Statistics 101 Sample questions: EXAM 21. A study was conducted of the effects of a special class designed to aid studentswith verbal skills. Each child was given a verbal skills test twice, both beforeand after completing a 4-week period in the class. Let Y = score on exam attime 2 - score on exam at time 1. Hence, if the population mean µ for Y is equalto 0, the class has no effect, on the average. For the four children in the study,the observed values of Y are 8-5=3, 10-3=7, 5-2=3, and 7-4=3 (e.g. for the firstchild, the scores were 5 on exam 1 and 8 on exam 2, so Y = 8-5=3). It is plannedto test the null hypothesis of no effect against the alternative hypothesis thatthe effect is positive, based on the following results from a computer softwarepackage:NumberVariable of Cases Mean Std. Dev. Std. ErrorY 4 4.000 2.000 1.000a. Set up the null and alternative hypotheses.b. Calculate the test statistic, and indicate whether the P-value was below0.05, based on using the appropriate table.c. Make a decision, using α = .05. Interpret.d. If the decision in (c) is actually (unknown to us) incorrect, what type oferror has been made? What could you do to reduce the chance of thattype of error?e. True or false? When we make a decision using α = .05, this means that ifthe special class is truly beneficial, there is only a 5% chance that we willconclude that it is not beneficial.2. For a random sample of Harvard University psychology majors, the responseson political ideology had a mean of 3.18 and standard deviation of 1.72 for 51nonvegetarian students and a mean of 2.22 and standard deviation of .67 forthe 20 vegetarian students.a. Defining appropriate notation, state the null and alternative hypothesesfor testing whether there is a difference between population mean ideology forvegetarian and nonvegetarian students.b. When we use software to compare the means with a significance test, weobtain the printoutVariances T DF Prob>|T|---------------------------------------Unequal 2.9146 41.9 0.0061Interpret the P -value, in context, based on its definition. (Note: You are notbeing asked to make a decision at some α-level.)3. A multiple-choice test question has four possible responses. The question isdesigned to be very difficult, with none of the four responses being obviouslywrong, yet with only one correct answer. It first occurs on an exam taken by 400students. The designers test whether more people answer the question correctlythan would be expected just due to chance (i.e., if everyone randomly guessedthe correct answer).a. Set up the hypotheses for the test.b. Of the 400 students, 125 correctly answer the question. Find the P -value,and interpret.c. Make a decision about H0, using α = .05. Based on this decision, what canyou conclude about the parameter?4. (13 pts.) A geographer conducts a study of the relationship between the levelof economic development of a nation (measured in thousands of dollars for percapita GDP) and the birth rate (average number of children per adult woman).For one analysis of the data, a part of the computer printout reportsStatistic Value SE--------------------------------------Correlation -0.460 0.170a. Explain how to interpret the reported value of the correlation.b. Can you tell whether the sign of the slope in the corresponding predictionequation would be positive or negative? Why?c. Suppose the prediction equation were ˆy = 5.6 − 0.1x. Interpret the slope,and show how to find the predicted birth rate for a nation that has x = 40.d. Sketch a scatterplot for which it would be inappropriate to use the correla-tion to describe the association.5. A study on educational aspirations of high school students (S. Crysdale, Inter-national Journal of Comparative Sociology, Vol. 16, 1975, pp. 19–36) measuredaspirations using the scale (some high school, high school graduate, some col-lege, college graduate). For students whose family income was low, the countsin these categories were (9, 44, 13, 10); when family income was middle, thecounts were (11, 52, 23, 22); when family income was high, the counts were (9,41, 12, 27). Software provides the results shown below.a. Find the sample conditional distribution on aspirations for those whosefamily income was high.2Table 1:Statistic DF Value Prob------------------------------------------------Chi-Square 6 8.871 0.181b. Give all steps of the chi-squared test, explaining how to interpret the P -value.c. Explain what further analyses you could do that would be more informativethan a chi-squared test.The following questions are true or false. Indicate T or F next to each.6. For a given set of data on two quantitative variables X and Y , theslope of the least squares prediction equation and the correlation must have thesame sign.7. For a given set of data on two quantitative variables X and Y , theprediction equation and the correlation do not depend on the units of measure-ment.8. The standardized residual that follows up the chi-squared test for acontingency table has value 3.2 in the cell in row 1 and column 1. This meansthat if the variables were independent, it would be very unusual to observe somany observations in that cell.9. Suppose that a study reports that a 95% confidence interval for thedifference µ1− µ2between the population mean annual incomes for whites (µ1)and for Hispanics (µ2) having jobs in home construction is ($5000, $5400). Then,a 95% confidence interval for the difference µ2−µ1between the population meanannual incomes for Hispanics and for whites having jobs in home constructionis (-$5400, -$5000).10. A study of medical utilization compares mean stay in the hospital forheart transplant operations in 2009 to the mean stay in 1995, for two separatesamples of such operations in the two years. In the comparison, since the samevariable (“length of stay in the hospital”) is measured for each sample, the3data should be analyzed using methods for dependent samples (such as thepaired-difference t test) rather than independent samples.11. Refer to the previous question. Suppose the data in 1995 were summa-rized by ¯y1= 10.5, s1= 8.9 (n1= 54), and the data in 2009 were summarizedby ¯y2= 8.0, s1= 7.8 (n2= 48). These statistics suggest that the variable“length of stay in the hospital” does not have a normal distribution. There-fore, even though the samples are relatively large, we cannot use the formula(¯y1− ¯y2) ± t(se), which assumes normal population distributions.12. For large