Week 7 Regression Testing the Hypothesis.pdfsafe
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University of Maryland, University College**We aren't endorsed by this school
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MATH 302
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Statistics
Date
Dec 26, 2024
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2
Uploaded by BailiffFangArmadillo
Recall, our Car Price dataCar Price:YearYearsOldObservation 1$20,00020154Observation 2$25,00020163Observation 3$30,00020181Observation 4$31,00020181Observation 5$22,50020163Observation 6$25,00020163Observation 7$29,50020181Observation 8$24,00020154Observation 9$24,50020172Observation 10$25,00020172With the Regression output,Next, we want to test the hypothesis and see if the results are significant.The hypothesis scenario looks like:Ho:𝜌= 0Ha:𝜌 ≠0If we look at the pvalue or theSignificance Fwe see the pvalue = .000673..000673 < .05, Yes this is significant. This means Years Old is a significantpredictor of the Price of a Car.SUMMARY OUTPUTRegression StatisticsMultiple R0.884606501R Square0.782528661Adjusted R Square0.755344744Standard Error1725.490814Observations10ANOVAdfSSMSFSignificance FRegression185706451.6185706451.6128.786460.000673381Residual823818548.392977318.548Total9109525000CoefficientsStandard Errort StatPvalueLower 95%Upper 95%Lower 95.0%Upper 95.0%Intercept31959.677421296.43524424.651965897.83E0928970.0923934949.2624528970.0923934949.26245Years Old2629.032258490.00646385.3653011790.0006733758.989191499.0753263758.989191499.075326
We can also comparer, the correlation to a critical value. Ifr< negative criticalvalue orr> positive critical value, thenris significant. Ifris significant and theline may be used for prediction.We knowr=.8846. There is a correlation critical value table in RealizeIT underTesting the Hypothesis in Week 7 but here is a link to a more detailed table.Correlation CV TableWe know alpha = .05, this is two tailed test from the hypothesis scenario and n =10. The critical value that corresponds to this in the table isrCV = 0.632. Weknow our correlation is negative, so we will use the negative value of this..8846 <0.632, this tells us thatris significant and you can use the line forprediction.This is the same conclusion we got with the pvalue from above.Lastly, we can run a ttest to see if the data is significant. From the regressionoutput the tStat for the slope is5.3653. But if we didn’t have the regressionoutput we can calculate this value using this equation.t =𝑟√𝑛−2√1−𝑟2Plugging in our correlation and sample size we get:t =−.8846√10−2√1−(−.8846)2= −2.50202.4663505=−5.3651t – Test Stat we calculated by hand is very close to the tstat in the output. It is alittle off because I did round some of my values.Then we can use the =T.DIST.2T function to find the pvalue. This Excel functionshould look familiar.=T.DIST.2T(ABS(5.3651),8)Remember if you have a negative value you will need to use the ABS function totake the absolute value of it.pvalue = 0.000673544 < .05, Yes, this is significant. This is the same conclusion aswe got above, and this is the same pvalue from the Regression Output.It does not matter what way you use to Test the Hypothesis of a Simple LinearRegression example, if done correctly you will get the same conclusion everytime.