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Basic Statistics Understanding Conventional Methods And Modern Insights 1st Edition Rand R. Wilcox - Solutions

Using a computer, determine what happens to the correlation between X and Y if the Y values are multiplied by 3.
Repeat the previous problem, only determine what happens to the slope of the least squares regression line.
Consider a least squares regression line Y = .5X +2+e, and where X and e are independent and both have a standard normal distribution. What happens to the correlation between X and Y if instead Y = .5X +2+2e? Hint: What happens to the residuals?
The numerator of the coefficient of determination is (Yi −Y¯)2 −(Yi −Yˆi)2.Based on the least squares principle, why is this value always greater than or equal to zero?
Imagine a study where the correlation between some amount of an experimental drug and liver damage yields a value for r close to zero and the hypothesis H0:ρ = 0 is not rejected. Why might it be unreasonable to conclude that the two variables under study are independent?
Suppose r2 = .95.(a) Explain why this does not provide convincing evidence that the least squares line provides a good fit to a scatterplot of the points.(b) If the least squares line provides a poor fit, what does this say about using Yˆ versus Y¯ to estimate Y .
Imagine a situation where points are removed for which the X values are judged to be outliers. Note that this restricts the range of X values. Without looking at the data, can you predict whether Pearson’s correlation will increase or decrease after these points are removed?
If the normality assumption is violated, what effect might this have when computing confidence intervals as described in box 8.1?
If the homoscedasticity assumption is violated, what effect might this have when computing confidence intervals as described in box 8.1?
Suppose that the sample means and variances are X¯ 1 = 15, X¯ 2 = 12, s 21 = 8, s 22 = 24 with sample sizes n1 = 20 and n2 = 10.Verify that s 2p = 13.14, T = 2.14 and that Student’s t-test rejects the hypothesis of equal means with α = .05.
For two independent groups of subjects, you get X¯ 1 = 45, X¯ 2 = 36, s 21 = 4, s 22 = 16 with sample sizes n1 = 20 and n2 = 30.Assume the population variances of the two groups are equal and verify that the estimate of this common variance is 11.25.
Still assuming equal variances, test the hypothesis of equal means using Student’s t-test and the data in the last problem. Use α = .05.
Repeat the last problem, only use Welch’s test for comparing means.
Comparing the results for the last two problems, what do they suggest regarding the power of Welch’s test versus Student’s t-test when the sample variances differ sufficiently.
For two independent groups of subjects, you get X¯ 1 = 86, X¯ 2 = 80, s 21 = s 22 = 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 = .01.
Repeat the last problem using Welch’s method.
Comparing the results of the last two problems, what do they suggest about using Student’s t versus Welch’s method when the sample variances are approximately equal?
For X¯ 1 = 10, X¯ 2 = 5, s 21 = 21, s 22 = 29, n1 = n2 = 16, compute a .95 confidence interval for the difference between the means using Welch’s method and state whether you would reject the hypothesis of equal means.
Repeat the last problem, only use Student’s t 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 amounts of the refunds reported by the students are Method 1 :
Responses to stress are governed by the hypothalamus. Imagine you have two groups of subjects. 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 subjects with no heart
The .95 confidence interval for the difference between the means, using Student’s t, is (2.2, 20.5). What are the practical concerns with this confidence interval?
For the first of two binomial distributions, there are 15 successes among 24 observations. For the second, there are 23 successes among 42 observations. Test H0: p1 = p2 with a Type I error probability of .05.
A film producer wants to know which of two versions of a particular scene is more likely to be viewed as disturbing. One group of 98 individuals views the first version and 40 say that it is disturbing. The other group sees the second version and 30 of 70 people say that it is disturbing. Test the
It is found that of 121 individuals who take a training program on investing in commodities, 20 make money during the next year and the rest do not. With another training program, 15 of 80 make money. Test the hypothesis that the probability of making money is the same for both training programs,
In a study dealing with violence between factions in the Middle East, one goal was to compare measures of depression for two groups of young males. In the first group, no family member was wounded or killed by someone belonging to the opposite faction, and the measures were
For the data in the last problem, the difference between the medians is −7.5 and the corresponding McKean–Schrader estimate of the standard error is S2 1 +S2 2 = 3.93. Verify that you do not reject the hypothesis of equal medians with α = .05.
Referring to the previous two problems, the hypothesis of equal means is rejected, but the hypothesis of equal medians is not. Comment on why this is not surprising.
Does the consumption of alcohol limit our attention span? An article in the July 10, 2006 Los Angeles Times described a study conducted at the University of Washington where 23 people were given enough alcohol to reach a blood alcohol level of 0.04% (half the legal limit in many states). A second
For 49 pairs of sisters, a researcher wanted to know whether the older sisters differ, on average, from the younger sisters, based on a test of social anxiety. It was found that D¯ = 3 and sD = 4.Test the hypothesis of equal means with a Type I error probability of .05. Discuss the interpretation
For the previous problem, compute a .95 confidence interval for the difference between the means. What concerns, if any, might there be about this confidence interval?
A course aimed at improving an understanding of good nutrition is attended by 28 students. Before the course began, students took a test on nutrition and got the following scores:72,60,56,41,32,30,39,42,37,33,32,63,54,47,91, 56,79,81,78,46,39,32,60,35,39,50,43,48.After the course was completed,
The median of 20 difference scores is found to be MD = 5, there are no tied values, and the McKean–Schrader estimate of the standard error is SD = 2.Verify that you would reject the hypothesis that the difference scores have a population median of zero with α = .05.
Referring to the results of the previous problem, suppose the medians corresponding to these two groups are 7 and 4, respectively. Is it reasonable to conclude that the hypothesis of equal medians would also be rejected with α = .05?
Despite any problems it might have, summarize how you would justify using Student’s t-test to compare two independent groups.
Summarize any practical concerns about Student’s t-test and comment on how they might be addressed.
Summarize the relative merits of comparing groups with medians.
For two independent groups, 1000 bootstrap samples are generated and it is found that there are 10 instances where the bootstrap sample trimmed mean for the first group is less than the bootstrap sample trimmed mean for the second. And there 2 instances where the bootstrap sample trimmed means are
Verify that for the self-awareness data in table 9.1, when applying Yuen’s method, the test statistic is Ty = 2.044.
For the following data, Group 1 Group 2 Group 3 34 6 54 7 23 8 48 6 87 7 44 9 3 2 10 95 9 X¯ 1 = 4.75 X¯ 2 = 4.62 X¯ 3 = 7.75 s2 1 = 6.214 s 22 = 3.982 s 23 = 2.214, assume that the three groups have a common population variance, σ2 p . Estimate σ2 p .
For the data in the previous problem, test the hypothesis of equal means using the ANOVA F. Use α = .05.
For the data in problem 1, verify that Welch’s test statistic is Fw = 7.7 with degrees of freedom ν1 = 2 and ν2 = 13.4. Then verify that you would reject the hypothesis of equal means with α = .01.
Construct an ANOVA summary table using the following data, as described in section 10.1, then test the hypothesis of equal means with α = .05.Group 1 Group 2 Group 3 Group 4 15 9 17 13 17 12 20 12 22 15 23 17
In the previous problem, what is your estimate of the assumed common variance?
For the data used in the last two problems, verify that for Welch’s test, Fw = 3.38 with ν1 = 3 and ν2 = 4.42.
Based on the results of the previous problem, would you reject the hypothesis of equal means with α = .1?
Why would you not recommend the strategy of testing for equal variances, and if not significant, using the ANOVA F test rather than Welch’s method?
Five independent groups are compared with n = 15 observations for each group.Fill in the missing values in the following summary table.Source of variation Degrees of freedom Sum of squares Mean square F Between groups – 50 – –Within groups – 150 –
Referring to box 10.2, verify that for the following data, MSBG = 14.4 and MSWG = 12.59.G1 G2 G3 9 16 7 10 8 6 15 13 9 6
Consider five groups ( J = 5) with population means 3, 4, 5, 6, and 7, and a common variance σ2 p =
If the number of observations is 10 (n = 10), indicate what is being estimated by MSBG, and based on the information given, determine its value. That is, if the population means and common variance were known, what would the value of MSBG be if it were giving perfectly accurate information? How
For the following data, verify that you do not reject with the ANOVA F testing with α = .05, but you do reject with Welch’s test.Group 1: 10 11 12987 Group 2: 10 66 15 32 22 51 Group 3: 1 12 42 31 55 19 What might explain the discrepancy between the two methods?
Consider the following ANOVA summary table:Source of Degrees of Sum of Mean square F variation freedom squares Between Groups 3 300 100 10 Within Groups 8 80 10 Total 11 428 Verify that the number of groups is J = 4, the total number of observations is N = 12 and that with α = .025 the critical
A researcher reports a p-value of .001 with the ANOVA F test. Describe what conclusions are reasonable based on this result.
Summarize the reasons you might fail to reject with the ANOVA F test.
Someone tests for equal variances and fails to reject. Does this justify the use of the ANOVA F test?
A researcher reports that a test for normality was performed, and that based on this test, no evidence of non-normality was found. Why might it be unreasonable to assume normality despite this result?
Outline how you might construct an example where sampling is from normal distributions, Welch’s test rejects, but the ANOVA F test does not.
Consider a 2-by-2 design with population means Factor B Level 1 Level 2 Level 1 μ1 = 110 μ2 = 70 Factor A Level 2 μ3 = 80 μ4 = 40 State whether there is a main effect for factor A, for factor B, and whether there is an interaction.
Consider a 2-by-2 design with population means Factor B Level 1 Level 2 Level 1 μ1 = 10 μ2 = 20 Factor A Level 2 μ3 = 40 μ4 = 10 State whether there is a main effect for factor A, for factor B, and whether there is an interaction.
A computer program reports that for a 2-by-3 ANOVA, with 15 observations in each group, MSA = 400, MSB = 200, MSINTER = 200, MSWG = 50.Perform the tests of no main effects and no interaction with α = .01.
A computer program reports that for a 4-by-5 ANOVA, with 10 observations in each group, MSA = 600, MSB = 400, MSINTER = 300, MSWG = 100.Perform the tests of no main effects and no interaction with α = .05.
Assuming normality and homoscedasticity, what problem occurs when comparing multiple groups with Student’s t-test?
For five independent groups, assume that you plan to do all pairwise comparisons of the means and you want FWE to be .05. Further assume that n1 = n2 = n3 =n4 = n5 = 20, X¯ 1 = 15, X¯ 2 = 10, s 21 = 4 and s 22 = 9, s 23 = s 24 = s 25 = 15, test H0 : μ1 = μ2 using Fisher’s method, assuming the
Repeat the previous problem, only use the Tukey-Kramer method
For four independent groups, assume that you plan to do all pairwise comparisons of the means and you want FWE to be .05. Assume n1 = n2 = n3 = n4 = n5 = 10 X¯ 1 = 20, X¯ 2 = 12, s 21 = 5, s 22 = 6, s 23 = 4, s 24 = 10, and s 25 = 15.Test H0 : μ1 = μ2 using Fisher’s method.
Repeat the previous problem, only use the Tukey-Kramer method
Imagine you compare four groups with Fisher’s method and you reject the hypothesis of equal means for the first two groups. If the largest observation in the fourth group is increased, what happens to MSWG? What does this suggest about power when comparing groups 1 and 2 with Fisher’s method?
Repeat the previous problem but with the Tukey-Kramer method.
For five independent groups, assume that you plan to do all pairwise comparisons of the means and you want FWE to be .05. Further assume that n1 = n2 = n3 = n4 = n5 = 20, X¯ 1 = 15, X¯ 2 = 10, s 21 = 4 and s 22 = 9, s 23 = s 24 = s 25 = 15, test H0 : μ1 = μ2 using Dunnett’s T3.
Repeat the previous problem, only use Games-Howell.
For four independent groups, assume that you plan to do all pairwise comparisons of the means and you want FWE to be .05. Further assume that n1 = n2 = n3 = n4 = n5 = 10 X¯ 1 = 20, X¯ 2 = 12, s 21 = 5, s 22 = 6, s 23 = 4, s 24 = 10, and s 25 = 15.Test H0 : μ1 = μ2 using Dunnett’s T3.
Repeat the previous problem, only use Games-Howell.
For four groups, you get sample medians M1 = 34, M2 = 16, M3 = 42, M4 = 22, S2 1 = 33, S2 2 = 64, S2 3 = 8, S2 4 = 5.Assuming that the goal is to test the hypothesis of equal medians for all pairs of groups such that FWE is .05, determine whether you would reject when comparing groups 2 and 4.
In the previous problem, comment on the results if there are tied values in the first group but not the other three.
Referring to example 8 in this section, compare ethnic groups A and B, ignoring type of medication. Assume this is the only hypothesis to be tested and that the goal is to have a Type I error probability of .05.
In the previous problem, imagine that the goal is to compare all pairs of groups for factor B and that the goal is to have the probability of at least one Type I error equal to .05. What is the critical value you would use when comparing ethnicity groups B and C?
Referring to example 8 in this section, imagine the goal is to check for interactions when dealing with ethnicity groups A and B. Test the hypothesis of no interaction for this special case assuming all other interactions are to be ignored and that the Type I error probability is to be .05.
You perform five tests and get the p-values .049, .048, .045, .047, and .042.Based on the Bonferroni inequality, which would be rejected with FWE equal to .05?
Referring to the previous problem, which would be rejected with Rom’s procedure?
What do the last two exercises illustrate?
Five tests are performed aimed at comparing the medians of dependent groups.The p-values are .24, .001, .005, .12, .04. Which should be rejected when using Rom’s method if FWE is to be .05?
For a one-way contingency table having four categories, the frequencies corresponding to each category are: 23, 14, 8, 32.Test the hypothesis that each category has the same probability. Use α = .05.
For a one-way contingency table, the following frequencies are observed: 23, 34, 43, 53, 16.Using α = .01, test the hypothesis that all five categories have the same probability.
A game show allows contestants to pick one of six boxes, one of which contains a large sum of money. To reduce the expected winnings among the contestants, the organizers speculate that contestants will not pick at random, but that some boxes are more likely to be chosen over others. To find out, a
Imagine that 54 individuals are asked whether they agree, disagree or have no opinion that persons with a college degree feel more satisfied with their lives. The observed frequencies are 9, 30, and 15.Test H0 : p1 = p2 = p3 with α = .05.
It is speculated that the probabilities associated with four categories are .1, .3, .5, and .1. The observed frequencies are 10, 40, 50, and
Test this speculation usingα = .05.
It is speculated that the probabilities associated with five categories are .2, .3, .3, .1 and .1. The observed frequencies are 20, 50, 40, 10 and 15.Test this hypothesis using α = .05.
Someone claims that the proportion of adults getting low, medium, and high amounts of exercise is .5, .3, and .2, respectively. To check this claim you sample 100 individuals and find that 40 get low amounts of exercise, 50 get medium amounts, and 10 get high amounts. Test the claim with α = .05.
A geneticist postulates that in the progeny of a certain dihybrid cross, the four phenotypes should be present in the ratio 9:3:3:1. So if p1, p2, p3, and p4 are the probabilities associated with these four phenotypes, the issue is whether there is empirical evidence indicating that H0: p1 = 9/16,
Does the likelihood of a particular crime vary depending on the day of the week?To find out, the number of crimes for Monday through Sunday were recorded and found to be 38, 31, 40, 39, 40, 44, and 48.Test the hypothesis that the likelihood of a crime is the same for each day of the week, using α
For a random sample of 200 adults, each adult was classified as having a high or low income, and each was asked whether they are optimistic about the future. The results were Yes No Total High 35 42 77 Low 80 43 123 Total 115 85 200 What is the estimated probability that a randomly sampled adult
Referring to the previous exercise, compute a .95 confidence interval forδ = p1+ −p+1. Also test H0 : p12 = p21 using Z with a Type I error probability ofα = .05.
In problem 10, would you reject the hypothesis that income and outlook are independent? Use α = .05. Would you use the φ coefficient to measure the association between income and outlook? Why?
In problem 10, estimate the odds ratio and interpret the results.
You observe Income (father)Income (daughter) High Medium Low Total High 30 50 20 100 Medium 50 70 30 150 Low 10 20 40 70 Total 90 140 90 320 Estimate the proportion of agreement and compute a .95 confidence interval.
For the values 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 5, 6, 6, 6, compute the midranks.
For two independent groups, you observe Group 1: 8,10,28,36,22,18,12 Group 2: 11,9,23,37,25,43,39,57.Compare these two groups with the Wilcoxon-Mann-Whitney test using α = .05.
For the data in the previous exercise, apply the Kolmogorov-Smirnov test, again using α = .05.

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