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Basic Statistics Understanding Conventional Methods And Modern Insights 1st Edition Rand R. Wilcox - Solutions
Summarize when it would and would not be reasonable to assume that the sampling distribution of M is normal.
For the values 59,106,174,207,219,237,313,365,458,497,515, 529,557,615,625,645,973,1065,3215, estimate the standard error of the 20% trimmed mean.
For the data in problem 25, why would you suspect that the standard error of the sample mean will be larger than the standard error of the 20% trimmed mean?Verify that this speculation is correct.
The ideal estimator of location would have a smaller standard error than any other estimator we might use. Explain why such an estimator does not exist.
Under normality, the sample mean has a smaller standard error than the 20%trimmed mean or median. If observations are sampled from a distribution that appears to be normal, does this suggest that the mean should be preferred over the trimmed mean and median?
If the sample mean and 20% trimmed mean are nearly identical, it might be thought that for future studies, it will make little difference which measure of location is used. Comment on why this is not necessarily the case.
Imagine that you are able to generate 25 observations on a computer from the distribution shown in the left panel of figure 5.4. Outline how you would determine the sampling distribution of the sample median, M. In particular, how would you determine the probability that M will have a value less
Explain the meaning of a .95 confidence interval.
If you want to compute a .80, or .92, or a .98 confidence interval for μ when σ is known, and sampling is from a normal distribution, what values for c should you use in equation (6.3)?
Assuming random sampling is from a normal distribution with standard deviationσ = 5, if you get a sample mean of X¯ = 45 based on n = 25 subjects, what is the.95 confidence interval for μ?
Repeat the previous example, only compute a .99 confidence interval instead.
A manufacturer claims that their light bulbs have an average life span that follows a normal distribution with μ = 1,200 hours and a standard deviation of σ = 25.If you randomly test 36 light bulbs and find that their average life span is X¯ = 1,150, does a .95 confidence interval for μ suggest
For the following situations, (a) n = 12, σ = 22, X¯ = 65, (b) n = 22, σ = 10, X¯ = 185, (c) n = 50, σ = 30, X¯ = 19, compute a .95 confidence interval for the mean assuming normality.
What happens to the length of a confidence interval for the mean of a normal distribution when the sample size is doubled? What happens if it is quadrupled?
The length of a bolt made by a machine parts company is a normal random variable with standard deviation σ equal to .01 mm. The lengths of four randomly selected bolts are: 20.01, 19.88, 20.00, 19.99. (a) Compute a .95 confidence interval of the mean. (b). Specifications require a mean length μ
The weight of trout sold at a trout farm has a standard deviation of .25. Based on a sample of 10 trout, the average weight is 2.10 lb. Compute a .99 confidence interval for the population mean, assuming normality.
A machine to measure the bounce of a ball is used on 45 randomly selected tennis balls. Experience has shown that the standard deviation of the bounce is .30. If X¯ = 1.70, and assuming normality, what is a .90 confidence interval for the average bounce?
Assuming the degrees of freedom are 20, find the value c for which(a) P(T >c) = .025,(b) P(T ≤c) = .995(c) P(−c ≤ T ≤c) = .90.
Compute a .95 confidence interval if(a) n = 10, X¯ = 26, s = 9,(b) n = 18, X¯ = 132, s = 20,(c) n = 25, X¯ = 52, s = 12.
Repeat the previous exercise, but compute a .99 confidence interval instead.
For a study on self-awareness, the observed values for one of the groups were 77,87,88,114,151,210,219,246,253,262,296, 299,306,376,428,515,666,1,310,2,611.Compute .95 confidence interval for the mean assuming normality.
Rats are subjected to a drug that might affect aggression. Suppose that for a random sample of rats, measures of aggression are found to be 5,12,23,24,18,9,18,11,36,15.Compute a .95 confidence interval for the mean assuming the scores are from a normal distribution.
Suppose M = 34 and the McKean–Schrader estimate of the standard error of M is SM = 3.Compute a .95 confidence interval for the population median.
For the data in problem 14 of section 6.2, the McKean–Schrader estimate of the standard error of M is SM = 77.8 and the sample median is 262.Compute a .99 confidence interval for the population median.
If n = 10 and you compute a confidence interval for the median using(X(k), X(n−k)), as described in box 6.1, what is the probability the probability that this interval contains the population median if k = 2?
Repeat the previous problem, only with n = 15 and k = 4.
For the data in problem 14 of section 6.2, if we use (88, 515) as a confidence interval for the median, what is the probability that this interval contains the population median?
You observe the following successes and failures: 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.Compute a .95 confidence interval for p using equation (6.10) as well as the Agresti-Coull method.
Given the following results for a sample from a binomial distribution, compute the squared standard error of pˆ when (a) n = 25, X = 5, (b) n = 48, X = 12,(c) n = 100, X = 80, (d) n = 300, X = 160.
Among 100 randomly sampled adults, 10 were found to be unemployed. Give a.95 confidence interval for the percentage of adults unemployed using equation (6.10). Compare this result to the Agresti-Coull confidence interval.
A sample of 1,000 fish was obtained from a lake. It was found that 290 were members of the bass family. Give a .95 confidence interval for the percentage of bass in the lake using equation (6.10).
Among a random sample of 1,000 adults, 60 reported never having any legal problems. Give a .95 confidence interval for the percentage of adults who never had legal problems using equation (6.10).
Among a sample of 600 items, only one was found to be defective. Explain why using equation (6.10) might be unsatisfactory when computing a confidence interval for the probability that an item is defective.
In the previous problem, compute a .90 confidence interval for the probability that an item is defective.
One-fourth of 300 persons in a large city stated that they are opposed to a certain political issue favored by the mayor. Calculate a 99% confidence interval for the fraction of people of individuals opposed to this issue using equation (6.10).
A test to detect a certain type of cancer has been developed and it is of interest to know the probability of a false-negative indication, meaning the test fails to detect cancer when it is present. The test is given to 250 patients known to have cancer and five tests fail to show its presence.
In the previous problem, imagine that 0 false-negative indications were found.Determine a .99 confidence interval for the probability of a false-negative indication.
A cosmetic company found that 180 of 1,000 randomly selected women in New York city have seen the company’s latest television advert. Compute a .95 confidence interval for the percentage of women in New York city who have seen the advert using equation (6.10).
Describe in general terms how non-normality can affect Student’s t-distribution.
Chapter 5 illustrated that when (randomly) sampling observations from a skewed distribution where outliers are rare, it is generally reasonable to assume that plots of sample means over many studies has, approximately, a normal distribution. This means that reasonably accurate confidence intervals
Listed here are the average LSAT scores for the 1973 entering classes of 15 American law schools.545 555 558 572 575 576 578 580 594 605 635 651 653 661 666(LSAT is a national test for prospective lawyers.) The .95 confidence interval for the population μ is (577.1,623.4). (a) Use a boxplot to
Compute a .95 confidence for the trimmed mean if (a) n = 24, s 2w = 12, X¯t = 52,(b) n = 36, s 2w = 30, X¯t = 10, (c) n = 12, s 2w = 9, X¯t = 16.
Repeat the previous exercise, but compute a .99 confidence interval instead.
Problem 14 (in this chapter) used data from a study of self-awareness. In another portion of the study, a group of participants had the following values.59,106,174,207,219,237,313,365,458,497,515, 529,557,615,625,645,973,1,065,3,215.Compute a .95 confidence interval for both the population mean and
The ideal estimator of location would have a smaller standard error than any other estimator we might use. Explain why such an estimator does not exist.
For the values 5,60,43,56,32,43,47,79,39,41, compute a .95 confidence interval for the trimmed mean and compare the results to the .95 confidence interval for the mean.
In the previous exercise, the confidence interval for the 20% trimmed mean is shorter than the confidence interval for the mean. Explain why this is not surprising
Given that X¯ = 78, σ2 = 25, n = 10 and α = .05, test H0 : μ > 80, assuming observations are randomly sampled from a normal distribution. Also, draw the standard normal distribution indicating where Z and the critical value are located.
Repeat the previous problem but test H0 : μ = 80.
For problem 2, compute a .95 confidence interval and verify that this interval is consistent with your decision about whether to reject the null hypothesis.
For problem 1, determine the p-value.
For problem 2, determine the p-value.
Given that X¯ = 120, σ = 5, n = 49 and α = .05, test H0 : μ > 130, assuming observations are randomly sampled from a normal distribution.
Repeat the previous problem but test H0 : μ = 130.
For the previous problem, compute a .95 confidence interval and compare the result with your decision about whether to reject H0.
If X¯ = 23 and α = .025, can you make a decision about whether to reject H0 : μ < 25 without knowing σ?
An electronics firm mass produces a component for which there is a standard measure of quality. Based on testing vast numbers of these components, the company has found that the average quality is μ = 232 with σ = 4.However, in recent years the quality has not been checked, so management asks you
An antipollution device for cars is claimed to have an average effectiveness of exactly 546.Based on a test of 20 such devices you find that X¯ = 565.Assuming normality and that σ = 40, would you rule out the claim with a Type I error probability of .05?
Comment on the relative merits of using a .95 confidence interval for addressing the effectiveness of the antipollution device in the previous problem.
For n = 25, α = .01, σ = 5 and H0 : μ ≥ 60, verify that power is .95 when μ = 56.
For n = 36, α = .025, σ = 8 and H0 : μ ≤ 100, verify that power is .61 whenμ = 103.
For n = 49, α = .05, σ = 10 and H0 : μ = 50, verify that power is approximately.56 when μ = 47.
A manufacturer of medication for migraine headaches knows that their product can damage the stomach if taken too often. Imagine that by a standard measuring process, the average damage is μ = 48.A modification of their product is being contemplated, and based on ten trials, it is found that X¯ =
For the previous problem, verify that power is .35 if μ = 46.
The previous problem indicates that power is relatively low with only n = 10 observations. Imagine that you want power to be at least .8. One way of getting more power is to increase the sample size, n. Verify that for sample sizes of 20, 30, and 40, power is .56, .71 and .81, respectively.
For the previous problem, rather than increase the sample size, what else might you do to increase power? What is a negative consequence of using this strategy?
Given the following values for X¯ and s: (a) X¯ = 44, s = 10, (b) X¯ = 43, s = 10,(c) X¯ = 43, s = 2, test the hypothesis H0: μ = 42 with α = .05 and n = 25.
For part b of the last problem you fail to reject but you reject for the situation in partc. What does this illustrate about power?
Given the following values for X¯ and s: (a) X¯ = 44, s = 10, (b) X¯ = 43, s = 10,(c) X¯ = 43, s = 2, test the hypothesis H0: μ < 42 with α = .05 and n = 16.
Repeat the previous problem only test H0: μ > 42.
A company claims that on average, when exposed to their toothpaste, 45% of all bacteria related to gingivitis is killed. You run 10 tests and find that the percentages of bacteria killed among these tests are 38, 44, 62, 72, 43, 40, 43, 42, 39, 41.The mean and standard deviation of these values are
A portion of a study by Wechsler (1958) reports that for 100 males taking the Wechsler Adult Intelligent Scale (WAIS), the sample mean and variance on picture completion are X¯ = 9.79 and s = 2.72. Test the hypothesis H0: μ ≥ 10.5 with α = .025.
Given that n = 16, X¯ = 40, and s = 4, test H0: μ ≤ 38 with α = .01.
Given that n = 9, X¯ = 76, and s = 4, test H0: μ = 32 with α = .05.
An engineer believes it takes an average of 150 man-hours to assemble a portion of an automobile. As a check, the time to assemble 10 such parts was ascertained yielding X¯ = 146 and s = 2.5. Test the engineer’s belief with α = .05.
In a study of court administration, the following times to disposition were determined for twenty cases and found to be 42,90,84,87,116,95,86,99,93,92 121,71,66,98,79,102,60,112,105,98.Test the hypothesis that the average time to disposition is less than or equal to 80, using α = .01.
Given the following values for X¯t and sw: (a) X¯t = 44, sw = 9, (b) X¯t = 43, sw = 9, (c) X¯t = 43, sw = 3.Assuming 20% trimming, test the hypothesis H0:μt = 42 with α = .05 and n = 20.
Repeat the previous problem, only test the hypothesis H0: μt < 42 with α = .05 and n = 16.
For the data in problem 24, the trimmed mean is X¯t = 42.17 with a Winsorized standard deviation of sw = 1.73. Test the hypothesis that the population trimmed mean is 45 with α = .05.
A standard measure of aggression in 7-year-old children has been found to have a 20% trimmed mean of 4.8 based on years of experience. A psychologist wants to know whether the trimmed mean for children with divorced parents differs from 4.8. Suppose X¯t = 5.1 with sw = 7 based on n = 25.Test the
Summarize the relative merits of using a percentile bootstrap method.
For the following pairs of points, verify that the least square regression line is Yˆ = 1.8X −8.5.X : 5,8,9,7,14 Y : 3,1,6,7,19.
Compute the residuals using the results from problem 1.Verify that if you square and sum the residuals, you get 47, rounding to the nearest integer.
Verify that for the data in problem 1, if you use Yˆ = 2X −9, the sum of the squared residuals is larger than 47.Why would you expect a value greater than 47?
Suppose that based on n = 25 values, s 2x = 12 and (Xi −X¯)(Yi −Y¯) = 144.What is the slope of least squares regression?
The following table reports breast cancer rates plus levels of solar radiation (in calories per day) for various cities in the United States. Fit a least squares regression to the data with the goal of predicting cancer rates and comment on what this line suggests.Daily Daily City Rate calories
For the following data, compute the least squares regression line for predicting gpa given SAT.SAT: 500 530 590 660 610 700 570 640 gpa: 2.3 3.1 2.6 3.0 2.4 3.3 2.6 3.5
Compute the residuals for the data used in the previous problem and verify that they sum to zero.
For the following data, compute the least squares regression line for predicting Y from X.X: 40 41 42 43 44 45 46 Y: 1.62 1.63 1.90 2.64 2.05 2.13 1.94
In problem 5, what would be the least squares estimate of the cancer rate given a solar radiation of 600? Indicate why this estimate might be unreasonable.
Maximal oxygen uptake (mou) is a measure of an individual’s physical fitness. You want to know how mou is related to how fast someone can run a mile. Suppose you randomly sample six athletes and get mou (milliliters/kilogram): 63.3 60.1 53.6 58.8 67.5 62.5 time (seconds): 241.5 249.8 246.1 232.4
Verify that for the following pairs of points, the least squares regression line has a slope of zero. Plot the points and comment on the assumption of that the regression line is straight.X : 123456 Y : 147741.
Repeat the last problem, only for the points X : 123456 Y : 456782.
Vitamin A is required for good health. However, one bite of polar bear liver results in death because it contains a high concentration of vitamin A. Comment on this fact in terms of extrapolation.
Sockett et al. (1987) report data related to patterns of residual insulin secretion in children. A portion of the study was concerned with whether age can be used to predict the logarithm of C-peptide concentrations at diagnosis. The observed values are Age (X): 5.2 8.8 10.5 10.6 10.4 1.8 12.7 15.6
For the data in the last problem, use a computer to verify that a least squares regression line using only X values (age) less than 7 yields b1 = 0.247 and b0 = 3.51. Verify that when using only the X values great than 7 you get b1 = .009 and b0 = 4.8. What does this suggest about using a linear
For the data in table 8.3, the sizes of the corresponding lots are:18,200 12,900 10,060 14,500 76,670 22,800 10,880 10,880 23,090 10,875 3,498 42,689 17,790 38,330 18,460 17,000 15,710 14,180 19,840 9,150 40,511 9,060 15,038 5,807 16,000 3,173 24,000 16,600.Use a computer to verify that the least
Given the following quantities, find the sample correlation coefficient, r, and test H0 : ρ = 0 at the indicated level.(a) n = 27, (Yi −Y¯)2 = 100, (Xi −X¯)2 = 625,(Xi −X¯)(Yi −Y¯) = 200, α = .01.(b) n = 5, (Yi −Y¯)2 = 16, (Xi −X¯)2 = 25,(Xi −X¯)(Yi −Y¯) = 10, α = .05.
The high school grade-point average (X) and college grade-point (Y ) for 29 randomly sampled college freshman yielded the following results:(Yi −Y¯)2 = 64, (Xi −X¯)2 = 100, (Xi −X¯)(Yi −Y¯) = 40.Test H0 : ρ = 0 at the .1 level and interpret the results.
For the previous problem, answer the following questions.(a) Is it reasonable to conclude that the least squares regression line has a positive slope?(b) Is it possible that despite the value for r, as high school grade-point averages increase, college grade-point averages decrease? Explain your
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