7.E: Estimation (Exercises)
These are homework exercises to accompany the Textmap created for "Introductory Statistics" by Shafer and Zhang.
7.1: Large Sample Estimation of a Population Mean
Basic
 A random sample is drawn from a population of known standard deviation \(11.3\). Construct a \(90\%\) confidence interval for the population mean based on the information given (not all of the information given need be used).
 \(n = 36,\; \bar{x}=105.2,\; s = 11.2\)
 \(n = 100,\; \bar{x}=105.2,\; s = 11.2\)
 A random sample is drawn from a population of known standard deviation \(22.1\). Construct a \(95\%\) confidence interval for the population mean based on the information given (not all of the information given need be used).
 \(n =121 ,\; \bar{x}=82.4,\; s = 21.9\)
 \(n =81 ,\; \bar{x}=82.4,\; s = 21.9\)
 A random sample is drawn from a population of unknown standard deviation. Construct a \(99\%\) confidence interval for the population mean based on the information given.
 \(n =49 ,\; \bar{x}=17.1,\; s = 2.1\)
 \(n =169 ,\; \bar{x}=17.1,\; s = 2.1\)
 A random sample is drawn from a population of unknown standard deviation. Construct a \(98\%\) confidence interval for the population mean based on the information given.
 \(n =225 ,\; \bar{x}=92.0,\; s = 8.4\)
 \(n =64 ,\; \bar{x}=92.0,\; s = 8.4\)
 A random sample of size \(144\) is drawn from a population whose distribution, mean, and standard deviation are all unknown. The summary statistics are \(\bar{x}=58.2\) and \(s = 2.6\).
 Construct an \(80\%\) confidence interval for the population mean \(\mu\).
 Construct a \(90\%\) confidence interval for the population mean \(\mu\).
 Comment on why one interval is longer than the other.
 A random sample of size \(256\) is drawn from a population whose distribution, mean, and standard deviation are all unknown. The summary statistics are \(\bar{x}=1011\) and \(s = 34\).
 Construct a \(90\%\) confidence interval for the population mean \(\mu\).
 Construct a \(99\%\) confidence interval for the population mean \(\mu\).
 Comment on why one interval is longer than the other.
Applications
 A government agency was charged by the legislature with estimating the length of time it takes citizens to fill out various forms. Two hundred randomly selected adults were timed as they filled out a particular form. The times required had mean \(12.8\) minutes with standard deviation \(1.7\) minutes. Construct a \(90\%\) confidence interval for the mean time taken for all adults to fill out this form.
 Four hundred randomly selected working adults in a certain state, including those who worked at home, were asked the distance from their home to their workplace. The average distance was \(8.84\) miles with standard deviation \(2.70\) miles. Construct a \(99\%\) confidence interval for the mean distance from home to work for all residents of this state.
 On every passenger vehicle that it tests an automotive magazine measures, at true speed \(55\) mph, the difference between the true speed of the vehicle and the speed indicated by the speedometer. For \(36\) vehicles tested the mean difference was \(1.2\) mph with standard deviation \(0.2\) mph. Construct a \(90\%\) confidence interval for the mean difference between true speed and indicated speed for all vehicles.
 A corporation monitors time spent by office workers browsing the web on their computers instead of working. In a sample of computer records of \(50\) workers, the average amount of time spent browsing in an eighthour work day was \(27.8\) minutes with standard deviation \(8.2\) minutes. Construct a \(99.5\%\) confidence interval for the mean time spent by all office workers in browsing the web in an eighthour day.
 A sample of \(250\) workers aged \(16\) and older produced an average length of time with the current employer (“job tenure”) of \(4.4\) years with standard deviation \(3.8\) years. Construct a \(99.9\%\) confidence interval for the mean job tenure of all workers aged \(16\) or older.
 The amount of a particular biochemical substance related to bone breakdown was measured in \(30\) healthy women. The sample mean and standard deviation were \(3.3\) nanograms per milliliter (ng/mL) and \(1.4\) ng/mL. Construct an \(80\%\) confidence interval for the mean level of this substance in all healthy women.
 A corporation that owns apartment complexes wishes to estimate the average length of time residents remain in the same apartment before moving out. A sample of \(150\) rental contracts gave a mean length of occupancy of \(3.7\) years with standard deviation \(1.2\) years. Construct a \(95\%\) confidence interval for the mean length of occupancy of apartments owned by this corporation.
 The designer of a garbage truck that lifts rollout containers must estimate the mean weight the truck will lift at each collection point. A random sample of \(325\) containers of garbage on current collection routes yielded \(\bar{x}=75.3 lb,\; s = 12.8 lb\). Construct a \(99.8\%\) confidence interval for the mean weight the trucks must lift each time.
 In order to estimate the mean amount of damage sustained by vehicles when a deer is struck, an insurance company examined the records of \(50\) such occurrences, and obtained a sample mean of \(\$2,785\) with sample standard deviation \(\$221\). Construct a \(95\%\) confidence interval for the mean amount of damage in all such accidents.
 In order to estimate the mean FICO credit score of its members, a credit union samples the scores of \(95\) members, and obtains a sample mean of \(738.2\) with sample standard deviation \(64.2\). Construct a \(99\%\) confidence interval for the mean FICO score of all of its members.
Additional Exercises
 For all settings a packing machine delivers a precise amount of liquid; the amount dispensed always has standard deviation \(0.07\) ounce. To calibrate the machine its setting is fixed and it is operated \(50\) times. The mean amount delivered is \(6.02\) ounces with sample standard deviation \(0.04\) ounce. Construct a \(99.5\%\) confidence interval for the mean amount delivered at this setting. Hint: Not all the information provided is needed.
 A power wrench used on an assembly line applies a precise, preset amount of torque; the torque applied has standard deviation \(0.73\) footpound at every torque setting. To check that the wrench is operating within specifications it is used to tighten \(100\) fasteners. The mean torque applied is \(36.95\) footpounds with sample standard deviation \(0.62\) footpound. Construct a \(99.9\%\) confidence interval for the mean amount of torque applied by the wrench at this setting. Hint: Not all the information provided is needed.
 The number of trips to a grocery store per week was recorded for a randomly selected collection of households, with the results shown in the table. \[\begin{matrix} 2 & 2 & 2 & 1 & 4 & 2 & 3 & 2 & 5 & 4\\ 2 & 3 & 5 & 0 & 3 & 2 & 3 & 1 & 4 & 3\\ 3 & 2 & 1 & 6 & 2 & 3 & 3 & 2 & 4 & 4 \end{matrix}\]Construct a \(95\%\) confidence interval for the average number of trips to a grocery store per week of all households.
 For each of \(40\) high school students in one county the number of days absent from school in the previous year were counted, with the results shown in the frequency table. \[\begin{array}{cc c c c c c} x &0 &1 &2 &3 &4 &5 \\ \hline f &24 &7 &5 &2 &1 &1\\ \end{array}\]Construct a \(90\%\) confidence interval for the average number of days absent from school of all students in the county.
 A town council commissioned a random sample of \(85\) households to estimate the number of fourwheel vehicles per household in the town. The results are shown in the following frequency table. \[\begin{array}{cc c c c c c} x &0 &1 &2 &3 &4 &5 \\ \hline f &1 &16 &28 &22 &12 &6\\ \end{array}\]Construct a \(98\%\) confidence interval for the average number of fourwheel vehicles per household in the town.
 The number of hours per day that a television set was operating was recorded for a randomly selected collection of households, with the results shown in the table. \[\begin{matrix} 3.7 & 4.2 & 1.5 & 3.6 & 5.9\\ 4.7 & 8.2 & 3.9 & 2.5 & 4.4\\ 2.1 & 3.6 & 1.1 & 7.3 & 4.2\\ 3.0 & 3.8 & 2.2 & 4.2 & 3.8\\ 4.3 & 2.1 & 2.4 & 6.0 & 3.7\\ 2.5 & 1.3 & 2.8 & 3.0 & 5.6 \end{matrix}\]Construct a \(99.8\%\) confidence interval for the mean number of hours that a television set is in operation in all households.
Large Data Set Exercises
Large Data Set missing from the original
 Large \(\text{Data Set 1}\) records the SAT scores of \(1,000\) students. Regarding it as a random sample of all high school students, use it to construct a \(99\%\) confidence interval for the mean SAT score of all students.
 Large \(\text{Data Set 1}\) records the GPAs of \(1,000\) college students. Regarding it as a random sample of all college students, use it to construct a \(95\%\) confidence interval for the mean GPA of all students.
 Large \(\text{Data Set 1}\) lists the SAT scores of \(1,000\) students.
 Regard the data as arising from a census of all students at a high school, in which the SAT score of every student was measured. Compute the population mean \(\mu\).
 Regard the first \(36\) students as a random sample and use it to construct a \(99\%\) confidence for the mean \(\mu\) of all \(1,000\) SAT scores. Does it actually capture the mean \(\mu\)?
 Large \(\text{Data Set 1}\) lists the GPAs of \(1,000\) students.
 Regard the data as arising from a census of all freshman at a small college at the end of their first academic year of college study, in which the GPA of every such person was measured. Compute the population mean \(\mu\).
 Regard the first \(36\) students as a random sample and use it to construct a \(95\%\) confidence for the mean \(\mu\) of all \(1,000\) GPAs. Does it actually capture the mean \(\mu\)?
Answers

 \(105.2\pm 3.10\)
 \(105.2\pm 1.86\)

 \(17.1\pm 0.77\)
 \(17.1\pm 0.42\)

 \(58.2\pm 0.28\)
 \(58.2\pm 0.36\)
 Asking for greater confidence requires a longer interval.
 \(12.8\pm 0.20\)
 \(1.2\pm 0.05\)
 \(4.4\pm 0.79\)
 \(3.7\pm 0.19\)
 \(2785\pm 61\)
 \(6.02\pm 0.03\)
 \(2.8\pm 0.48\)
 \(2.54\pm 0.30\)
 \((1511.43,1546.05)\)

 \(\mu = 1528.74\)
 \((1428.22,1602.89)\)
7.2: Small Sample Estimation of a Population Mean
Basic
 A random sample is drawn from a normally distributed population of known standard deviation \(5\). Construct a \(99.8\%\) confidence interval for the population mean based on the information given (not all of the information given need be used).
 \(n = 16,\; \bar{x}=98,\; s = 5.6\)
 \(n = 9,\; \bar{x}=98,\; s = 5.6\)
 A random sample is drawn from a normally distributed population of known standard deviation \(10.7\). Construct a \(95\%\) confidence interval for the population mean based on the information given (not all of the information given need be used).
 \(n = 25,\; \bar{x}=103.3,\; s = 11.0\)
 \(n = 4,\; \bar{x}=103.3,\; s = 11.0\)
 A random sample is drawn from a normally distributed population of unknown standard deviation. Construct a \(99\%\) confidence interval for the population mean based on the information given.
 \(n = 18,\; \bar{x}=386,\; s = 24\)
 \(n = 7,\; \bar{x}=386,\; s = 24\)
 A random sample is drawn from a normally distributed population of unknown standard deviation. Construct a \(98\%\) confidence interval for the population mean based on the information given.
 \(n = 8,\; \bar{x}=58.3,\; s = 4.1\)
 \(n = 27,\; \bar{x}=58.3,\; s = 4.1\)
 A random sample of size \(14\) is drawn from a normal population. The summary statistics are \(\bar{x}=933,\; and\; s = 18\).
 Construct an \(80\%\) confidence interval for the population mean \(\mu\).
 Construct a \(90\%\) confidence interval for the population mean \(\mu\).
 Comment on why one interval is longer than the other.
 A random sample of size \(28\) is drawn from a normal population. The summary statistics are \(\bar{x}=68.6,\; and\; s = 1.28\).
 Construct a \(95\%\) confidence interval for the population mean \(\mu\).
 Construct a \(99.5\%\) confidence interval for the population mean \(\mu\).
 Comment on why one interval is longer than the other.
Application Exercises
 City planners wish to estimate the mean lifetime of the most commonly planted trees in urban settings. A sample of \(16\) recently felled trees yielded mean age \(32.7\) years with standard deviation \(3.1\) years. Assuming the lifetimes of all such trees are normally distributed, construct a \(99.8\%\) confidence interval for the mean lifetime of all such trees.
 To estimate the number of calories in a cup of diced chicken breast meat, the number of calories in a sample of four separate cups of meat is measured. The sample mean is \(211.8\) calories with sample standard deviation \(0.9\) calorie. Assuming the caloric content of all such chicken meat is normally distributed, construct a \(95\%\) confidence interval for the mean number of calories in one cup of meat.
 A college athletic program wishes to estimate the average increase in the total weight an athlete can lift in three different lifts after following a particular training program for six weeks. Twentyfive randomly selected athletes when placed on the program exhibited a mean gain of \(47.3\) lb with standard deviation \(6.4\) lb. Construct a \(90\%\) confidence interval for the mean increase in lifting capacity all athletes would experience if placed on the training program. Assume increases among all athletes are normally distributed.
 To test a new tread design with respect to stopping distance, a tire manufacturer manufactures a set of prototype tires and measures the stopping distance from \(70\) mph on a standard test car. A sample of \(25\) stopping distances yielded a sample mean \(173\) feet with sample standard deviation \(8\) feet. Construct a \(98\%\) confidence interval for the mean stopping distance for these tires. Assume a normal distribution of stopping distances.
 A manufacturer of chokes for shotguns tests a choke by shooting \(15\) patterns at targets \(40\) yards away with a specified load of shot. The mean number of shot in a \(30\)inch circle is \(53.5\) with standard deviation \(1.6\). Construct an \(80\%\) confidence interval for the mean number of shot in a \(30\)inch circle at \(40\) yards for this choke with the specified load. Assume a normal distribution of the number of shot in a \(30\)inch circle at \(40\) yards for this choke.
 In order to estimate the speaking vocabulary of threeyearold children in a particular socioeconomic class, a sociologist studies the speech of four children. The mean and standard deviation of the sample are \(\bar{x}=1120\) and \(s = 215\) words. Assuming that speaking vocabularies are normally distributed, construct an \(80\%\) confidence interval for the mean speaking vocabulary of all threeyearold children in this socioeconomic group.
 A thread manufacturer tests a sample of eight lengths of a certain type of thread made of blended materials and obtains a mean tensile strength of \(8.2\) lb with standard deviation \(0.06\) lb. Assuming tensile strengths are normally distributed, construct a \(90\%\) confidence interval for the mean tensile strength of this thread.
 An airline wishes to estimate the weight of the paint on a fully painted aircraft of the type it flies. In a sample of four repaintings the average weight of the paint applied was \(239\) pounds, with sample standard deviation \(8\) pounds. Assuming that weights of paint on aircraft are normally distributed, construct a \(99.8\%\) confidence interval for the mean weight of paint on all such aircraft.
 In a study of dummy foal syndrome, the average time between birth and onset of noticeable symptoms in a sample of six foals was \(18.6\) hours, with standard deviation \(1.7\) hours. Assuming that the time to onset of symptoms in all foals is normally distributed, construct a \(90\%\) confidence interval for the mean time between birth and onset of noticeable symptoms.
 A sample of \(26\) women’s size \(6\) dresses had mean waist measurement \(25.25\) inches with sample standard deviation \(0.375\) inch. Construct a \(95\%\) confidence interval for the mean waist measurement of all size \(6\) women’s dresses. Assume waist measurements are normally distributed.
Additional Exercises
 Botanists studying attrition among saplings in new growth areas of forests diligently counted stems in six plots in fiveyearold new growth areas, obtaining the following counts of stems per acre: \[\begin{matrix} 9,432 & 11,026 & 10,539\\ 8,773 & 9,868 & 10,247 \end{matrix}\]
Construct an \(80\%\) confidence interval for the mean number of stems per acre in all fiveyearold new growth areas of forests. Assume that the number of stems per acre is normally distributed.  Nutritionists are investigating the efficacy of a diet plan designed to increase the caloric intake of elderly people. The increase in daily caloric intake in \(12\) individuals who are put on the plan is (a minus sign signifies that calories consumed went down): \[\begin{matrix} 121 & 284 & 94 & 295 & 183 & 312\\ 188 & 102 & 259 & 226 & 152 & 167 \end{matrix}\]
Construct a \(99.8\%\) confidence interval for the mean increase in caloric intake for all people who are put on this diet. Assume that population of differences in intake is normally distributed.  A machine for making precision cuts in dimension lumber produces studs with lengths that vary with standard deviation \(0.003\) inch. Five trial cuts are made to check the machine’s calibration. The mean length of the studs produced is \(104.998\) inches with sample standard deviation \(0.004\) inch. Construct a \(99.5\%\) confidence interval for the mean lengths of all studs cut by this machine. Assume lengths are normally distributed. Hint: Not all the numbers given in the problem are used.
 The variation in time for a baked good to go through a conveyor oven at a large scale bakery has standard deviation \(0.017\) minute at every time setting. To check the bake time of the oven periodically four batches of goods are carefully timed. The recent check gave a mean of \(27.2\) minutes with sample standard deviation \(0.012\) minute. Construct a \(99.8\%\) confidence interval for the mean bake time of all batches baked in this oven. Assume bake times are normally distributed. Hint: Not all the numbers given in the problem are used.
 Wildlife researchers tranquilized and weighed three adult male polar bears. The data (in pounds) are: \(926, 742, 1109\). Assume the weights of all bears are normally distributed.
 Construct an \(80\%\) confidence interval for the mean weight of all adult male polar bears using these data.
 Convert the three weights in pounds to weights in kilograms using the conversion \(1\; lb = 0.453\; kg\) (so the first datum changes to \((926)(0.453)=419\)). Use the converted data to construct an \(80\%\) confidence interval for the mean weight of all adult male polar bears expressed in kilograms.
 Convert your answer in part (a) into kilograms directly and compare it to your answer in (b). This illustrates that if you construct a confidence interval in one system of units you can convert it directly into another system of units without having to convert all the data to the new units.
 Wildlife researchers trapped and measured six adult male collared lemmings. The data (in millimeters) are: \(104, 99, 112, 115, 96, 109\). Assume the lengths of all lemmings are normally distributed.
 Construct a \(90\%\) confidence interval for the mean length of all adult male collared lemmings using these data.
 Convert the six lengths in millimeters to lengths in inches using the conversion \(1\; mm = 0.039\; in\) (so the first datum changes to \((104)(0.039) = 4.06\)). Use the converted data to construct a \(90\%\) confidence interval for the mean length of all adult male collared lemmings expressed in inches.
 Convert your answer in part (a) into inches directly and compare it to your answer in (b). This illustrates that if you construct a confidence interval in one system of units you can convert it directly into another system of units without having to convert all the data to the new units.
Answers

 \(98\pm 3.9\)
 \(98\pm 5.2\)

 \(386\pm 16.4\)
 \(386\pm 33.6\)

 \(933\pm 6.5\)
 \(933\pm 8.5\)
 Asking for greater confidence requires a longer interval.
 \(32.7\pm 2.9\)
 \(47.3\pm 2.19\)
 \(53.5\pm 0.56\)
 \(8.2\pm 0.04\)
 \(18.6\pm 1.4\)
 \(9981\pm 486\)
 \(104.998\pm 0.004\)

 \(926\pm 200\)
 \(419\pm 90\)
 \(419\pm 91\)
7.3: Large Sample Estimation of a Population Proportion
Basic
 Information about a random sample is given. Verify that the sample is large enough to use it to construct a confidence interval for the population proportion. Then construct a \(90\%\) confidence interval for the population proportion.
 \(n = 25, \hat{p}=0.7\)
 \(n = 50, \hat{p}=0.7\)
 Information about a random sample is given. Verify that the sample is large enough to use it to construct a confidence interval for the population proportion. Then construct a \(95\%\) confidence interval for the population proportion.
 \(n = 2500, \hat{p}=0.22\)
 \(n = 1200, \hat{p}=022\)
 Information about a random sample is given. Verify that the sample is large enough to use it to construct a confidence interval for the population proportion. Then construct a \(98\%\) confidence interval for the population proportion.
 \(n = 80, \hat{p}=0.4\)
 \(n = 325, \hat{p}=0.4\)
 Information about a random sample is given. Verify that the sample is large enough to use it to construct a confidence interval for the population proportion. Then construct a \(99.5\%\) confidence interval for the population proportion.
 \(n = 200, \hat{p}=0.85\)
 \(n = 75, \hat{p}=0.85\)
 In a random sample of size \(1,100,338\) have the characteristic of interest.
 Compute the sample proportion \(\hat{p}\) with the characteristic of interest.
 Verify that the sample is large enough to use it to construct a confidence interval for the population proportion.
 Construct an \(80\%\) confidence interval for the population proportion \(p\).
 Construct a \(90\%\) confidence interval for the population proportion \(p\).
 Comment on why one interval is longer than the other.
 In a random sample of size \(2,400,420\) have the characteristic of interest.
 Compute the sample proportion \(\hat{p}\) with the characteristic of interest.
 Verify that the sample is large enough to use it to construct a confidence interval for the population proportion.
 Construct a \(90\%\) confidence interval for the population proportion \(p\).
 Construct a \(99\%\) confidence interval for the population proportion \(p\).
 Comment on why one interval is longer than the other.
Applications
Q7.3.7
A security feature on some web pages is graphic representations of words that are readable by human beings but not machines. When a certain design format was tested on \(450\) subjects, by having them attempt to read ten disguised words, \(448\) subjects could read all the words.
 Give a point estimate of the proportion \(p\) of all people who could read words disguised in this way.
 Show that the sample is not sufficiently large to construct a confidence interval for the proportion of all people who could read words disguised in this way.
Q7.3.8
In a random sample of \(900\) adults, \(42\) defined themselves as vegetarians.
 Give a point estimate of the proportion of all adults who would define themselves as vegetarians.
 Verify that the sample is sufficiently large to use it to construct a confidence interval for that proportion.
 Construct an \(80\%\) confidence interval for the proportion of all adults who would define themselves as vegetarians.
Q7.3.9
In a random sample of \(250\) employed people, \(61\) said that they bring work home with them at least occasionally.
 Give a point estimate of the proportion of all employed people who bring work home with them at least occasionally.
 Construct a \(99\%\) confidence interval for that proportion.
Q7.3.10
In a random sample of \(1,250\) household moves, \(822\) were moves to a location within the same county as the original residence.
 Give a point estimate of the proportion of all household moves that are to a location within the same county as the original residence.
 Construct a \(98\%\) confidence interval for that proportion.
Q7.3.11
In a random sample of \(12,447\) hip replacement or revision surgery procedures nationwide, \(162\) patients developed a surgical site infection.
 Give a point estimate of the proportion of all patients undergoing a hip surgery procedure who develop a surgical site infection.
 Verify that the sample is sufficiently large to use it to construct a confidence interval for that proportion.
 Construct a \(95\%\) confidence interval for the proportion of all patients undergoing a hip surgery procedure who develop a surgical site infection.
Q7.3.12
In a certain region prepackaged products labeled \(500\) g must contain on average at least \(500\) grams of the product, and at least \(90\%\) of all packages must weigh at least \(490\) grams. In a random sample of \(300\) packages, \(288\) weighed at least \(490\) grams.
 Give a point estimate of the proportion of all packages that weigh at least \(490\) grams.
 Verify that the sample is sufficiently large to use it to construct a confidence interval for that proportion.
 Construct a \(99.8\%\) confidence interval for the proportion of all packages that weigh at least \(490\) grams.
Q7.3.13
A survey of \(50\) randomly selected adults in a small town asked them if their opinion on a proposed “no cruising” restriction late at night. Responses were coded \(1\) for in favor, \(0\) for indifferent, and \(2\) for opposed, with the results shown in the table.\[\begin{matrix} 1 & 0 & 2 & 0 & 1 & 0 & 0 & 1 & 1 & 2\\ 0 & 2 & 0 & 0 & 0 & 1 & 0 & 2 & 0 & 0 \\ 0 & 2 & 1 & 2 & 0 & 0 & 0 & 2 & 0 & 1\\ 0 & 2 & 0 & 2 & 0 & 1 & 0 & 0 & 2 & 0\\ 1 & 0 & 0 & 1 & 2 & 0 & 0 & 2 & 1 & 2 \end{matrix}\]
 Give a point estimate of the proportion of all adults in the community who are indifferent concerning the proposed restriction.
 Assuming that the sample is sufficiently large, construct a \(90\%\) confidence interval for the proportion of all adults in the community who are indifferent concerning the proposed restriction.
Q7.3.14
To try to understand the reason for returned goods, the manager of a store examines the records on \(40\) products that were returned in the last year. Reasons were coded by \(1\) for “defective,” \(2\) for “unsatisfactory,” and \(0\) for all other reasons, with the results shown in the table. \[\begin{matrix} 0 & 2 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2\\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{matrix}\]
 Give a point estimate of the proportion of all returns that are because of something wrong with the product, that is, either defective or performed unsatisfactorily.
 Assuming that the sample is sufficiently large, construct an \(80\%\) confidence interval for the proportion of all returns that are because of something wrong with the product.
Q7.3.15
In order to estimate the proportion of entering students who graduate within six years, the administration at a state university examined the records of \(600\) randomly selected students who entered the university six years ago, and found that \(312\) had graduated.
 Give a point estimate of the sixyear graduation rate, the proportion of entering students who graduate within six years.
 Assuming that the sample is sufficiently large, construct a \(98\%\) confidence interval for the sixyear graduation rate.
Q7.3.16
In a random sample of \(2,300\) mortgages taken out in a certain region last year, \(187\) were adjustablerate mortgages.
 Give a point estimate of the proportion of all mortgages taken out in this region last year that were adjustablerate mortgages.
 Assuming that the sample is sufficiently large, construct a \(99.9\%\) confidence interval for the proportion of all mortgages taken out in this region last year that were adjustablerate mortgages.
Q7.3.17
In a research study in cattle breeding, \(159\) of \(273\) cows in several herds that were in estrus were detected by means of an intensive once a day, onehour observation of the herds in early morning.
 Give a point estimate of the proportion of all cattle in estrus who are detected by this method.
 Assuming that the sample is sufficiently large, construct a \(90\%\) confidence interval for the proportion of all cattle in estrus who are detected by this method.
Q7.3.18
A survey of \(21,250\) households concerning telephone service gave the results shown in the table.
Landline  No Landline  

Cell phone  12,474  5,844 
No cell phone  2,529  403 
 Give a point estimate for the proportion of all households in which there is a cell phone but no landline.
 Assuming the sample is sufficiently large, construct a \(99.9\%\) confidence interval for the proportion of all households in which there is a cell phone but no landline.
 Give a point estimate for the proportion of all households in which there is no telephone service of either kind.
 Assuming the sample is sufficiently large, construct a \(99.9\%\) confidence interval for the proportion of all all households in which there is no telephone service of either kind.
Additional Exercises
 In a random sample of \(900\) adults, \(42\) defined themselves as vegetarians. Of these \(42\), \(29\) were women.
 Give a point estimate of the proportion of all selfdescribed vegetarians who are women.
 Verify that the sample is sufficiently large to use it to construct a confidence interval for that proportion.
 Construct a \(90\%\) confidence interval for the proportion of all all selfdescribed vegetarians who are women.
 A random sample of \(185\) college soccer players who had suffered injuries that resulted in loss of playing time was made with the results shown in the table. Injuries are classified according to severity of the injury and the condition under which it was sustained.
Minor Moderate Serious Practice 48 20 6 Game 62 32 17  Give a point estimate for the proportion \(p\) of all injuries to college soccer players that are sustained in practice.
 Construct a \(95\%\) confidence interval for the proportion \(p\) of all injuries to college soccer players that are sustained in practice.
 Give a point estimate for the proportion \(p\) of all injuries to college soccer players that are either moderate or serious.
 Construct a \(95\%\) confidence interval for the proportion \(p\) of all injuries to college soccer players that are either moderate or serious.
 The body mass index (BMI) was measured in \(1,200\) randomly selected adults, with the results shown in the table.
BMI Under 18.5 18.5–25 Over 25 Men 36 165 315 Women 75 274 335  Give a point estimate for the proportion of all men whose BMI is over \(25\).
 Assuming the sample is sufficiently large, construct a \(99\%\) confidence interval for the proportion of all men whose BMI is over \(25\).
 Give a point estimate for the proportion of all adults, regardless of gender, whose BMI is over \(25\).
 Assuming the sample is sufficiently large, construct a \(99\%\) confidence interval for the proportion of all adults, regardless of gender, whose BMI is over \(25\).
 Confidence intervals constructed using the formula in this section often do not do as well as expected unless \(n\) is quite large, especially when the true population proportion is close to either \(0\) or \(1\). In such cases a better result is obtained by adding two successes and two failures to the actual data and then computing the confidence interval. This is the same as using the formula \[\tilde{p}\pm z_{\alpha /2}\sqrt{\frac{\tilde{p}(1\tilde{p})}{\tilde{n}}}\\ \text{where}\\ \tilde{p}=\frac{x+2}{n+4}\; \text{and}\; \tilde{n}=n+4\]
Suppose that in a random sample of \(600\) households, \(12\) had no telephone service of any kind. Use the adjusted confidence interval procedure just described to form a \(99.9\%\) confidence interval for the proportion of all households that have no telephone service of any kind.
Large Data Set Exercises
Large Data Set missing from the original
 Large \(\text{Data Sets 4 and 4A}\) list the results of \(500\) tosses of a die. Let \(p\) denote the proportion of all tosses of this die that would result in a four. Use the sample data to construct a \(90\%\) confidence interval for \(p\).
 Large \(\text{Data Set 6}\) records results of a random survey of \(200\) voters in each of two regions, in which they were asked to express whether they prefer Candidate \(A\) for a U.S. Senate seat or prefer some other candidate. Use the full data set (\(400\) observations) to construct a \(98\%\) confidence interval for the proportion \(p\) of all voters who prefer Candidate \(A\).
 Lines \(2\) through \(536\) in \(\text{Data Set 11}\) is a sample of \(535\) real estate sales in a certain region in \(2008\). Those that were foreclosure sales are identified with a \(1\) in the second column.
 Use these data to construct a point estimate \(\hat{p}\) of the proportion \(p\) of all real estate sales in this region in \(2008\) that were foreclosure sales.
 Use these data to construct a \(90\%\) confidence for \(p\).
 Lines \(537\) through \(1106\) in Large \(\text{Data Set 11}\) is a sample of \(570\) real estate sales in a certain region in \(2010\). Those that were foreclosure sales are identified with a \(1\) in the second column.
 Use these data to construct a point estimate \(\hat{p}\) of the proportion \(p\) of all real estate sales in this region in \(2010\) that were foreclosure sales.
 Use these data to construct a \(90\%\) confidence for \(p\).
Answers

 \((0.5492, 0.8508)\)
 \((0.5934, 0.8066)\)

 \((0.2726, 0.5274)\)
 \((0.3368, 0.4632)\)

 \(0.3073\)
 \(\hat{p}\pm 3\sqrt{\frac{\hat{p}\hat{q}}{n}}=0.31\pm 0.04\; \text{and}\; [0.27,0.35]\subset [0,1]\)
 \((0.2895, 0.3251)\)
 \((0.2844, 0.3302)\)
 Asking for greater confidence requires a longer interval.

 \(0.9956\)
 \((0.9862, 1.005)\)

 \(0.244\)
 \((0.1740, 0.3140)\)

 \(0.013\)
 \((0.01, 0.016)\)
 \((0.011, 0.015)\)

 \(0.52\)
 \((0.4038, 0.6362)\)

 \(0.52\)
 \((0.4726, 0.5674)\)

 \(0.5824\)
 \((0.5333, 0.6315)\)

 \(0.69\)
 \(\hat{p}\pm 3\sqrt{\frac{\hat{p}\hat{q}}{n}}=0.69\pm 0.21\; \text{and}\; [0.48,0.90]\subset [0,1]\)
 \(0.69\pm 0.12\)

 \(0.6105\)
 \((0.5552, 0.6658)\)
 \(0.5583\)
 \((0.5214, 0.5952)\)
 \((0.1368,0.1912)\)

 \(\hat{p}=0.2280\)
 \((0.1982,0.2579)\)
7.4: Sample Size Considerations
Basic
 Estimate the minimum sample size needed to form a confidence interval for the mean of a population having the standard deviation shown, meeting the criteria given.
 \(\sigma = 30, 95\%\) confidence, \(E = 10\)
 \(\sigma = 30, 99\%\) confidence, \(E = 10\)
 \(\sigma = 30, 95\%\) confidence, \(E = 5\)
 Estimate the minimum sample size needed to form a confidence interval for the mean of a population having the standard deviation shown, meeting the criteria given.
 \(\sigma = 4, 95\%\) confidence, \(E = 1\)
 \(\sigma = 4, 99\%\) confidence, \(E = 1\)
 \(\sigma = 4, 95\%\) confidence, \(E = 0.5\)
 Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given.
 \(p\approx 0.37, 80\%\) confidence, \(E = 0.05\)
 \(p\approx 0.37, 90\%\) confidence, \(E = 0.05\)
 \(p\approx 0.37, 80\%\) confidence, \(E = 0.01\)
 Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given.
 \(p\approx 0.81, 95\%\) confidence, \(E = 0.02\)
 \(p\approx 0.81, 99\%\) confidence, \(E = 0.02\)
 \(p\approx 0.81, 95\%\) confidence, \(E = 0.01\)
 Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given.
 \(80\%\) confidence, \(E = 0.05\)
 \(90\%\) confidence, \(E = 0.05\)
 \(80\%\) confidence, \(E = 0.01\)
 Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given.
 \(95\%\) confidence, \(E = 0.02\)
 \(99\%\) confidence, \(E = 0.02\)
 \(95\%\) confidence, \(E = 0.01\)
Applications
 A software engineer wishes to estimate, to within \(5\) seconds, the mean time that a new application takes to start up, with \(95\%\) confidence. Estimate the minimum size sample required if the standard deviation of start up times for similar software is \(12\) seconds.
 A real estate agent wishes to estimate, to within \(\$2.50\), the mean retail cost per square foot of newly built homes, with \(80\%\) confidence. He estimates the standard deviation of such costs at \(\$5.00\). Estimate the minimum size sample required.
 An economist wishes to estimate, to within \(2\) minutes, the mean time that employed persons spend commuting each day, with \(95\%\) confidence. On the assumption that the standard deviation of commuting times is \(8\) minutes, estimate the minimum size sample required.
 A motor club wishes to estimate, to within \(1\) cent, the mean price of \(1\) gallon of regular gasoline in a certain region, with \(98\%\) confidence. Historically the variability of prices is measured by \(\sigma =\$0.03\)Estimate the minimum size sample required.
 A bank wishes to estimate, to within \(\$25\), the mean average monthly balance in its checking accounts, with \(99.8\%\) confidence. Assuming \(\sigma =\$250\), estimate the minimum size sample required.
 A retailer wishes to estimate, to within \(15\) seconds, the mean duration of telephone orders taken at its call center, with \(99.5\%\) confidence. In the past the standard deviation of call length has been about \(1.25\) minutes. Estimate the minimum size sample required. (Be careful to express all the information in the same units.)
 The administration at a college wishes to estimate, to within two percentage points, the proportion of all its entering freshmen who graduate within four years, with \(90\%\) confidence. Estimate the minimum size sample required.
 A chain of automotive repair stores wishes to estimate, to within five percentage points, the proportion of all passenger vehicles in operation that are at least five years old, with \(98\%\) confidence. Estimate the minimum size sample required.
 An internet service provider wishes to estimate, to within one percentage point, the current proportion of all email that is spam, with \(99.9\%\) confidence. Last year the proportion that was spam was \(71\%\). Estimate the minimum size sample required.
 An agronomist wishes to estimate, to within one percentage point, the proportion of a new variety of seed that will germinate when planted, with \(95\%\) confidence. A typical germination rate is \(97\%\). Estimate the minimum size sample required.
 A charitable organization wishes to estimate, to within half a percentage point, the proportion of all telephone solicitations to its donors that result in a gift, with \(90\%\) confidence. Estimate the minimum sample size required, using the information that in the past the response rate has been about \(30\%\).
 A government agency wishes to estimate the proportion of drivers aged \(1624\) who have been involved in a traffic accident in the last year. It wishes to make the estimate to within one percentage point and at \(90\%\) confidence. Find the minimum sample size required, using the information that several years ago the proportion was \(0.12\).
Additional Exercises
 An economist wishes to estimate, to within six months, the mean time between sales of existing homes, with \(95\%\) confidence. Estimate the minimum size sample required. In his experience virtually all houses are resold within \(40\) months, so using the Empirical Rule he will estimate \(\sigma\) by onesixth the range, or \(40/6=6.7\).
 A wildlife manager wishes to estimate the mean length of fish in a large lake, to within one inch, with \(80\%\) confidence. Estimate the minimum size sample required. In his experience virtually no fish caught in the lake is over \(23\) inches long, so using the Empirical Rule he will estimate \(\sigma\) by onesixth the range, or \(23/6=3.8\).
 You wish to estimate the current mean birth weight of all newborns in a certain region, to within \(1\) ounce (\(1/16\) pound) and with \(95\%\) confidence. A sample will cost \(\$400\) plus \(\$1.50\) for every newborn weighed. You believe the standard deviations of weight to be no more than \(1.25\) pounds. You have \(\$2,500\) to spend on the study.
 Can you afford the sample required?
 If not, what are your options?
 You wish to estimate a population proportion to within three percentage points, at \(95\%\) confidence. A sample will cost \(\$500\) plus \(50\) cents for every sample element measured. You have \(\$1,000\) to spend on the study.
 Can you afford the sample required?
 If not, what are your options?
Answers

 \(35\)
 \(60\)
 \(139\)

 \(154\)
 \(253\)
 \(3832\)

 \(165\)
 \(271\)
 \(4109\)
 \(23\)
 \(62\)
 \(955\)
 \(1692\)
 \(22,301\)
 \(22,731\)
 \(5\)

 no
 decrease the confidence level