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probability statistics
Modern Statistics For The Social And Behavioral Sciences 1st Edition Rand Wilcox - Solutions
Two methods for reducing shoulder pain after laparoscopic surgery were compared by Jorgensen et al. (1995). The data were Group 1: 1 2 1 1 1 1 1 1 1 1 2 4 1 1 Group 2: 3 3 4 3 1 2 3 1 1 5 4 Verify that both the Wilcoxon-Mann-Whitney test and Cliff’s method reject at the 0.05 level. Although the
The last example in Section 8.5.5 dealt with comparing males to females regarding the desired number of sexual partners over the next 30 years.Using Student’s T, we fail to reject which is not surprising because there is an extreme outlier among the responses given by males. If we simply discard
For the data in Table 8.4, verify that the 0.95 confidence interval for the difference between the biweight midvariances is (−159234, 60733).
Describe a general situation where comparing medians will have more power than comparing means or 20% trimmed means.
For the data in Table 8.5, use the R function comvar2 to verify that the 0.95 confidence interval for the difference between the variances is(−1165766.8, 759099.7).
For the self-awareness data in Table 8.5, verify that the R function yuenbt, with the argument tr set to 0, returns (−571.4, 302.5) as a .95 confidence interval for the difference between the means.
Create a boxplot of the data in Table 8.6 and comment on why the probability coverage, based on Student’s T or Welch’s method, might differ from the nominal α level.
The 20%Winsorized standard deviation (sw) for the first group in Table 8.6 is 1.365 and for the second group it is 4.118. Verify that the 0.95 confidence interval for the difference between the 20% trimmed means, using Yuen’s method, is (5.3, 22.85).
In the previous exercise you do not reject the hypothesis of equal variances.Why is this not convincing evidence that the assumption of equal variances, when using Student’s T, is justified?
For the data in Table 8.6, the sample variances are 116.04 and 361.65, respectively. Verify that based on Welch’s method, the .95 confidence interval for the difference between the means is (1.96, 20.83). Check this result with the R function yuen.
Student’s T rejects the hypothesis of equal means based on the data in Table 8.6. Interpret what this means.
Verify that the 0.95 confidence interval for the difference between the means, based on the data in Table 8.6 and Student’s T, is (2.2, 20.5).What are the practical exercises with this confidence interval?
The sample means for the data in Table 8.6 are 22.4 and 11. If we test the hypothesis of equal means using Student’s T, verify that T = 2.5 and that you would reject with α = 0.05.
For the data in Table 8.6, if we assume that the groups have a common variance, verify that the estimate of this common variance is s2 p = 236.
In the previous exercise, you rejected the hypothesis of equal means.What does this imply about the accuracy of the confidence interval for the difference between the population means based on Student’s T?
Published studies indicate that generalized brain dysfunction may predispose someone to violent behavior. Of interest is determining which brain areas might be dysfunctional in violent offenders. In a portion of such a study conducted by Raine, Buchsbaum, and LaCasse (1997), glucose metabolism
Use Cohen’s d to measure effect size using the data in the previous two exercises.
Repeat the previous exercise, only use Yuen’s test with 20% trimmed means.
Responses to stress are governed by the hypothalamus. Imagine you have two groups of participants. The first shows signs of heart disease and the other does not. You want to determine whether the groups differ in terms of the weight of the hypothalamus. For the first group of participants with no
You compare lawyers versus professors in terms of job satisfaction and fail to reject the hypothesis of equal means or equal trimmed means.Does this mean it is safe to conclude that the typical lawyer has about the same amount of job satisfaction as the typical professor?
Repeat the last exercise, only compare 20% trimmed means instead.
Two methods for training accountants are to be compared. Students are randomly assigned to one of the two methods. At the end of the course, each student is asked to prepare a tax return for the same individual.The returns reported by the students are Method Returns 1 132 204 603 50 125 90 185 134
Repeat the last exercise, only use Student’s T instead.
For ¯X1 = 10, ¯X2 = 5, s21= 21, s22= 29, n1 = n2 = 16, compute a 0.95 confidence interval for the difference between the means using Welch’s method and state whether you would reject the hypothesis of equal means.
Referring to the last exercise, compute a 0.99 confidence interval for the difference between the trimmed means.
If for two independent groups, you get ¯Xt1 = 42, ¯Xt2 = 36, s2 w1 = 25, s2 w2 = 36, n1 = 24, and n2 = 16, test the hypothesis of equal trimmed means with α = 0.05.
Comparing the results of the last two exercises, what do they suggest about using Student’s T versus Welch’s method when the sample variances are approximately equal?
Repeat the last exercise using Welch’s method.
For two independent groups of subjects, you get ¯X1 = 86, ¯X2 = 80, s21= s22= 25, with sample sizes n1 = n2 = 20. Assume the population variances of the two groups are equal and verify that Student’s T rejects with α = 0.01.
Comparing the test statistics for the last two exercises, what do they suggest regarding the power of Welch’s test versus Student’s T test for the data being examined?
Repeat the last exercise, only use Welch’s test for comparing means.
Still assuming equal variances, test the hypothesis of equal means using the data in the last exercise assuming random sampling from normal distributions. Use α = 0.05.
Suppose that the sample means and variances are ¯X1 = 15, ¯X2 = 12, s21= 8, s22= 24 with sample sizes n1 = 20 and n2 = 10. Verify that s2 p = 13.14, T = 2.14 and that Student’s T test rejects the hypothesis of equal means with α = 0.05.
For Hubble’s data on the recession velocity of galaxies, shown in Table 6.1 (in Chapter 6), verify that the modified bootstrap method yields a 0.95 confidence interval of (310.1, 630.1) for the slope.
Using the data in Exercise 6 of Chapter 6, verify that the R function lsfitci returns a 0.95 confidence interval for the slope (when predicting cancer rates given solar radiation) of (−0.049, −0.25). The classic 0.95 confidence interval based on Equation (6.8) is(−0.047, −0.24). What does
Restricting the range as was done in the previous exercise, verify that the R function pcorb returns a 0.95 confidence interval of (−0.63, −0.19).Compare this to the result in Exercise 15.
Create a scatterplot of the data in Table 6.6 and note that six points having X values greater than 125 are visibly separated from the bulk of the observations. Now compute a 0.95 confidence interval for the slope of the least squares regression line with the six points having X values greater than
The previous exercise indicates that the variables in Table 6.6 are dependent, but the results in Exercises 14 and 15 failed to detect any association. Describe a possible reason for the discrepant results.
Using the data in Table 6.6, verify that R function indt rejects the hypothesis that these two variables are independent.
Again using the data in Table 6.6, verify that you do not reject H0:ρ = 0 using the R function pcorb.
Verify that for the data in Table 6.6, the 0.95 confidence interval for the slope of the least squares regression line is (−0.27, 0.11) using the R function lsfitci.
For the data in Exercise 8, compute a 0.95 confidence interval for the median using the R function sint described in Chapter 4. What does this suggest about using a median versus a one-step M-estimator?
For the data in Exercise 8, what practical problem might occur if the standard error of the median is estimated using a bootstrap method?
Repeat the last exercise, only now trim 40%. Verify that the 0.95 confidence interval is now (7.0, 14.5). Why is this confidence interval so much shorter than the confidence interval in the last exercise?
Use trimpb on the data used in the previous two exercises, but this time trim 30%. Verify that the 0.95 confidence interval for the trimmed mean is (7.125, 35.500). Why do you think this confidence interval is shorter versus the confidence interval in the last exercise?
For the data in the previous exercise, verify that the 0.95 confidence interval for the 20% trimmed mean returned by the R function trimpb is (7.25, 63). Why would you expect this confidence interval to be substantially longer than the confidence interval based on a one-step Mestimator?
For the following observations 2, 4, 6, 7, 8, 9, 7, 10, 12, 15, 8, 9, 13, 19, 5, 2, 100, 200, 300, 400 verify that the 0.95 confidence interval, based on a one-step M-estimator and returned by onesampb, is (7.36, 19.77).
Which of the two confidence intervals given in the last two exercises is likely to have probability coverage closer to 0.95?
For the data in Exercise 2, verify that the 0.95 confidence interval for the population 20% trimmed mean using trimpb is (11.2, 22.0).
For the data in Exercise 1, verify that the equal-tailed .95 confidence interval for the population 20% trimmed mean using trimcibt is (8.17, 15.75).
Referring to the previous two exercises, which confidence interval for the mean is more likely to have probability coverage at least 0.95?
For the data in the previous exercise, verify that the 0.95 confidence interval for the mean returned by the R function trimcibt is (12.40, 52.46). Note that the length of this interval is much higher than the length of the confidence interval for the mean using trimci.
Rats are subjected to a drug that might cause liver damage. Suppose that for a random sample of rats, measures of liver damage are found to be 5, 12, 23, 24, 6, 58, 9, 18, 11, 66, 15, 8.Verify that trimci returns a 0.95 confidence interval for the 20%trimmed mean, using R, equal to (7.16, 22.84).
For the following 10 bootstrap sample means, determine a p-value when testing the hypothesis H0: μ = 8.5.7.6, 8.1, 9.6, 10.2, 10.7, 12.3, 13.4, 13.9, 14.6, 15.2.What would be an appropriate 0.8 confidence interval for the population mean?
R has a built-in data set called leuk. The third column indicates survival times of patients diagnosed with acute myelogenous leukaemia. The first column indicates the patient’s white blood cell count at the time of diagnosis. Using least squares regression, assuming homoscedasticity, test the
To illustrate a point, assume that for the data in Table 6.7, the goal is to predict X given Y, rather than predict Y given X. Use the R functions ols and rqfit to test the hypothesis H0 :β1 = 0 with α = .05. Verify that ols fails to reject, its p-value is approximately 0.97, but rqfit rejects.
Verify that for the data in Table 6.6, when testing the hypothesis of homoscedasticity using the R function khomreg, you fail to reject withα = 0.05. Explain why this result is not a satisfactory reason for assuming homoscedasticity.
The data in Table 6.7 are from a study, conducted by L. Doi where the goal is to understand how well certain measures predict reading ability in children. Verify that the 0.95 confidence interval for the slope is(−0.16, .12) based on Equation (6.8).
Suppose you observe X: 12.2, 41, 5.4, 13, 22.6, 35.9, 7.2, 5.2, 55, 2.4, 6.8, 29.6, 58.7 Y : 1.8, 7.8, 0.9, 2.6, 4.1, 6.4, 1.3, 0.9, 9.1, 0.7, 1.5, 4.7, 8.2 TABLE 6.7: Reading Data X: 34 49 49 44 66 48 49 39 54 57 39 65 43 43 44 42 71 40 41 38 42 77 40 38 43 42 36 55 57 57 41 66 69 38 49 51 45 141
You measure stress (X) and performance (Y ) on some task and get X : 18 20 35 16 12 Y : 36 29 48 64 18 Verify that you do not reject H0 : β1 = 0 using α = 0.05. Is this result consistent with what you get when testing H0 : ρ = 0? Why would it be incorrect to conclude that X and Y are independent?
Imagine two scatterplots, where in each scatterplot the points are clustered around a line having slope 0.3. If for the first scatterplot r = 0.8, does this mean that points are more tightly clustered around the line versus the other scatterplot where r = 0.6?
For the data in Exercise 17, 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 3498 42,689 17,790 38,330 18,460 17,000 15,710 14,180 19,840 9150 40,511 9060 15,038 5807 16,000 3173 24,000 16,600.Verify that the least squares regression
The selling price and size of a home for a suburb of Los Angeles in the year 1997 are shown in Table 6.6. At the time, even a small empty lot would cost at least $200,000. Verify that based on the least squares regression line for these data, if we estimate the cost of an empty lot by setting the
For the data in Exercise 15, 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 greater than 7 you get b1 =0.009 and b0 = 4.8. What does this suggest about using a linear rule for all of the data?
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
Vitamin A is required for good health. You conduct a study and find that as vitamin A intake increases, certain health problems decrease.However, for levels of vitamin A not included in your study, a sufficiently high dose can result in death. Comment on what this illustrates in the context of
Repeat Exercise 12, only for the points X: 1 2 3 4 5 6 Y : 4 5 6 7 8 2
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 that the regression line is straight.X: 1 2 3 4 5 6 Y : 1 4 7 7 4 1
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
In Exercise 6, what would be the least squares estimate of the cancer rate given a solar radiation of 600? Indicate why this estimate might be unreasonable.
For the data in the last exercise, verify that the coefficient of determination is 0.36 and interpret what this tells you.
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.City Rate Daily City Rate Daily
Verify that for the data in Table 6.3, the least squares regression line isˆY = −0.0405X + 4.581.
Suppose that based on n = 25 values, s2 x = 12, s2 y = 25, and r = 0.6.What is the slope of the least squares regression?
Verify that for the data in Exercise 1, if you use ˆY = 2X −9, the sum of the squared residuals is larger than 47. Why would you expect a value greater than 47?
Compute the residuals using the results from Exercise 1. Verify that if you square and sum the residuals, you get 47, rounding to the nearest integer.
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 ¯Xt = 5.1 with sw = 7 based on n = 25. Test the
For the data in Exercise 24, the trimmed mean is ¯Xt = 42.17 with a Winsorized standard deviation of sw = 1.73. Test the hypothesis that the population trimmed mean is 45 with α = 0.05.
Repeat the previous exercise, only test the hypothesis H0: μt < 42 withα = 0.05 and n = 16.
Given the following values for ¯Xt and sw: (a) ¯Xt = 44, sw = 9, (b)¯X t = 43, sw = 9, (c) ¯Xt = 43, sw = 3, and assuming 20% trimming, test the hypothesis H0: μt = 42 with α = 0.05 and n = 20.
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 = 0.025.
A company claims that on average, when exposed to its 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
Repeat the previous exercise only test H0: μ > 42.
For part b of the last exercise you fail to reject but you reject for the situation in partc. What does this illustrate about power?
For the previous exercise, rather than increase the sample size, what else might you do to increase power? What is a negative consequence of using this strategy?
The previous exercise indicates that power is relatively low with only n = 10 observations. Imagine that you want power to be at least 0.8.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 0.56, 0.71, and 0.81, respectively.
For the previous exercise, verify that power is 0.35 if μ = 46.
A manufacturer of medication for migraine headaches knows that its product can cause liver damage if taken too often. Imagine that by a standard measuring process, the average liver damage is μ = 48. A modification of the product is being contemplated, and based on n = 10 trials, it is found that
For n = 49, α = 0.05, σ = 10, and H0: μ = 50, verify that power is approximately 0.56 when μ = 47.
For n = 36, α = 0.025, σ = 8, and H0: μ ≤ 100, verify that power is 0.61 when μ = 103.
For n = 25, α = .01, σ = 5, and H0: μ ≥ 60, verify that power is 0.95 when μ = 56.
Comment on the relative merits of using a 0.95 confidence interval for addressing the effectiveness of the antipollution device in the previous exercise.
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
If ¯X = 23 and α = 0.025, can you make a decision about whether to reject H0: μ < 25 without knowing σ?
For the previous exercise, compute a 0.95 confidence interval and compare the result with your decision about whether to reject H0.
Repeat the previous exercise but test H0: μ = 130.
Given that ¯X = 120, σ = 5, n = 49 and α = 0.05, test H0: μ > 130, assuming observations are randomly sampled from a normal distribution.
For Exercise 2, determine the p-value.
For Exercise 1, determine the p-value.
For the previous problem, compute a 0.95 confidence interval and verify that this interval is consistent with your decision about whether to reject the null hypothesis.
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