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7. The data comparing GPA with starting salaries for recent college graduates: (24 points) GPA, X:3.26 2.60 3.35 2.86 3.82 2.

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We should use the function as = CORREL(x,y)D7 A 1 GPA fx =CORREL(A2:A8,B2:58) D E F C B Salary 3.26 33.8 2.6 29.8 3.35 33.5 30.4 3.82 36.4 2.21 27.6 3.47 35.3 2.86 0.98Rejection rule: If t>4.032 ort<-4.032, then reject the null hypothesis H. 0.988/7-2 V1-0.9882 2.2092 0.1545 = 14.30 The valueConclusion: Use a significance level, a = 0.05. Here the value of the test statistic 14.30, which is greater than the criticaSUMMARY OUTPUT Regression Statistics Multiple R 0.988215 R Square 0.976569 Adjusted R Square 0.971883 Standard Error 0.536321û = 14.8156+5.7066x = 14.8156+ 5.7066(3.75) = 14.8156+21.3998 = 36.2154 The estimated salary of a person with a GPA of 3.75 iSalary 十十十十十十十十十十十 25 ++ ++ ++ + ++ + | + ++ 5 2. ++ + + 3 GPA ++ 3.5

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