2197 Words9 Pages

An AJ Davis store manager is requesting to identify some concerns of the customers’ database by using the Credit Balance (X) vs the Income (Y).
1) The scatterplot is to show the relationship between the Credit Balance (X) vs the Income (Y). The line in the middle is to show the ‘best fit”. There is a strong positive relation between the two variables as we can see from the line. Furthermore, we can state as the Credit Balance (X) increases that there is an increase in the Income level (Y), as the slope of the graph also indicates.
2) From the above scatter plot, we can see that there is a positive relationship between the two variables of Income (Y), which is also the dependent variable, and the Credit Balance (X), which is the independent*…show more content…*

the Credit Balance (X) we saw that there is a strong positive slope, meaning that customers who carry a larger credit balance usually will also have higher income as shown in the best-fit line. For linear regression equation we have the following; Income ($1000) = -3.51589 + 0.0119264 Credit Balance($). From this, we can see the B0= -3.51589, which is also the Y intercept, while B1 is 0.0119264 is the slope for this variable. The coefficient of correlation and the coefficient of determination both showed a positive strong relationship between the two variable of Credit Balance (X) and Income (Y). The R^2=86.52%, while R=0.80. From the Analysis of Variance, we saw the P Value= 0.000 the F Value= 85.6452. When we tested with the significance level of 5% we concluded the P value was less than, therefore we concluded to reject the null hypothesis for this level. We also performed the 95% Confidence level to be ($0.009and $0.015) for B1. In addition, we can estimated that a customer with a $4000 credit balance to have an income in between (41.7665, 46.6130) in $1000 using the 95% CI confidence levels to calculate the income level. At the same time, the average mean income is (41.7665+46.6130)/2= 44.19 or $44,190 rounded up to nearest dollar. We cannot really predict the credit balance of $10,000 as it is out of*…show more content…*

Credit Balance($) 0.01193 0.00129 9.25447 0.000 Summary of Model S = 8.40667 R-Sq = 64.08% R-Sq(adj) = 63.34% PRESS = 3613.50 R-Sq(pred) = 61.74% Analysis of Variance Source DF Seq SS Adj SS Adj MS F P Regression 1 6052.72 6052.72 6052.72 85.6452 0.000000 Credit Balance($) 1 6052.72 6052.72 6052.72 85.6452 0.000000 Error 48 3392.26 3392.26 70.67 Lack-of-Fit 47 3390.26 3390.26 72.13 36.0666 0.131532 Pure Error 1 2.00 2.00 2.00 Total 49 9444.98 Correlations: Income ($1000), Credit Balance($) Pearson correlation of Income ($1000) and Credit Balance($) = 0.801 P-Value = 0.000 To calculate the for the confidence levels at α=0.05 General Regression Analysis: Income ($1000) versus Credit Balance($) Regression Equation Income ($1000) = -3.51589 + 0.0119264 Credit Balance($) Coefficients Term Coef SE Coef T P Constant -3.51589 5.48309 -0.64123 0.524 Credit Balance($) 0.01193 0.00129 9.25447 0.000 Summary of Model S = 8.40667 R-Sq = 64.08% R-Sq(adj) =

the Credit Balance (X) we saw that there is a strong positive slope, meaning that customers who carry a larger credit balance usually will also have higher income as shown in the best-fit line. For linear regression equation we have the following; Income ($1000) = -3.51589 + 0.0119264 Credit Balance($). From this, we can see the B0= -3.51589, which is also the Y intercept, while B1 is 0.0119264 is the slope for this variable. The coefficient of correlation and the coefficient of determination both showed a positive strong relationship between the two variable of Credit Balance (X) and Income (Y). The R^2=86.52%, while R=0.80. From the Analysis of Variance, we saw the P Value= 0.000 the F Value= 85.6452. When we tested with the significance level of 5% we concluded the P value was less than, therefore we concluded to reject the null hypothesis for this level. We also performed the 95% Confidence level to be ($0.009and $0.015) for B1. In addition, we can estimated that a customer with a $4000 credit balance to have an income in between (41.7665, 46.6130) in $1000 using the 95% CI confidence levels to calculate the income level. At the same time, the average mean income is (41.7665+46.6130)/2= 44.19 or $44,190 rounded up to nearest dollar. We cannot really predict the credit balance of $10,000 as it is out of

Credit Balance($) 0.01193 0.00129 9.25447 0.000 Summary of Model S = 8.40667 R-Sq = 64.08% R-Sq(adj) = 63.34% PRESS = 3613.50 R-Sq(pred) = 61.74% Analysis of Variance Source DF Seq SS Adj SS Adj MS F P Regression 1 6052.72 6052.72 6052.72 85.6452 0.000000 Credit Balance($) 1 6052.72 6052.72 6052.72 85.6452 0.000000 Error 48 3392.26 3392.26 70.67 Lack-of-Fit 47 3390.26 3390.26 72.13 36.0666 0.131532 Pure Error 1 2.00 2.00 2.00 Total 49 9444.98 Correlations: Income ($1000), Credit Balance($) Pearson correlation of Income ($1000) and Credit Balance($) = 0.801 P-Value = 0.000 To calculate the for the confidence levels at α=0.05 General Regression Analysis: Income ($1000) versus Credit Balance($) Regression Equation Income ($1000) = -3.51589 + 0.0119264 Credit Balance($) Coefficients Term Coef SE Coef T P Constant -3.51589 5.48309 -0.64123 0.524 Credit Balance($) 0.01193 0.00129 9.25447 0.000 Summary of Model S = 8.40667 R-Sq = 64.08% R-Sq(adj) =

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