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statistics principles and methods

Questions and Answers of Statistics Principles And Methods

=+1.1 Give a brief definition of the terms descriptive statistics and inferential statistics.
=+■ Are the conclusions drawn by the researchers supported by the data analysis?
=+■ Was an appropriate method of analysis used, given the type of data and how the data were collected?
=+■ Were the data summarized in an appropriate way?
=+■ Were the data collected in a sensible way?
=+■ Was relevant information collected? Were the right things measured?
=+■ What were the researchers trying to learn? What questions motivated their research?
=+d. If fewer than 20 in the sample favor the ban, is this at odds with the assertion that (at least) 90% of the populace favors the ban? (Hint: Consider P(x , 20) when p 5 .9.)
=+c. What are the mean value and standard deviation of the number who favor the ban?
=+b. What is the probability that at least 20 will favor the ban?
=+a. What is the probability that more than 20 will favor the ban?
=+A.16 Exit polling has been a controversial practice in recent elections, because early release of the resulting information appears to affect whether or not those who have not yet voted will do
=+c. How do the “error probabilities” of Parts (a) and (b)change if the value 15 in the decision rule is changed to 14?
=+b. What is the probability that the program is not implemented if p 5 .70? if p 5 .60?
=+a. What is the probability that the program is implemented when p 5 .80?
=+strongly suggests that p , .80 (fewer than 80% have detectors), as opposed to p $ .80, the program will be implemented. Let x 5 the number of residences among the 25 that have a detector, and
=+A.15 A city ordinance requires that a smoke detector be installed in all residential housing. There is concern that too many residences are still without detectors, so a costly inspection program
=+d. What happens to the “error probabilities” of Parts (a)and (b) if the decision rule is changed so that the coin is judged fair if 7 # x # 18 and judged unfair otherwise? Is this a better rule
=+c. What is the probability of judging the coin to be fair when P(H) 5 .6? when P(H) 5 .4? Why are the probabilities so large compared to the probabilities in Part (b)?702 A p p e n d i x A ■ The
=+b. What is the probability of judging the coin to be fair when P(H) 5 .9, so that there is a substantial bias? Repeat for P(H) 5 .1.
=+a. What is the probability of judging the coin to be biased when it is actually fair?
=+A.14 A coin is to be spun 25 times. Let x 5 the number of spins that result in heads (H). Consider the following rule for deciding whether or not the coin is fair:Judge the coin to be fair if 8 # x
=+A.13 Suppose that 20% of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of size 1000 have (approximately) a binomial
=+d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.
=+c. Compute the variance and standard deviation of x.
=+b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the x distribution.)
=+a. What kind of probability distribution does x have?
=+A.12 You are to take a multiple-choice exam consisting of 100 questions with 5 possible responses to each question.Suppose that you have not studied and so must guess (select 1 of the 5 answers in
=+d. What is the probability that among 25 randomly selected cars, the number that pass is within 1 standard deviation of the mean value?
=+c. Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection?
=+b. Among 15 randomly selected cars, what is the probability that between 5 and 10 (inclusive) fail to pass inspection?
=+a. Among 15 randomly selected cars, what is the probability that at most 5 fail the inspection?
=+A.11 Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection.
=+A.10 If the temperature in Florida falls below 32°F during certain periods of the year, there is a chance that the citrus crop will be damaged. Suppose that the probability is .1 that any given
=+A.9 An experiment was conducted to investigate whether a graphologist (handwriting analyst) could distinguish a normal person’s handwriting from the handwriting of a psychotic. A well-known
=+to find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part):a. 5% defective partsb. 10% defective partsc. 20% defective parts
=+A.8 Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random
=+d. The headline indicates that fewer than 25 of the 500 agents tested failed the test. Is this a surprising result if all 500 are trustworthy? Answer based on the values of the mean and standard
=+. The article indicated that 500 FBI agents were tested.Consider the random variable x 5 number of the 500 tested who fail. If all 500 agents tested are trustworthy, what are the mean and the
=+b. What is the probability that more than 2 fail, even though all are trustworthy?
=+a. What is the probability that all 10 pass?
=+A.7 The article “FBI Says Fewer Than 25 Failed Polygraph Test” (San Luis Obispo Tribune, July 29, 2001) described the impact of a new program that requires top FBI officials to pass a
=+A.6 fema sult tribu a litter of size 5.
=+c. What is P(2 # x)? (Hint: Make use of your computation in Part (b).)d. W
=+a. What is p(2), that is, P(x 5 2)?b. What is P(x # 1)?
=+A.5 Twenty-five percent of the customers entering a grocery store between 5 P M. and 7 P M. use an express checkout. Consider five randomly selected customers, and let x denote the number among
=+c. Calculate the probability that more than half of the selected passengers rested or slept.
=+a. What is p(8)?b. Calculate P(x # 7).
=+A.4 Refer to Exercise A.3, and suppose that 10 rather than 6 passengers are selected (n 5 10, p 5 .8), so that Appendix Table 9 can be used.
=+b. Calculate p(6), the probability that all six selected passengers rested or slept.c. Determine P(x $ 4).
=+a. Calculate p(4), and interpret this probability.
=+route, the actual percentage is exactly 80%, and consider randomly selecting six passengers. Then x, the number among the selected six who rested or slept, is a binomial random variable with n 5 6
=+A.3 The Los Angeles Times (December 13, 1992)reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost 80% rested or slept.
=+b. Calculate p(4), the probability that all four selected households have a VCR.c. Determine P(x # 3).
=+a. Calculate p(2) 5 P(x 5 2), and interpret this probability.
=+A.2 Suppose that in a certain metropolitan area, 9 out of 10 households have a VCR. Let x denote the number among four randomly selected households that have a VCR, so x is a binomial random
=+b. In a binomial experiment consisting of 20 trials, how many outcomes have exactly 10 successes? exactly 15 successes? exactly 5 successes?
=+a. In a binomial experiment consisting of six trials, how many outcomes have exactly one success, and what are these outcomes?
=+A.1 Consider the following two binomial experiments.
=+rejected. It happens because the test and the multiple comparison method are based on different distributions.Consult your friendly neighborhood statistician for more information.)
=+b. What happens when the T–K procedure is applied?(Note: This “contradiction” can occur when H0 is “barely”
=+a. Carry out the F test at level a 5 .05.
=+15.34 ● Consider the accompanying data on plant growth after the application of different types of growth hormone.1 13 17 7 14 2 21 13 20 17 Hormone 3 18 14 17 21 4 7 11 18 10 5 6 11 15 8
=+15.33 Exercise 15.8 presented the accompanying summary information on satisfaction levels for employees on three different work schedules:n1 5 24 n2 5 24 n3 5 20 5 6.60 5 5.37 5 5.20 MSE was
=+percentages for the different brands. Use a 5 .05.b. Use the T–K procedure to compute 95% simultaneous confidence intervals for all differences between means and interpret the resulting intervals.
=+a. Test for differences among the true average PAPUFA
=+Parkay 12.8 12.5 13.4 13.0 12.3 Blue Bonnet 13.5 13.4 14.1 14.3 Chiffon 13.2 12.7 12.6 13.9 Mazola 16.8 17.2 16.4 17.3 18.0 Fleischmann’s 18.1 17.2 18.7 18.4
=+15.32 ● Samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids(PAPUFA, in percent), resulting
=+detect differences among the mean water losses for the different fumigation durations. How would you interpret this pattern?Duration of fumigation 16 0 8 2 4 Sample mean water loss 27.57 28.23
=+15.31 The accompanying underscoring pattern appeared in the article “Effect of SO2 on Transpiration, Chlorophyll Content, Growth, and Injury in Young Seedlings of Woody Angiosperms” (Canadian
=+the three size classes. Suppose that 95% simultaneous confidence intervals for m1 2 m2, m1 2 m3, and m2 2 m3 are(210, 290), (150, 450), and (10, 310), respectively. How would you interpret these
=+34–40). The energy content (cal/g) of three sizes (4 mm or less, 5–7 mm, and 8–10 mm) of serviceberries was studied. Let m1, m2, and m3 denote the true energy content for
=+15.30 The nutritional quality of shrubs commonly used for feed by rabbits was the focus of a study summarized in the article “Estimation of Browse by Size Classes for Snowshoe Hare” (Journal of
=+article states, “Sequential exposure to the two pollutants had no effect on growth compared to the control. Simultaneous exposure to the gases significantly reduced plant growth.” Let and
=+15.29 The article “Growth Response in Radish to Sequential and Simultaneous Exposures of NO2 and SO2” (Environmental Pollution [1984]: 303–325) compared a control group (no exposure), a
=+b. Is there evidence that seeds eaten and then excreted by lizards germinate at a higher rate than those eaten and then excreted by birds? Give statistical evidence to support your answer.
=+a. Construct the appropriate ANOVA table, and test the hypothesis that there is no difference between mean number of seeds germinating for the four treatments.
=+in batches of 5, and for each group of 5 they recorded how many of the seeds germinated. This resulted in 20 observations for each treatment. The treatment means and standard deviations are given
=+15.28 Do lizards play a role in spreading plant seeds?Some research carried out in South Africa would suggest so(“Dispersal of Namaqua Fig (Ficus cordata cordata) Seeds by the Augrabies Flat
=+15.27 ▼ Sample mean chlorophyll concentrations for the four Jerusalem artichoke varieties introduced in Exercise 15.9 were .30, .24, .41, and .33, with corresponding sample sizes of 5, 5, 4, and
=+The accompanying output gives the T–K intervals as calculated by MINITAB. Identify significant differences and give the underscoring pattern.Individual error rate = 0.00750 Critical value = 4.37
=+15.26 ● The accompanying data resulted from a flammability study in which specimens of five different fabrics were tested to determine burn times.1 17.8 16.2 15.9 15.5 2 13.2 10.4 11.3 Fabric 3
=+(1) visual contact and imagery, (2) nonvisual contact and imagery, (3) visual contact, and (4) control. There were 20 subjects assigned to each method. Calculate the 99%T–K intervals, indicate
=+b. Calculate the 95% T–K intervals, and then use the underscoring procedure described in this section to identify significant differences among the learning methods.15.25 The article referenced
=+a. Is there sufficient evidence to conclude the mean putting performance score is not the same for the four methods?
=+There were 20 subjects randomly assigned to each method.The following summary information on putting performance score was reported:Method 1 2 3 4 16.30 15.25 12.05 9.30 5 13.23 s 2.03 3.23 2.91
=+15.24 The degree of success at mastering a skill often depends on the method used to learn the skill. The article“Effects of Occluded Vision and Imagery on Putting Golf Balls” (Perceptual and
=+a. m1 5 m2, and m3 differs from m1 and m2.b. m1 5 m3, and m2 differs from m1 and m3.m2 and m3.m one another.
=+2, and 3, respectively. The given 95% simultaneous confidence intervals are based on summary quantities that appear in the article:Difference m1 2 m2 m1 2 m3 m2 2 m3 Interval (23.11, 21.11)
=+15.23 gas-ex of Conifer Needles with Glass Beads for Determination of Leaf Surface Area” (Forest Science [1980]: 29–32)included an analysis of dry matter per unit surface area(mg/cm3) for trees
=+x Assume that each of the seven samples studied can be viewed as a random sample for the respective group.Is there sufficient evidence to conclude that the mean value of highest number of drinks
=+Beer only 7.52 6.41 1256 Wine only 2.69 2.66 1107 Spirits only 5.51 6.44 759 Beer and wine 5.39 4.07 1334 Beer and spirits 9.16 7.38 1039 Wine and spirits 4.03 3.03 1057 Beer, wine, and spirits
=+15.22 The article “Heavy Drinking and Problems Among Wine Drinkers” (Journal of Studies on Alcohol [1999]:467–471) analyzed drinking problems among Canadians.For each of several different
=+b. State and test the relevant hypotheses using a significance level of .05.
=+a. Verify the entries in the ANOVA table.
=+Analysis of Variance Source df Sum of sq. Mean Sq. F Ratio F Prob Between Groups 3 14.778 4.926 2.142 .129 Within Groups 19 43.690 2.299 Total 22 58.467
=+age (in months) when the children first walked are shown in the accompanying table. Also given is the ANOVA table, obtained from the SPSS computer package.Age n Total Treatment 1 9.00 9.50 9.75 6
=+1 week and lasting 7 weeks. The second group of children received daily exercises but not the walking exercises administered to the first group. The third and fourth groups were control groups:
=+15.21 ● Parents are frequently concerned when their child seems slow to begin walking (although when the child finally walks, the resulting havoc sometimes has the parents wishing they could
=+Construct an ANOVA table, and use it to test the null hypothesis of no difference in mean antigen concentrations for the three groups.Asymptomatic infants 1.56 1.06 0.87 1.39 0.71 0.87 Infants

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