






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 0.988215! Null hypothesis: Ho:p=0 There is no correlation between GPA and Salary. Alternative hypothesis: H:20 There is a correlation between GPA and Salary.
Rejection 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 value of the test statistic is t=14.30.
Conclusion: Use a significance level, a = 0.05. Here the value of the test statistic 14.30, which is greater than the critical value 4.032. That is, t(=14.30) > 4.032. Therefore, by the rejection rule, reject the null hypothesis H.. Interpretation: There is sufficient evidence at level of significance a = 0.05. There is a correlation between GPA and Salary. EXCEL (Data Analysis) output:
SUMMARY OUTPUT Regression Statistics Multiple R 0.988215 R Square 0.976569 Adjusted R Square 0.971883 Standard Error 0.536321 Observations ANOVA df Regression Residual Total SSMSF gnificance F 1 59.9418 59.9418 208.3918 2.88E-05 5 1.438199 0.28764 6 61.38 Intercept GPA Coefficientsandard Errt Stat p-value Lower 95%Upper 95% 14.81562 1.234863 11.99778 7.1E-05 11.6413 17.98993 5.706569 0.395307 14.43578 2.88E-05 4.690399 6.722739 The regression equation is 9 = 14.8156+5.7066x.
û = 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 is $36.2154. The value of the correlation coefficient is r =0.988. Yes, there is an evidence of a significant correlation between GPA and salary.
Salary 十十十十十十十十十十十 25 ++ ++ ++ + ++ + | + ++ 5 2. ++ + + 3 GPA ++ 3.5