mean When Is Unknown 7-3. Confidence Intervals and Sample Size for Proportions 7-4. Confidence Intervals for Variances and Standard Deviations Chapter 7 Ending 7-4 Confidence Intervals for Variances and Standard Deviations Lecture: Finding a confidence interval for a variance and a standard deviation. OBJECTIVE Find a confidence interval for a variance and a standard deviation. In Sections 7-1 through 7-3 confidence intervals were calculated for means and proportions. This section will explain how to find confidence intervals for variances and standard deviations. In statistics, the variance and standard deviation of a variable are as important as the mean. For example, when products that fit together (such as pipes) are manufactured, it is important to keep the variations of the diameters of the products as small as possible otherwise, they will not fit together properly and will have to be scrapped. In the manufacture of medicines, the variance and standard deviation of the medication in the pills play an important role in making sure patients receive the proper dosage. For these reasons, confidence intervals for variances and standard deviations are necessary. To calculate these confidence intervals, a new statistical distribution is needed. It is called the chisquare distribution. Historical Note The 2 distribution with 2 degrees of freedom was formulated by a mathematician named Hershel in 1869 while he was studying the accuracy of shooting arrows at a target. Many other mathematicians have since contributed to its development. The chisquare variable is similar to the t variable in that its distribution is a family of curves based on the number of degrees of freedom. The symbol for chisquare is 2 (Greek letter chi, pronounced \"ki\"). Several of the distributions are shown in Figure 7-9, along with the corresponding degrees of freedom. The chisquare distribution is obtained from the values of (n - 1)s2/2 when random samples are selected from a normally distributed population whose variance is 2. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 1/15 4/8/2017 eBook with Connect Math the values of (n - 1)s2/2 when random samples are selected from a normally distributed population whose variance is 2. Figure 7-9 The ChiSquare Family of Curves Page 400 A chisquare variable cannot be negative, and the distributions are skewed to the right. At about 100 degrees of freedom, the chisquare distribution becomes somewhat symmetric. The area under each chisquare distribution is equal to 1.00, or 100%. A summary of the characteristics of the chisquare distribution is given next. Characteristics of the ChiSquare Distribution 1. All chisquare values are greater than or equal to 0. 2. The chisquare distribution is a family of curves based on the degrees of freedom. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 2/15 4/8/2017 eBook with Connect Math 2. The chisquare distribution is a family of curves based on the degrees of freedom. 3. The area under each chisquare distribution curve is equal to 1. 4. The chisquare distributions are positively skewed. Table G in Appendix A gives the values for the chisquare distribution. These values are used in the denominators of the formulas for confidence intervals. Two different values are used in the formula because the distribution is not symmetric. One value is found on the left side of the table, and the other is on the right. See Figure 7-10. Figure 7-10 ChiSquare Distribution for d.f. = n - 1 Table G is set up similarly to the table used for the t distribution. The left column denotes the degrees of freedom, and the top row represents the area to the right of the critical value. For example, to find the table values corresponding to the 95% confidence interval, you must first change 95% to a decimal and subtract it from 1 (1 - 0.95 = 0.05). Then divide the answer by 2(/2 = 0.05/2 = 0.025). This is the column on the right side of the table, used to get the values for 2right. To get the value for 2left, subtract the value of /2 from 1 (1 - 0.05/2 = 0.975). Finally, find the appropriate row corresponding to the degrees of freedom n - 1. A similar procedure is used to find the values for a 90 or 99% confidence interval. Page 401 EXAMPLE 7-13 Find the values for 2right and 2left for a 90% confidence interval when n = 25. SOLUTION To find 2right, subtract 1 - 0.90 = 0.10 then divide 0.10 by 2 to get 0.05. To find 2left, subtract 1 - 0.05 = 0.95. Then use the 0.95 and 0.05 columns with d.f. = n - 1 = 25 - 1 = 24. See Figure 7-11. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 3/15 4/8/2017 eBook with Connect Math Figure 7-11 2 Table for Example 7-13 The values are See Figure 7-12. Figure 7-12 2 Table for Example 7-13 If the number for the degrees of freedom is not given in the table, use the closest lower value in the table. For example, for d.f. = 53, use d.f. = 50. This is a conservative approach. Useful estimates for 2 and are s2 and s, respectively. To find confidence intervals for variances and standard deviations, you must assume that the variable is normally distributed. Page 402 The formulas for the confidence intervals are shown here. Formula for the Confidence Interval for a Variance https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 4/15 4/8/2017 Formula for the Confidence Interval for a Variance eBook with Connect Math Formula for the Confidence Interval for a Standard Deviation Recall that s2 is the symbol for the sample variance and s is the symbol for the sample standard deviation. If the problem gives the sample standard deviation s, be sure to square it when you are using the formula. But if the problem gives the sample variance s2, do not square it when you are using the formula, since the variance is already in square units. Assumptions for Finding a Confidence Interval for a Variance or Standard Deviation 1. The sample is a random sample. 2. The population must be normally distributed. In this text, the assumptions will be stated in the exercises however, when encountering statistics in other situations, you must check to see that these assumptions have been met before proceeding. Rounding Rule for a Confidence Interval for a Variance or Standard Deviation When you are computing a confidence interval for a population variance or standard deviation by using raw data, round off to one more decimal place than the number of decimal places in the original data. When you are computing a confidence interval for a population variance or standard deviation by using a sample variance or standard deviation, round off to the same number of decimal places as given for the sample variance or standard deviation. Example 7-14 shows how to find a confidence interval for a variance and standard deviation. EXAMPLE 7-14Nicotine Content Find the 95% confidence interval for the variance and standard deviation of the nicotine content of cigarettes manufactured if a random sample of 20 cigarettes has a standard deviation of 1.6 milligrams. Assume the variable is normally distributed. SOLUTION Since = 0.05, the two critical values, respectively, for the 0.025 and 0.975 levels for 19 degrees of freedom are 32.852 and 8.907. The 95% confidence interval for the variance is found by substituting in the formula. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 5/15 4/8/2017 eBook with Connect Math Page 403 Hence, you can be 95% confident that the true variance for the nicotine content is between 1.5 and 5.5. For the standard deviation, the confidence interval is Hence, you can be 95% confident that the true standard deviation for the nicotine content of all cigarettes manufactured is between 1.2 and 2.3 milligrams based on a sample of 20 cigarettes. EXAMPLE 7-15Named Storms Find the 90% confidence interval for the variance and standard deviation for the number of named storms per year in the Atlantic basin. A random sample of 10 years has been used. Assume the distribution is approximately normal. 10 5121113 1519181416 Source: Atlantic Oceanographic and Meteorological Laboratory. SOLUTION Step 1 Find the variance for the data. Use the formulas in Chapter 3 or your calculator. The variance s2 = 16.9. Step 2 Find 2right and 2left from Table G in Appendix A, using 10 - 1 = 9 degrees of freedom. In this case, use = 0.05 and 0.95 2right = 3.325 2left = 16.919. Step 3 Substitute in the formula. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 6/15 4/8/2017 eBook with Connect Math Hence, you can be 90% confident that the standard deviation for the number of named storms is between 3.0 and 6.8 based on a random sample of 10 years. Note: If you are using the standard deviation instead (as in Example 7-14) of the variance, be sure to square the standard deviation when substituting in the formula. Applying the Concepts 7-4 Confidence Interval for Standard Deviation Shown are the ages (in years) of the Presidents at the times of their deaths. 6790838573807879 6871536574647756 6663704957716771 5860726757609063 88784664819393 Page 404 1. Do the data represent a population or a sample? 2. Select a random sample of 12 ages and find the variance and standard deviation. 3. Find the 95% confidence interval of the standard deviation. 4. Find the standard deviation of all the data values. 5. Does the confidence interval calculated in question 3 contain the standard deviation? 6. If it does not, give a reason why. 7. What assumption(s) must be considered for constructing the confidence interval in step 3? See page 412 for the answers. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 7/15 4/8/2017 eBook with Connect Math Exercises 7-4 1. What distribution must be used when computing confidence intervals for variances and standard deviations? 2. What assumption must be made when computing confidence intervals for variances and standard deviations? 3. Using Table G, find the values for 2left and 2right. 1. = 0.05, n = 12 2. = 0.10, n = 20 3. = 0.05, n = 27 4. = 0.01, n = 6 5. = 0.10, n = 41 4. Lifetimes of Wristwatches Find the 90% confidence interval for the variance and standard deviation for the lifetimes of inexpensive wristwatches if a random sample of 24 watches has a standard deviation of 4.8 months. Assume the variable is normally distributed. Do you feel that the lifetimes are relatively consistent? 5. Carbohydrates in Yogurt The number of carbohydrates (in grams) per 8ounce serving of yogurt for each of a random selection of brands is listed below. Estimate the true population variance and standard deviation for the number of carbohydrates per 8ounce serving of yogurt with 95% confidence. Assume the variable is normally distributed. 174241203941351543 2538334223172534 6. Carbon Monoxide Deaths A study of generationrelated carbon monoxide deaths showed that a random sample of 6 recent years had a standard deviation of 4.1 deaths per year. Find the 99% confidence interval of the variance and standard deviation. Assume the variable is normally distributed. Source: Based on information from Consumer Protection Safety Commission. 7. Cost of Knee Replacement Surgery U.S. insurers' costs for knee replacement surgery range from $17,627 to $25,462. Estimate the population variance (standard deviation) in cost with 98% confidence based on a random sample of 10 persons who have had this surgery. The retail costs (for uninsured persons) for the same procedure range from $40,640 to $58,702. Estimate the population variance and standard deviation in cost with 98% confidence based on a sample of 10 persons, and compare your two intervals. Assume the variable is normally distributed. Source: Time Almanac. 8. Age of College Students Find the 90% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park College if a random sample of 24 students has a standard deviation of 2.3 years. Assume the variable is normally distributed. 9. NewCar Lease Fees A newcar dealer is leasing various brandnew randomly selected models for the monthly rates (in dollars) listed below. Estimate the true population variance (and standard deviation) in leasing rates with 90% confidence. Assume the variable is normally distributed. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 8/15 4/8/2017 eBook with Connect Math population variance (and standard deviation) in leasing rates with 90% confidence. Assume the variable is normally distributed. Exercise: NewCar Lease Fees 169169199239239249 10. Stock Prices A random sample of stock prices per share (in dollars) is shown. Find the 90% confidence interval for the variance and standard deviation for the prices. Assume the variable is normally distributed. 26.6913.8828.3712.00 75.377.50 47.5043.00 3.81 53.8113.6245.12 6.94 28.2528.0060.50 40.2510.8746.1214.75 Source: Pittsburgh Tribune Review. 11. Number of Homeless Individuals A researcher wishes to find the confidence interval of the population standard deviation for the number of homeless people in a large city. A random sample of 25 months had a standard deviation of 462. Find the 95% confidence interval. Assume the variable is normally distributed. Page 405 12. Home Ownership Rates The percentage rates of home ownership for 8 randomly selected states are listed below. Estimate the population variance and standard deviation for the percentage rate of home ownership with 99% confidence. Assume the variable is normally distributed. 66.075.870.973.963.468.573.365.9 Source: World Almanac. 13. Calories in a Standard Size Candy Bar Estimate the standard deviation in calories for these randomly selected standardsize candy bars with 95% confidence. (The number of calories is listed for each.) Assume the variable is normally distributed. 220220210230275260240 220240240280230280260 14. SAT Scores Estimate the variance in mean mathematics SAT scores by state, using the randomly selected scores listed below. Estimate with 99% confidence. Assume the variable is normally distributed. 490502211209499565 469543572550515500 Source: World Almanac 2012. 15. Daily Cholesterol Intake The American Heart Association recommends a daily cholesterol intake of less than 300 mg. Here are the cholesterol amounts in a random sample of single servings of grilled meats. Estimate the standard deviation in cholesterol with 95% confidence. Assume the variable is normally distributed. 90200 80105 95 85 70105115110 100225125130145 https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxRt... 9/15 4/8/2017 eBook with Connect Math Extending the Concepts 16. Calculator Battery Lifetimes A confidence interval for a standard deviation for large samples taken from a normally distributed population can be approximated by Find the 95% confidence interval for the population standard deviation of calculator batteries. A random sample of 200 calculator batteries has a standard deviation of 18 months. Technology Step by Step TI84 Plus Step by Step The TI84 Plus does not have a builtin confidence interval for the variance or standard deviation. However, the downloadable program named SDINT is available in your online resources. Follow the instructions online for downloading the program. Finding a Confidence Interval for the Variance and Standard Deviation (Data) 1. Enter the data values into L1. 2. Press PRGM, move the cursor to the program named SDINT, and press ENTER twice. 3. Press 1 for Data. 4. Type L1 for the list and press ENTER. 5. Type the confidence level and press ENTER. 6. Press ENTER to clear the screen. Example TI7-4 Find the 90% confidence interval for the variance and standard deviation for the data: 59545352513949464948 https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxR... 10/15 4/8/2017 eBook with Connect Math Page 406 Finding a Confidence Interval for the Variance and Standard Deviation (Statistics) 1. Press PRGM, move the cursor to the program named SDINT, and press ENTER twice. 2. Press 2 for Stats. 3. Type the sample standard deviation and press ENTER. 4. Type the sample size and press ENTER. 5. Type the confidence level and press ENTER. 6. Press ENTER to clear the screen. Example TI7-5 This refers to Example 7-14 in the text. Find the 95% confidence interval for the variance and standard deviation, given n = 20 and s = 1.6. https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxR... 11/15 4/8/2017 eBook with Connect Math https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxR... 12/15 4/8/2017 eBook with Connect Math https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxR... 13/15 4/8/2017 eBook with Connect Math https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxR... 14/15 4/8/2017 eBook with Connect Math https://wwwawd.connectmath.com/alekscgi/x/Isl.exe/1o_uIgNsIkr7j8P3jHlQ1HXBxM9hozYnmeRn5Cb56mtPrkRS4nerzDGG2avolTiK5JrnDDR2too6Ggg7s0f0aPz_m4k4eP8WNsPPpKqo?1oiPvmAdTNWxR... 15/15