Statistical Inference and Hypothesis Testing: Applications in Real-World Scenarios

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Statistical Inference and Hypothesis Testing: Applications in Real-World Scenarios1.Tocombat red-light-running crashes, many states are installing red light camerasat dangerousintersections.Thesecameras photograph the license plates ofvehicles that run red lights andautomatically issue tickets.Howeffective are these photo enforcement programs?The VirginiaDepartment of Transportation (VDOT) conducted acomprehensive study of its newly adoptedprogram and published the results in a2012 study.In one portion of the study,VDOT provided crashdata both before and after installation of the cameras atseveral intersections.The data,measured asthe number of crashes caused by red light running per intersectionper year, for 13 intersections inFairfax County are given in the first columnsin the Minitab file.a.What is (are) the parameter(s) we are conducting inference on?Choose one of thefollowing symbols (m; mD(the mean difference from a matched pairs design);m1-m2(the meandifference from independent samples); p, or p1–p2) and describe it in words.Since the same intersections are observed before and after the installations of cameras, use paired t-test for this situation. The parameter of interest ismD(themeandifference from a matched pairsdesignb.Depending on your answer to part (a), construct one or two probability plots and one ortwo box-plots to visualize the distribution(s) of your sample data.If you construct twoprobability plots and two box-plots, please construct two separate Minitab probability plotsand one Minitab box-plot displaying both boxes on the same graph.Also, properly title andlabel your graphs. Copy and paste these graphs into your assignment.Below the graphs,answer the following questions.Since the difference between the number of crashes Before and after installations are considered, theprobability plot and the box plots for the differences are constructed and are shown as follows:All points are close to straight line indicating the parent distribution of differences follows normality.The box plot:

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The box-plot of the difference seem to be approximately normal.i.Are there any major deviations from normality?There are no major deviations fromnormality as all points are close to straight line in probability plotand the box-plot is approximately symmetric.ii.Are there any outliers present?No, no outliers are observed.iii.Is it appropriate to conduct statistical inference procedures, why or why not?Yes, it is appropriate to conduct the statistical inference procedures as normality assumption issatisfied.c.At the 0.01 significance level, test the claim that the installation of the camerasdecreased the mean number of crashes at these particular intersections.i.State the null and alternative hypotheses.:0HThe installation of thecamerasnotdecreased the mean number of crashes at these particularintersections,0dm£:1HThe installation of thecameras decreased the mean number of crashes at these particularintersections,0dm>Note: Here d = before–afterii.State the significance level for this problem.The significance level is 0.01. If the P-value is less than the significance level reject the nullhypothesis

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iii.Calculate the test statistic (subtract after–before)The MINITAB output for the paired t-test is as follows:From theoutput, the test statistic value =3.45iv.Calculate the P-value and include the probability notation statement.The P-value()(), ...13132 0 013 450 002 From outputP t+-=>=v.State whether you reject or do not reject the null hypothesis.Since theP-value is less than the significance level, reject the null hypothesis.vi.State your conclusion in context of the problem(i.e. interpret your results).There is sufficient evidence to support the claim thatinstallation of thecameras decreased the meannumber of crashes at these particular intersectionsd.For the above situation, construct a 96.2% confidence interval for the abovedata.Interpret the confidence interval as we learned in class.The 96.2% of the confidence interval is shown in the following output.From the output, the 96.2% CI for the mean difference = (0.36, 1.86).1 11ddsxtna/ 2±´=±

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