Stat 303: Section 204Practice Exam-4Instructor: Priya KohliUse the following for the next two questions. The weight efficiency of vehicles was the subject of a recent study; when weight was measured in pounds and used to predict fuel efficiency measured in miles per gallon, the linear regression equation was found to be = 47 – 0.007 x. 1.) What does the number -.007 represent in this equation?A.) If a car increases by one pound, it will get 0.007 fewer miles per gallon.B.) If a car increases by one mile per gallon, it will weigh 0.007 fewer pounds.C.) If a car increases by one pound, it will get 0.007 more miles per gallon.D.) If a car weighs 0 lbs, our equation predicts that it will get -0.007 miles to the gallon.E.) The correlation between car weight and fuel efficiency is -0.007.2.) Suppose your car weighs 2500 lbs. How many miles per gallon would this equation predict it to get? (We’re just looking for the point estimate here; don’t worry about confidence intervals here.)A.) -0.007B.) 47C.) 29.5 D.) 32.4E.) 34.2This chart is from a study in which doctors were randomly placed into two groups: one to take aspirin every day, and the other to take a placebo. (Yes, the doctors were the patients in this study!) Neither the doctors taking the drugs nor the physicians examining them knew whether the doctor was taking the placebo or the aspirin. The chi-squared test statistic was 28.3, with a corresponding p-value less than 0.001.Heart Attack Group Yes No TotalPlacebo 190 10,840 11,030Aspirin 100 10,930 11,030Total 290 21,770 22,0603.) What is true about this study?A.) We can say that the aspirin caused the doctors taking it to have a lower rate of myocardial infarctions.B.) We cannot say that there is a cause-and-effect relationship between aspirin and myocardial infarction, because lurking variables are a huge problem with observational studies like this one.C.) Since my p-value was less than 0.001, I cannot reject, concluding that whether a person has a heart attack is not related to which medicine is taken.D.) A and C aboveE.) B and C aboveUse the following for the next two questions. A French teacher grouped her high school students into three groups, one who had never studied a foreign language before but had good English skills, one who had never studied a foreign language and had poor English skills, and one who had studied at least one foreign language. She compared the first quiz scores of the three groups; the ANOVA table is shown below.Sum of Degrees of MeanF-Ratio p-ValueSquares Freedom SquaresBetween Variation 30.000 2.500 0.1768Within Variation 30.000Total Variation 60.0004.) How many degrees of freedom are there for the Between Variation?A.) 2B.) 5C.) 7D.) 8E.) The answer cannot be determined from the information given.5.) What null and alternative hypotheses did the French teacher set up?A.) Ho: Prior Language Skills and French Quiz Score are not related,Ha: Prior Language Skills and French Quiz Score are dependent.B.) Ho: β1 = 0, Ha: β1 ≠ 0C.) Ho: π1 = π2 = π3, Ha: at least one proportion is different from the othersD.) Ho: μ1 = μ2 = μ3 , Ha: μ1 ≠ μ2 ≠ μ3E.) Ho: μ1 = μ2 = μ3 , Ha: at least one mean is different from the others.6.) What is the basic definition of p-value?A.) The probability of finding your test statistic or one more in favor of the alternative hypothesis assuming the null hypothesis is actually true.B.) The probability of finding your test statistic or one more in favor of the hypothesis assuming that actually the alternative hypothesis is true.C.) The probability that the null hypothesis is true.D.) The probability of rejecting the null hypothesis when it is actually false.E.) The probability of rejecting the null hypothesis when it is actually true.MultipleR-SquareAdjusted StErr ofSummary R R-Square Estimate0.8853 0.8771 20.71547269Degrees of Sum of Mean of F-Ratio p-ValueANOVA Table Freedom Squares SquaresExplained 46385.91867 46385.91867 < 0.0001Unexplained 6007.831325 429.130809CoefficientStandardt-Value p-ValueConfidence Interval 95%Regression Table Error Lower UpperConstant -9.277108434 12.09620168 0.4559 -35.22088077 16.66666391age 21.14457831 2.033765121 < 0.0001 16.78258595 25.50657067Use the following to answer the next 6 problems. A department store chain wanted to know when to replace its old registers, so it kept records on maintenance costs and age of the registers. Using a random sample of 16 registers, they ran a linear regression; below is the output.Scatterplot of Residual vs Fit-40.0-30.0-20.0-10.00.010.020.030.040.00.0 50.0 100.0 150.0 200.0 250.0FitResidualStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyStatTools Student VersionFor Academic Use OnlyQ-Q Normal Plot of Residual / resids-3.5-2.5-1.5-0.50.51.52.53.5-3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5Z-ValueStandardized Q-Value7.) What conclusions can we draw from the output?A.) Age and maintenance costs have a positive relationship.B.) The slope is not significantly different from 0.C.) The correlation between age and maintenance costs is -0.94.D.) The estimate of the slope is -9.2771.E.) Two of the above are true.8.) Do you think the assumptions for running a linear regression are met?A.) It looks like the data is definitely not normally distributed; it seems to have a heavy right skew.B.) The constant variance assumption is definitely not met.C.) The mean zero assumption is not met; they should have used a higher-order polynomial.D.) We know for sure that they should not have run a linear regression with this data; the residual plot shows that there is definitely not a linear relationship between the variables.E.) It looks like all the assumptions are met!9.) What is the test statistic for testing whether the slope is 0 for this problem? (hint: t statistics for slope)A.) 108.0927B.) -0.7669C.) -9.277D.) 10.3968E.) 21.144510.) What is a description of Type I error for the hypothesis test for the slope in this problem?A.) Concluding that the slope is positive when actually it is negative.B.) Concluding that the slope is not zero when actually it
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