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In this assignment, we compute the consumer characteristics to predict the amount charged by the users of credit card. The data for Consumer information is given below: -

 Income (\$1000s) Household Size Amount Charged (\$) Income (\$1000s) Household Size Amount Charged (\$) 54 3 4016 54 6 5573 30 2 3159 30 1 2583 32 4 5100 48 2 3866 50 5 4742 34 5 3586 31 2 1864 67 4 5037 55 2 4070 50 2 3605 37 1 2731 67 5 5345 40 2 3348 55 6 5370 66 4 4764 52 2 3890 51 3 4110 62 3 4705 25 3 4208 64 2 4157 48 4 4219 22 3 3579 27 1 2477 29 4 3890 33 2 2514 39 2 2972 65 3 4214 35 1 3121 63 4 4965 39 4 4183 42 6 4412 54 3 3720 21 2 2448 23 6 4127 44 1 2995 27 2 2921 37 5 4171 26 7 4603 62 6 5678 61 2 4273 21 3 3623 30 2 3067 55 7 5301 22 4 3074 42 2 3020 46 5 4820 41 7 4828 66 4 5149

The data comprises of household size, annual income and annual charges of credit card for a sample of 50 consumers. Now we move on to the analysis part: -

The descriptive statistics of the data is given below:

 Descriptive statistics Income (\$1000s) Household Size Amount Charged (\$) Mean 43.48 3.42 3963.86 Standard Error 2.057785614 0.245930138 132.023387 Median 42 3 4090 Mode 54 2 3890 Standard Deviation 14.55074162 1.738988681 933.5463219 Sample Variance 211.7240816 3.024081633 871508.7351 Kurtosis -1.247719422 -0.722808552 -0.742482171 Skewness 0.095855639 0.527895977 -0.128860064 Range 46 6 3814 Minimum 21 1 1864 Maximum 67 7 5678 Sum 2174 171 198193 Count 50 50 50 Largest(1) 67 7 5678 Smallest(1) 21 1 1864 Confidence Level (95.0%) 4.135274935 0.494215106 265.3109241

The equation for credit card charges can be given as: -

Yt = βXt + ui  ......Eq(1)

Here Yt is our dependent variable which is annual charges on credit card and Xt is our independent variable which is annual income (\$1000s). The regression results are given below: -

 SUMMARY OUTPUT Regression Statistics Multiple R 0.630781 R Square 0.397884 Adjusted R Square 0.38534 Standard Error 731.9025 Observations 50 ANOVA df SS MS F Significance F Regression 1 16991229 16991229 31.71892 9.1E-07 Residual 48 25712699 535681.2 Total 49 42703928 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 2204.241 329.134 6.697091 0.00 1542.472 2866.009 1542.472207 2866.0088 Income (\$1000s) 40.46963 7.185716 5.631955 0.00 26.02178 54.91748 26.02177931 54.917479

From the regression results, we can say that 38.5% of the variation in annual charges on credit card is explained by the variable annual income (Adjusted R2). The coefficients imply that if there is \$1000 (1 unit of the variable annual income) increase in annual income, then there is an increase of 40.47 units in annual credit card charges.

Another equation for credit card charges can be given as: -

Yt = βZt + ui  ......Eq(2)

Here Yt is our dependent variable which is annual charges on credit card and Xt is our independent variable which is household size. The regression results are given below:-

 SUMMARY OUTPUT Regression Statistics Multiple R 0.752854 R Square 0.566789 Adjusted R Square 0.557764 Standard Error 620.8163 Observations 50 ANOVA df SS MS F Significance F Regression 1 24204112 24204112 62.80048 2.86E-10 Residual 48 18499816 385412.8 Total 49 42703928 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 2581.644 195.2699 13.2209 0.00 2189.028 2974.26 2189.027669 2974.2605 Household Size 404.1567 50.99978 7.924676 0.00 301.6148 506.6986 301.6147764 506.69863

From the regression results, we can say that 55.8% of the variation in annual charges on credit card is explained by the variable household size (Adjusted R2). The coefficients imply that if there is 1 unit increase in the number of household members, then there is an increase of 404.2 units in annual credit card charges.

After viewing the above two variables, we can say that household size is better than annual income in predicting annual credit card charges.

The equation for credit card charges taking both the variables household size and annual income can be given as: -

Yt = β1Xt + β2Zt + ui  .....Eq(3)

The regression results are given below:-

 SUMMARY OUTPUT Regression Statistics Multiple R 0.908502 R Square 0.825376 Adjusted R Square 0.817945 Standard Error 398.3249 Observations 50 ANOVA df SS MS F Significance F Regression 2 35246779 17623389 111.0745 1.55E-18 Residual 47 7457149 158662.8 Total 49 42703928 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 1305.034 197.771 6.598712 0.00 907.17 1702.898 907.17 1702.898 Income (\$1000s) 33.12196 3.970237 8.342563 0.00 25.13487 41.10904 25.13487 41.10904 Household Size 356.3402 33.2204 10.72655 0.00 289.5094 423.171 289.5094 423.171

From the regression results, we can say that 81.8% of the variation in annual charges on credit card is explained by the variables household size and annual income (Adjusted R2). The coefficients imply that if there is 1 unit increase in the number of household members, then there is an increase of 356.34 units in annual credit card charges whereas if there is \$1000 (1 unit of the variable annual income) increase in annual income, then there is an increase of 33.12 units in annual credit card charges.

Hence the fitted regression equation can be given as: -

Yt = 33.12Xt + 356.34Zt

The predicted annual credit card charge for a three-person household with an annual income of \$40,000 would be: -

Yt = 33.12*40 + 356.34*3 = \$2393.82.

The other factors that affect the annual charges of credit card are interest rate, level of education of the individual and past credit history of the individual. If the interest rate is high, the annual charges will increase proportionately and this would tend the individual to decrease the frequency of credit card usage.

The descriptive statistics of the variables are given below: -

 Descriptive Statistics HI001 FINAL EXAM HI001 ASSIGNMENT 01 HI001 ASSIGNMENT 02 Mean 31.90909091 17.34343434 15.50505051 Standard Error 0.700162085 0.237298066 0.23564704 Median 32 17 16 Mode 29 18 17 Standard Deviation 6.966524782 2.361085949 2.344658442 Sample Variance 48.53246753 5.57472686 5.497423212 Kurtosis 7.67534849 10.3018632 0.698973651 Skewness -1.753036803 0.803185137 -0.464616962 Range 50 22 13 Minimum 0 8 8 Maximum 50 30 21 Sum 3159 1717 1535 Count 99 99 99 Largest(1) 50 30 21 Smallest(1) 0 8 8 Confidence Level(95.0%) 1.389448835 0.470910278 0.467633869

 Descriptive Statistics HI003 FINAL EXAM HI003 ASSIGNMENT 01 HI003 ASSIGNMENT 02 Mean 26.23232323 18.31313131 13.60606061 Standard Error 0.861918907 0.408537639 0.187651228 Median 25 19 13 Mode 25 20 13 Standard Deviation 8.57598484 4.064898183 1.867106141 Sample Variance 73.54751598 16.52339724 3.486085343 Kurtosis 0.474751131 1.51303057 3.505251459 Skewness -0.027305979 -0.236180187 1.121313851 Range 46 20 12 Minimum 4 10 8 Maximum 50 30 20 Sum 2597 1813 1347 Count 99 99 99 Largest(1) 50 30 20 Smallest(1) 4 10 8 Confidence Level(95.0%) 1.710449975 0.810729628 0.372387745

 Descriptive Statistics HI002 FINAL EXAM HI002 ASSIGNMENT 01 HI002 ASSIGNMENT 02 Mean 26.73737374 17.93939394 12.49494949 Standard Error 0.636870612 0.365435664 0.213139666 Median 27 19 13 Mode 27 20 14 Standard Deviation 6.336782578 3.636038947 2.120712902 Sample Variance 40.15481344 13.22077922 4.497423212 Kurtosis 3.924830269 3.372356549 5.049593179 Skewness -0.312442386 -1.183845155 -1.204878419 Range 50 26 16 Minimum 0 4 4 Maximum 50 30 20 Sum 2647 1776 1237 Count 99 99 99 Largest(1) 50 30 20 Smallest(1) 0 4 4 Confidence Level(95.0%) 1.26384897 0.725195163 0.42296872

The 10 different correlations between the pairs of variables are given below: -

The variables HI003 FINAL EXAM and HI002 FINAL EXAM are positively correlated with a correlation coefficient of 0.207867. The p-value is 0.039 and hence the correlation coefficient is statistically significant. It is a weak correlation.

The variables HI001 FINAL EXAM and HI002 FINAL EXAM are positively correlated with a correlation coefficient of 0.142303. The p-value is 0.1600 and hence the correlation coefficient is statistically insignificant. It is a weak correlation.

The variables HI001 ASSIGNMENT 01 and HI003 ASSIGNMENT 01 are positively correlated with a correlation coefficient of 0.155602. The p-value is 0.1241 and hence the correlation coefficient is statistically insignificant. It is a weak correlation.

The variables HI003 ASSIGNMENT 01 and HI003 ASSIGNMENT 02 are positively correlated with a correlation coefficient of 0.567657. The p-value is 0.000 and hence the correlation coefficient is statistically significant. It is a strong correlation.

The variables HI001 FINAL EXAM and HI003 FINAL EXAM are positively correlated with a correlation coefficient of 0.187035. The p-value is 0.0638 and hence the correlation coefficient is statistically significant. It is a weak correlation.

The variables HI001 ASSIGNMENT 01 and HI001 ASSIGNMENT 02 are positively correlated with a correlation coefficient of 0.648505. The p-value is 0.000 and hence the correlation coefficient is statistically significant. It is a strong correlation.

The variables HI001 ASSIGNMENT 02 and HI002 ASSIGNMENT 02 are positively correlated with a correlation coefficient of 0.035405. The p-value is 0.7279 and hence the correlation coefficient is statistically insignificant. It is a weak correlation.

The variables HI002 ASSIGNMENT 01 and HI002 ASSIGNMENT 02 are positively correlated with a correlation coefficient of 0.603392. The p-value is 0.000 and hence the correlation coefficient is statistically significant. It is a strong correlation.

The variables HI002 ASSIGNMENT 01 and HI003 ASSIGNMENT 01 are negatively correlated with a correlation coefficient of -0.11055. The p-value is 0.2760 and hence the correlation coefficient is statistically insignificant. It is a weak correlation.

The variables HI003 ASSIGNMENT 02 and HI002 ASSIGNMENT 02 are positively correlated with a correlation coefficient of 0.031706. The p-value is 0.7554 and hence the correlation coefficient is statistically insignificant. It is a weak correlation.

The Descriptive Statistics of the first group (Med 1) is given below: -

 Descriptive Statistics Florida New York North Carolina Mean 5.55 8 7.05 Standard Error 0.478347 0.492041932 0.634428877 Median 6 8 7.5 Mode 7 8 8 Standard Deviation 2.139233 2.200478417 2.837252192 Sample Variance 4.576316 4.842105263 8.05 Kurtosis -1.06219 0.626431669 -0.904925496 Skewness -0.27356 0.625687389 -0.056188269 Range 7 9 9 Minimum 2 4 3 Maximum 9 13 12 Sum 111 160 141 Count 20 20 20 Largest(1) 9 13 12 Smallest(1) 2 4 3 Confidence Level(95.0%) 1.001192 1.029855598 1.327874898

The Descriptive Statistics of the second group (Med 2) is given below: -

 Descriptive Statistics Florida New York North Carolina Mean 14.5 15.25 13.95 Standard Error 0.708965146 0.923024 0.65884668 Median 14.5 14.5 14 Mode 17 14 12 Standard Deviation 3.170588522 4.12789 2.946451925 Sample Variance 10.05263158 17.03947 8.681578947 Kurtosis -0.340799481 -0.03014 -0.592052134 Skewness 0.280721497 0.525352 -0.041733773 Range 12 15 11 Minimum 9 9 8 Maximum 21 24 19 Sum 290 305 279 Count 20 20 20 Largest(1) 21 24 19 Smallest(1) 9 9 8 Confidence Level (95.0%) 1.483881102 1.931912 1.378981946

By viewing the descriptive statistics, we can say that the depression scores of the healthy group is far less than that of the group suffering from chronic health condition such as arthritis, hypertension, and/or heart ailment. We can also say that according to the sample, the individuals from Florida possess far better health conditions than individuals from New York and North Carolina.

Here the hypothesis that needs to be tested is whether there is a significant difference in the depression scores among various regions. The ANOVA test for both the groups are given below: -

For Med 1: -

 Anova: Single Factor SUMMARY Groups Count Sum Average Variance Florida 20 111 5.55 4.576316 New York 20 160 8 4.842105 North Carolina 20 141 7.05 8.05 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 61.03333 2 30.51667 5.240886 0.00814 3.158843 Within Groups 331.9 57 5.822807 Total 392.9333 59

In this case the F value is higher than the F critical value. So we reject the null hypothesis and state that there is significant difference in depression scores among the healthy individuals.

For Med 2: -

 Anova: Single Factor SUMMARY Groups Count Sum Average Variance Florida 20 290 14.5 10.05263 New York 20 305 15.25 17.03947 North Carolina 20 279 13.95 8.681579 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 17.03333 2 8.516667 0.714212 0.493906 3.158843 Within Groups 679.7 57 11.92456 Total 696.7333 59

In this case the F value is lower than the F critical value. So we accept the null hypothesis and state that there is no significant difference in depression scores among the non-healthy individuals.

According to me, the best way to treat depression is to arrange for a good setup of counselling. This cannot be cured in the short run and hence requires time to develop. We also need to see to the health conditions of the individuals for controlling their depression scores.
References

Field, A. (2012). Discovering statistics using SPSS (and sex and drugs and rock 'n' roll). 1st ed. Los Angeles [Calif.]: SAGE.

Hastie, T., Friedman, J. and Tibshirani, R. (2013). The elements of statistical learning. 1st ed. New York [u.a.]: Springer.

Huff, D. and Geis, I. (2006). How to lie with statistics. 1st ed. New York: W.W. Norton & Co.

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