Sampling Distribution HW Answer Key
Name _______________________________ Period _____________ Chapter 9 Sampling Distributions Homework 1. 2. 3. 4. 5. Identify each boldface number as a parameter or a statistic and use the appropriate notation. A carload lot of ball bearings has a mean diameter of 2.5003 cm. An inspector chooses 100 bearings from the lot that and finds they have a mean diameter of 2.5009 cm. ππ = 2.5003 π₯π₯Μ = 2.5006 population parameter sample statistic ππΜ = .072 sample statistic The Bureau of Labor Statistics last month interviewed 60,000 members of the U.S. labor force, of whom 7.2% were unemployed. A telemarketing firm in Los Angeles uses a device that dials residential telephone numbers in that city at random. Of the first 100 numbers dialed, 48% are unlisted. This is not surprising because 52% of all Los Angeles residential phones are unlisted. ππΜ = .48 ππ = .52 sample statistic population parameter π₯π₯Μ = 335 π₯π₯Μ = 289 sample statistic sample statistic A researcher carries out a randomized comparative experiment with young rats. After 8 weeks, the mean weight gain is 335 grams for the control group and 289 grams for the experimental group. The figure below shows histograms of four sampling distributions of statistics intended to estimate the same parameter. Label each distribution relative to the others as having large or small bias and as having large or small variability. Large bias; large variability Small bias; large variability Small bias; small variability Large bias; small variability 6. The IRS plans to examine an SRS of individual federal income tax returns from each state. One variable of interest is the proportion of returns claiming itemized deductions. The total number of tax returns in each state varies from almost 14 million in California to fewer than 210,000 in Wyoming. (a) Will the sampling variability of the sample proportion change from state to state if an SRS of 2000 tax returns is selected from each state? Explain. 10 β 2000 = 20000 NO, the 10% condition has not been met; both populations are greater than 20000 people. (b) Will the sampling variability of the sample proportion change from state to state if an SRS of 1% of all tax returns is selected in each state? Explain. 7. . 01 β 14000000 = 140000 ππππππ .01 β 210000 = 2100 Yes. A larger sample (140000) will have less variability. A USA Today poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Of the respondents, 67% said that they drink it. Suppose that 70% of U.S. adults actually drink the cereal milk. (a) Find the mean and the standard deviation of the proportion ππΜ of the sample that say they drink the cereal milk. πππποΏ½ = .7 .7β.3 πππποΏ½ = οΏ½1012 = .014405 (b) Explain why you can use the formula for the standard deviation of ππΜ in this setting (rule of thumb #1). 10 β 1012 = 10120 The 10% condition has been met; there are more than 10120 US adults. (c) Check that you can use the normal approximation for the distribution of ππΜ (rule of thumb #2). 1012 β .7 β₯ 10, 708.4 β₯ 10 1012 β .3 β₯ 10, 303.6 β₯ 10 (d) Find the probability of obtaining a sample of 1012 adults in which 67% or fewer say the drink the cereal milk. Do you have any doubts about the result of the poll? normCdf(-9E999, .67, .7, .014405) = .018643. There is a very small chance (< 2%) that 67% or fewer drink the milk. (e) What sample size would be required to reduce the standard deviation of the sample proportion to one-half the value you found in part (a)? 8. Four times the sample in (a); 4 β 1012 = 4048 adults in the sample The Gallop Poll asked a probability sample of 1785 adults whether they attended church or synagogue during the past week. Suppose that 40% of the adult population did attend. We would like to know the probability that an SRS of size 1785 would come within plus or minus 3 percentage points of this true value. οΏ½ is the proportion of the sample who did attend church or synagogue, what is the mean of the (a) If ππ οΏ½? What is the standard deviation? sampling distribution of ππ .4β.6 πππποΏ½ = .4 πππποΏ½ = οΏ½1785 = .011595 (b) Explain why you can use the formula for the standard deviation of ππΜ in this setting (rule of thumb #1). 10 β 1785 = 17850 The 10% condition has been met; there are more than 17850 adults. οΏ½ (rule of thumb #2). (c) Check that you can use the normal approximation for the distribution of ππ 1785 β .4 β₯ 10, 714 β₯ 10 1785 β .6 β₯ 10, 1071 β₯ 10 (d) Find the probability that ππΜ takes a value between 0.37 and 0.43. Will an SRS of size 1785 usually οΏ½ within plus or minus 3 percentage points of the true population proportion? Explain. give a result ππ 9. normCdf(.37, .43, .4, .011595) = .990327. Yes, over 99% of all samples should give ππΜ within ±3% of the true population proportion. Problem #8 asks the probability that ππΜ from an SRS estimates p = 0.4 within 3 percentage points. Find this probability for SRSs of size 300, 1200, and 4800. What general fact do your results illustrate? For n = 300 For n = 1200 For n = 4800 πππποΏ½ = οΏ½.4β.6 = .0283 300 .4β.6 πππποΏ½ = οΏ½1200 = .0141 .4β.6 πππποΏ½ = οΏ½4800 = .00707 normCdf(.37, .43, .4, .0283) ππ(. 37 β€ ππΜ β€ .43) = .7109 normCdf(.37, .43, .4, .0141) ππ(. 37 β€ ππΜ β€ .43) = .9666 normCdf(.37, .43, .4, .00707) ππ(. 37 β€ ππΜ β€ .43) = .99998 For larger sample sizes, the sample proportions are more likely to be close to the true population proportion. 10. Your mail-order company advertises that it ships 90% of its orders within three working days. You select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals that 86 of these orders were shipped on time. (a) What is the sample proportion of orders shipped on time? 86 ππΜ = 100 = .86 (b) If the company really ships 90% of its orders on time, what is the probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller? 10 β 100 = 1000 The 10% condition has been met; there are more than 1000 orders. 100 β .9 β₯ 10, 90 β₯ 10 ; 100 β .1 β₯ 10, 10 β₯ 10; πππποΏ½ = .9 11. πππποΏ½ = οΏ½.9β.1 = .03 100 normCdf(-9E999, .86, .9, .03) = .091211. ππ(ππΜ β€ .86) = .091211 It happens about 9% of the time, a small probability, but not unlikely. Explain why you can not use the methods you learned in 9.2 to find the following probabilities: (a) A factory employs 3000 unionized workers, of whom 30% are Hispanic. The 15-member union executive committee contains 3 Hispanics. What would be the probability of 3 or fewer Hispanics if the executive committee were chosen at random from all the workers? 15 β .3 β₯ 10, 4.5 β₯ 10 Not Normal, try binomial probability (b) A university is concerned about the academic standing of its intercollegiate athletes. A study committee chooses an SRS of 50 of the 316 athletes to interview in detail. Suppose that in fact 40% of the athletes have been told by coaches to neglect their studies on at least one occasion. What is the probability that at least 15 in the sample are among this group? 10 β 50 = 500 The 10% condition has not been met; there are only 316 athletes. (c) How could you find the probability for (a)? Find it. 12. ππ(π₯π₯ β€ 3) binomialCdf(15, .3, 0, 3) = .296868 Taxi fares are normally distributed with mean fare $22.27 and standard deviation $2.20. a) Which should have the greater probability of falling between $21 and $24 β the mean of 10 random taxi fares or the amount of a single random taxi fare? Why? b) The mean of 10 random taxi fares. Larger sample size has less variability; more likely to be closer to the true mean Which should have a greater probability of being greater than $24 β the mean of 10 random taxi fares or the amount of a single taxi fare? Why? A single taxi fare. It is more likely that 1 value will be farther away from the mean as opposed to the mean of a sample of 10 taxi fares.. 13. Suppose a sample of n = 50 items is drawn from a population of manufactured products and the weight, x, of each item is recorded. Prior experience has shown that the weight has a mean of 6 ounces and standard deviation of 2.5 ounces. a) What is the shape of the sampling distribution of x ? The distribution will be approximately Normal due to the large sample size. b) c) d) What is the mean and standard deviation of the sampling distribution? πππ₯π₯Μ = 6 ; πππ₯π₯Μ = 2.5 β50 = .353553; 10(50) = 500; There are more than 500 manufactured products What is the probability that the manufacturerβs sample has a mean weight of less than 5 ounces? If you did get a sample mean of 5 or less, does this indicate that the manufacturing process may be faulty? Explain. Use CLT n = 50 > 30. normCdf(-9E999, 5, 6, .353553); ππ(π₯π₯Μ β€ 5) = .002339. This will happen less than 1% of the time, the manufacturing process is most likely faulty. How would the sampling distribution of x change if the sample size, n, were increased from 50 to 100? 2.5 πππ₯π₯Μ = β100 = .25; It would have less variability; the sampling distribution would be closer to the mean. 14. A soft-drink bottle vendor claims that its production process yields bottles with a mean internal strength of 157 psi (pounds per square inch) and a standard deviation of 3 psi and is normally distributed. As part of its vendor surveillance, a bottler strikes an agreement with the vendor that permits the bottler to sample from the vendorβs production to verify the vendorβs claim. N(157, 3) a) b) c) Suppose the bottler randomly selects a single bottle to sample. What is the mean and 3 standard deviation? πππ₯π₯Μ = 157 ; πππ₯π₯Μ = β1 =3 What is the probability that the psi of the single bottle is 1.3 psi or more below the process mean? normCdf(-9E999, 155.7, 157, 3); ππ(π₯π₯ β€ 155.7) = .332386. Suppose the bottler randomly selected 40 bottles from the last 10,000 produced. Describe the sampling distribution of the sample means? 10(40) = 400 and we are looking at the last 10,000. N(157, .474342) d) e) πππ₯π₯Μ = 157 ; πππ₯π₯Μ = 3 β40 = .474342 What is the probability that the sample mean of the 40 bottles is 1.3 psi or more below the process mean? normCdf(-9E999, 155.7, 157, .474342); ππ(π₯π₯Μ β€ 155.7) = .003066. In order to reduce the standard deviation 50% (half), how large would the sample size need to be? 4 times as large; 4(40) = 160. πππ₯π₯Μ = 3 β160 = .237171 15. The distribution of weights of chips in these bags is normally distributed with mean 10 ounces and standard deviation 0.12 ounces. N(10, .12) a) b) If a bag of chips is selected at random, what is the probability that it weighs less than 9.95 ounces? normCdf(-9E999, 9.95, 10, .12); ππ(π₯π₯ β€ 9.95) = .338461. Describe the sampling distribution of sample means for samples of size 12 (n = 12). 10(12) = 120; there are more than 120 bags of chips. πππ₯π₯Μ = 10; πππ₯π₯Μ = N(10, .034641) c) d) normCdf(-9E999, 9.95, 10, .034641); ππ(π₯π₯Μ β€ 9.95) = .074457. For random samples of 5 bags, what mean weight is at the 60th percentile? invNorm(.6, 10, .053666) π₯π₯Μ = 10.0136 .12 β5 = .053666 You randomly select 20 mortgage institutions and determine the current mortgage rate at each. The current mean mortgage interest rate is known to be 6.93% with a standard deviation of 0.42% and is known to be approximately normal. a) b) c) 17. = .034641 Whatβs the probability that the mean weight of a random sample of 12 bags weighs less than 9.95 ounces? 10(5) = 50; there are more than 50 bags of chips. πππ₯π₯Μ = 10; πππ₯π₯Μ = 16. .12 β12 Construct the sampling distribution for the sample of 20 mortgage institutions. 10(20) = 200; there are more than 200 mortgage institutions. πππ₯π₯Μ = 6.93; .42 πππ₯π₯Μ = = .093915 N(6.93, .093915) or N(.0693, .00093915) β20 What is the probability that a sample of mortgage rates is above 7%? normCdf(7, 9E999, 6.93, .09315); ππ(π₯π₯Μ β₯ 7) = .228029. What is the probability that a sample of mortgage rates is between 6% and 6.5%? normCdf(6, 6.5, 6.93, .09315); ππ(6 < π₯π₯Μ < 6.5) = .000002. You randomly select 50 newly constructed houses in a particular area of Oregon. The mean construction cost in this area is $181,000 with a standard deviation of $28,000. a) b) c) Construct a sampling distribution for a sample of size 50. 10(50) = 500; there are more than 500 newly constructed home in a particular area of 28000 Oregon. πππ₯π₯Μ = 181000; πππ₯π₯Μ = = 3959.8 N(181000, 3959.8) β50 What is the probability that a sample of new houses averages less than $170,000? normCdf(-9E999, 170000, 181000, 3959.8); ππ(π₯π₯Μ β€ 170000) = .002735. What is the minimum cost of a house in the 80th percentile? Canβt do the problem. n = 1 and it is not stated that the population is approximately normally distributed. 18. You randomly select 18 adult male athletes and measure the resting heart rate of each. The mean heart rate is 64 beats per minute with a standard deviation of 2.5 beats per minute. It is safe to assume the resting heart rates are approximately normal. a) b) c) What range of resting heart beats mark the middle 95% of all the heart rates? Within 2ππ; 64 ± 2(2.5) (59, 69) What is the probability that an adult male athlete has resting heart rate below 60 beats per minute? normCdf(-9E999, 60, 64, 2.5); ππ(π₯π₯ < 60) = .054799. Would you consider it unusual if your sample of athletes had a resting heart rate above 65.4 beats per minute? 10(18) = 180; there are more than 180 adult male athletes. πππ₯π₯Μ = 64; 2.5 πππ₯π₯Μ = = .589256 N(64, .589256) β18 19. normCdf(65.4, 9E999, 64, .589256); ππ(π₯π₯Μ β₯ 65.4) = .008754. Yes because this happens less than 1% of the time. A football filled with helium was kicked on a windless day at Katy High School. The distances (in yards) are listed below. The average distance kicked of helium footballs is known to be 27.5 yards with a standard deviation of 6.3 yards. (Note: you arenβt told that the population distribution is normal, and the n < 30, you can infer normality from sample data. Check graphically!). 11 29 a) b) 12 30 14 30 22 30 23 30 24 31 26 31 26 31 26 32 27 32 28 33 28 34 29 35 29 39 29 What is the probability that a sample this size will have a mean distance of more than 30 yards? What is the probability that a sample this size will have a mean distance between 25 and 30 yards? The histogram does not appear to be Normally distributed. The boxplot shows outliers on both sides. We canβt use Normal calculations for this distribution. 20. Fourteen randomly selected people with a bachelorβs degree in economics were asked their monthly salary. According to the U.S. Bureau of Labor Statistics, the average monthly salary is approximately $4418 with a standard deviation of $389. The monthly salaries are listed below: 4450.66 4527.64 4596.73 4407.34 4366.66 5036.64 Check normality graphically! a) b) 4455.40 5083.73 4151.70 3946.47 3727.08 4023.41 4283.76 4806.22 The histogram, boxplot and NPP do not have unusual features; we can use Normal calculations for this distribution. What is the probability that a sample of 14 economic graduates will have an average monthly salary of over $4600? 10(14) = 140; there are more than 140 economic graduates. πππ₯π₯Μ = 4418; 389 πππ₯π₯Μ = = 103.965 N(4418, 103.965) β14 normCdf(4600, 9E999, 4418, 103.965); ππ(π₯π₯Μ > 4600) = .040008. How would the sampling distributionβs standard deviation change if the sample size is quadrupled? Be specific. ππβππππ ππ = 14, πππ₯π₯Μ = 389 β14 β4 = 2, π π π π ππππ π€π€π€π€π€π€ππππ ππππ ππππππ ππππ βππππππ. 389 = 103.965, π€π€βππππ ππ = 56, πππ₯π₯Μ = = 51.9823 β56 21. According to the EPA, the mean waste generated per person per day was 4.5 pounds with a standard deviation of 1.2 pounds. a) What is the probability that a sample of 20 people would have a mean waste of more than 5.3 pounds per day? n = 20 we donβt have the data and we were not told anything about the population so we canβt verify Normality. b) What is the probability that a sample of 45 people would have a mean waste of more than 5 pounds per day? n = 45 > 30, Central Limit Theorem can be used to verify Normality. 10(45) = 450; there are more than 450 people 1.2 πππ₯π₯Μ = 4.5; πππ₯π₯Μ = = .178885 N(4.5, .178885) β45 normCdf(5, 9E999, 4.5, .178885); ππ(π₯π₯Μ > 5) = .002594.
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