In Today’s world, eCommerce becoming very popular in all over the world. In Australia, eCommerce market generates about US $ 11789 Million revenue in 2018 and expected to rise in near future.
eCommerce provides the online platform to the end user to buy the products directly. This service is very useful for the people who are very busy in their work. eCommerce also provides offers to the customers which attract the more people. In eCommerce platform, consumers have more choice in quality, brand, color, price etc.
As the people got interest in the buying through eCommerce platform, it increases the challenges to the service provider. It mainly consist of customer satisfaction and increased competition. For the service provider, it is mandatory to satisfy the customer and compete in the market.
In this case study, we develop the data sets regarding to computer and computer accessories for the 1000 products. We considered the following attributes
We define following variables as
Total Monthly sale amount= Sale Price × Number of customers
Total monthly profit= Profit × Number of customers
Profit Percentage= Total monthly profit /
We are interested to know the following things
Statistical tools and techniques are the necessary for analysis. In the analysis, selection of proper statistical tools and techniques is very important. For profit analysis, we summarised the profit percentage on total sale and profit per item for shipping type, customer type and region. We have used the Chi-square test of association for testing the associations of attributes. We have used independent two sample t-test for the testing the mean price, sale price, profit and rating for shipping type and customer type. We analyzed the mean price, sale price, profit and rating of region using one way ANOVA technique. We test the significance of correlation between two variable. We tried to fit regression model for the monthly profit and monthly sales.
Profit Analysis:
In profit analysis, we have given the total sales amount, total profit, profit percentage on total sale amount and profit per item. Larger the value of profit percentage on total sale amount and profit per item would be considered better for increasing the profit.
Table 1 gives the profit analysis according to shipping type. Both the shipping type customer paid and free have no much difference in the average profit percentage. Still total monthly sale and profit seems to be large for paid shipping customer than free shipping customer.
Table 1: Profit analysis according to shipping type
SHIPPING_TYPE |
Total Monthly Sale (Amount) |
Total Monthly Profit |
Profit Percentage |
Customer paid |
12543201 |
8973541 |
61.73% |
Free |
11502245 |
8275595 |
61.76% |
Table 2 represent the profit analysis according to customer type. New customers produces more total monthly sale and profit than existing customer. Also profit earned in case of new customer is more than existing customer.
Table 2: Profit analysis according to customer type
CUSTOMER_TYPE |
Total Monthly Sale (Amount) |
Total Monthly Profit |
Profit Percentage |
Existing |
11518008 |
8313334 |
61.36% |
New |
12527438 |
8935802 |
62.10% |
Table 3 represent the profit analysis according to region. ACT region shows less profit percentage than other region. Total monthly sale was significantly large for NSW region and small for VIC and SA region.
Table 3: Profit analysis according to region
REGION |
Total Monthly Sale (Amount) |
Total Monthly Profit |
Profit Percentage |
ACT |
3948523 |
2789840 |
56.7% |
NSW |
9063314 |
6460887 |
62.1% |
QLD |
5714404 |
4118342 |
60.6% |
SA |
2939639 |
2132391 |
65.6% |
VIC |
2379566 |
1747676 |
66.5% |
From Table 4, we have observed the summary statistics for profit and rating variable. One can study the properties of variables by observing the summary statistics. For example, as skewness are negative for both the variables suggest that they are negatively skewed.
Table 4: Descriptive statistics for profit and rating variable
PROFIT |
Rating |
|
Mean |
969.775 |
2.83622 |
Standard Error |
17.62128 |
0.047632 |
Median |
1023.5 |
2.72 |
Mode |
1333 |
5 |
Standard Deviation |
557.2338 |
1.506243 |
Sample Variance |
310509.5 |
2.268768 |
Kurtosis |
-0.50686 |
-1.00682 |
Skewness |
-0.40549 |
-0.22308 |
Range |
2686 |
5 |
Minimum |
-531 |
0 |
Maximum |
2155 |
5 |
Sum |
969775 |
2836.22 |
Count |
1000 |
1000 |
Here we test the
H0: There is no association between two attributes.
Vs
H1: There is association between two attributes.
We studied the association between two variables by Chi-square test for association. We have three pair of attributes for possible association
Shipping type and region.
Contingency Table for shipping type and region
Observed Frequencies |
REGION |
|||||
SHIPPING_TYPE |
ACT |
NSW |
QLD |
SA |
VIC |
Grand Total |
Customer paid |
80 |
207 |
124 |
67 |
49 |
527 |
Free |
81 |
180 |
113 |
53 |
46 |
473 |
Grand Total |
161 |
387 |
237 |
120 |
95 |
1000 |
Expected Frequencies |
REGION |
||||
SHIPPING_TYPE |
ACT |
NSW |
QLD |
SA |
VIC |
Customer paid |
84.847 |
203.949 |
124.899 |
63.24 |
50.065 |
Free |
76.153 |
183.051 |
112.101 |
56.76 |
44.935 |
P-value= 0.8754
As P-value is larger (>0.05) we fail to reject H0 i.e. there is no any association between region and shipping type.
Customer type and region.
Contingency Table for customer type and region
Observed Frequencies |
REGION |
|||||
CUSTOMER_TYPE |
ACT |
NSW |
QLD |
SA |
VIC |
Grand Total |
Existing |
76 |
185 |
121 |
53 |
45 |
480 |
New |
85 |
202 |
116 |
67 |
50 |
520 |
Grand Total |
161 |
387 |
237 |
120 |
95 |
1000 |
Expected Frequencies |
REGION |
||||
CUSTOMER_TYPE |
ACT |
NSW |
QLD |
SA |
VIC |
Existing |
77.28 |
185.76 |
113.76 |
57.6 |
45.6 |
New |
83.72 |
201.24 |
123.24 |
62.4 |
49.4 |
P-value= 0.7990
As P-value is larger (>0.05) we fail to reject H0 i.e. there is no any association between region and customer type.
Contingency Table for customer type and shipping type
Observed Frequencies |
SHIPPING_TYPE |
||
CUSTOMER_TYPE |
Customer paid |
Free |
Grand Total |
Existing |
247 |
233 |
480 |
New |
280 |
240 |
520 |
Grand Total |
527 |
473 |
1000 |
Expected Frequencies |
SHIPPING_TYPE |
|
CUSTOMER_TYPE |
Customer paid |
Free |
Existing |
252.96 |
227.04 |
New |
274.04 |
245.96 |
P-value= 0.4499
As P-value is larger (>0.05) we fail to reject H0 i.e. there is no any association between customer type and shipping type.
We test whether there is significant difference between two groups. Here our null and alternative hypothesis are as follows
H0: There is no significant difference between the two groups (two independent samples).
Vs
H1: There is significant difference between the two groups (two independent samples)
In shipping type, we have two groups customer paid and free shipping. We run the Two Sample t-test for testing the difference of mean of two groups customer paid and free shipping.
Table 5a shows the result of two sample t-test by assuming unequal variances for testing the difference of mean of two groups customer paid and free shipping of product price. From Table 5a, we compare the P value=0.409583 > 0.05, so we fail to reject null hypothesis i.e. there is no significant difference between the mean price of products for which customer paid shipping charges and free shipping.
Table 5a: t-Test: Two-Sample Assuming Unequal Variances for testing the difference of mean of two groups customer paid and free shipping of product price
|
Customer paid |
Free |
Mean |
387.1404 |
377.7104 |
Variance |
36877.6 |
28703.86 |
Observations |
527 |
473 |
Hypothesized Mean Difference |
0 |
|
df |
998 |
|
t Stat |
0.824976 |
|
P(T<=t) one-tail |
0.204791 |
|
t Critical one-tail |
1.646382 |
|
P(T<=t) two-tail |
0.409583 |
|
t Critical two-tail |
1.962344 |
Table 5b displays the result of two sample t-test by assuming unequal variances for testing the difference of mean of two groups customer paid and free shipping of sale price. From Table 5b, we can see that the P value=0.856732 > 0.05, so we fail to reject null hypothesis i.e. there is no significant difference between the mean sale price of products for which customer paid shipping charges and free shipping.
Table 5b: t-Test: t-Test: Two-Sample Assuming Unequal Variances for testing the difference of mean of two groups customer paid and free shipping of sale price
|
Customer paid |
Free |
Mean |
1355.313 |
1349.271 |
Variance |
286744.9 |
272226.6 |
Observations |
527 |
473 |
Hypothesized Mean Difference |
0 |
|
df |
991 |
|
t Stat |
0.180583 |
|
P(T<=t) one-tail |
0.428366 |
|
t Critical one-tail |
1.646393 |
|
P(T<=t) two-tail |
0.856732 |
|
t Critical two-tail |
1.962361 |
Table 5c demonstrate the result of two sample t-test by assuming unequal variances for testing the difference of mean of two groups customer paid and free shipping of number of customers. From Table 5c, we can see that the P value=0.174658 > 0.05, so we fail to reject null hypothesis i.e. there is no significant difference between the mean number of customers for which customer paid shipping charges and free shipping.
Table 5c: t-Test: Two-Sample Assuming Unequal Variances for testing the difference of mean of two groups customer paid and free shipping of number of customers
|
Customer paid |
Free |
Mean |
17.37002 |
17.86258 |
Variance |
27.62518 |
37.39421 |
Observations |
527 |
473 |
Hypothesized Mean Difference |
0 |
|
df |
936 |
|
t Stat |
-1.35842 |
|
P(T<=t) one-tail |
0.087329 |
|
t Critical one-tail |
1.646483 |
|
P(T<=t) two-tail |
0.174658 |
|
t Critical two-tail |
1.962502 |
Table 5d shows the result of two sample t-test by assuming unequal variances for testing the difference of mean of two groups customer paid and free shipping of profit. From Table 5d, we can see that the P value=0.923549 > 0.05, so we do not reject null hypothesis i.e. there is no significant difference between the mean profit for which customer paid shipping charges and free shipping.
Table 5d: t-Test: Two-Sample Assuming Unequal Variances for testing the difference of mean of two groups customer paid and free shipping of profit
|
Customer paid |
Free |
Mean |
968.1727 |
971.5603 |
Variance |
313939.3 |
307339.2 |
Observations |
527 |
473 |
Hypothesized Mean Difference |
0 |
|
df |
989 |
|
t Stat |
-0.09599 |
|
P(T<=t) one-tail |
0.461774 |
|
t Critical one-tail |
1.646396 |
|
P(T<=t) two-tail |
0.923549 |
|
t Critical two-tail |
1.962366 |
Table 5e presents the result of two sample t-test by assuming unequal variances for testing the difference of mean of two groups customer paid and free shipping of rating. From Table 5e, we can see that the P value < 0.05, so we reject the null hypothesis i.e. there is significant difference between the mean rating for which customer paid shipping charges and free shipping.
Table 5e: t-Test: Two-Sample Assuming Unequal Variances for testing the difference of mean of two groups customer paid and free shipping of rating
|
Customer paid |
Free |
Mean |
1.996262 |
3.772072 |
Variance |
1.858913 |
1.064905 |
Observations |
527 |
473 |
Hypothesized Mean Difference |
0 |
|
df |
971 |
|
t Stat |
-23.3604 |
|
P(T<=t) one-tail |
1.97E-96 |
|
t Critical one-tail |
1.646424 |
|
P(T<=t) two-tail |
3.95E-96 |
|
t Critical two-tail |
1.96241 |
In customer type, we have two groups’ new customer and existing customer. We run the Two Sample t-test for testing difference between the new customer’s and existing customer’s mean of
Table 6a shows the result of two sample t-test by assuming unequal variances for mean of product price. From Table 6a, we compare the P value=0.75979 > 0.05, so we fail to reject null hypothesis i.e. there is no significant difference between the mean price of products for the new customer and existing customer.
Table 6a: t-Test: Two-Sample Assuming Unequal Variances for testing the difference of mean of new customer’s and exiting customer’s product price
|
Existing Customer |
New Customer |
Mean |
380.8521 |
384.3673 |
Variance |
32279.71 |
33724.35 |
Observations |
480 |
520 |
Hypothesized Mean Difference |
0 |
|
df |
995 |
|
t Stat |
-0.30584 |
|
P(T<=t) one-tail |
0.379895 |
|
t Critical one-tail |
1.646386 |
|
P(T<=t) two-tail |
0.75979 |
|
t Critical two-tail |
1.962351 |
Table 6b displays the result of two sample t-test by assuming unequal variances for testing the difference of mean of new customer’s and exiting customer’s product price. From Table 6b, we can see that the P value=0.345394> 0.05, so we fail to reject null hypothesis i.e. there is no significant difference between the mean sale price of products for the new customer and existing customer.
Table 6b: Two-Sample Assuming Unequal Variances for testing the difference of mean of new customer’s and exiting customer’s sale price
|
Existing Customer |
New Customer |
Mean |
1368.904 |
1337.271 |
Variance |
287795.1 |
272108.4 |
Observations |
480 |
520 |
Hypothesized Mean Difference |
0 |
|
df |
986 |
|
t Stat |
0.944013 |
|
P(T<=t) one-tail |
0.172697 |
|
t Critical one-tail |
1.6464 |
|
P(T<=t) two-tail |
0.345394 |
|
t Critical two-tail |
1.962373 |
Table 6c demonstrate the result of two sample t-test by assuming unequal variances for testing the difference of mean of new customer’s and exiting customer’s number of customers. From Table 6c, we can see that the P value=0.431685> 0.05, so we fail to reject null hypothesis i.e. there is no significant difference between the mean number of customer for the new customer and existing customer.
Table 6c: Two-Sample Assuming Unequal Variances for testing the difference of mean of new customer’s and exiting customer’s number of customer
|
Existing Customer |
New Customer |
Mean |
17.45625 |
17.73846 |
Variance |
30.20267 |
34.20892 |
Observations |
480 |
520 |
Hypothesized Mean Difference |
0 |
|
df |
998 |
|
t Stat |
-0.78663 |
|
P(T<=t) one-tail |
0.215842 |
|
t Critical one-tail |
1.646382 |
|
P(T<=t) two-tail |
0.431685 |
|
t Critical two-tail |
1.962344 |
Table 6d presents the result of two sample t-test by assuming unequal variances for testing the difference of mean of new customer’s and exiting customer’s profit. From Table 6d, we can see that the P value=0.319876> 0.05, so we fail to reject null hypothesis i.e. there is no significant difference between the mean profit for the new customer and existing customer.
Table 6d: Two-Sample Assuming Unequal Variances for testing the difference of mean of new customer’s and exiting customer’s profit
|
Existing Customer |
New Customer |
Mean |
988.0521 |
952.9038 |
Variance |
321110 |
300730.3 |
Observations |
480 |
520 |
Hypothesized Mean Difference |
0 |
|
df |
985 |
|
t Stat |
0.995215 |
|
P(T<=t) one-tail |
0.159938 |
|
t Critical one-tail |
1.646402 |
|
P(T<=t) two-tail |
0.319876 |
|
t Critical two-tail |
1.962375 |
Table 6e presents the result of two sample t-test by assuming unequal variances for testing the difference of mean of new customer’s and exiting customer’s rating. From Table 6e, we can see that the P value < 0.05, so we reject null hypothesis i.e. there is significant difference between the mean rating for the new customer and existing customer.
Table 6e: Two-Sample Assuming Unequal Variances for testing the difference of mean of new customer’s and exiting customer’s rating
|
Existing Customer |
New Customer |
Mean |
1.875688 |
3.722865 |
Variance |
1.770462 |
1.092094 |
Observations |
480 |
520 |
Hypothesized Mean Difference |
0 |
|
df |
908 |
|
t Stat |
-24.2784 |
|
P(T<=t) one-tail |
4.8E-101 |
|
t Critical one-tail |
1.646534 |
|
P(T<=t) two-tail |
9.7E-101 |
|
t Critical two-tail |
1.96258 |
In this section, we test whether the mean of different region is not significantly different. We run the one way ANOVA for testing whether the different region have different mean or not for
Table 7 represent the descriptive statistics for the Product Price, Sale Price, Number of customers, Profit and Rating for different region
Table 7: Summary statistics for the Product Price, Sale Price, Number of customers, Profit and Rating for different region.
Variable |
REGION |
N |
Mean |
Median |
StDev |
Min |
Max |
Q1 |
Q3 |
Product PRICE |
ACT |
161 |
392.1 |
321 |
184.7 |
102 |
789 |
258 |
540 |
NSW |
387 |
385.34 |
321 |
178.16 |
102 |
854 |
258 |
540 |
|
QLD |
237 |
379.7 |
321 |
185.7 |
102 |
987 |
254 |
456 |
|
SA |
120 |
368.2 |
321 |
168 |
102 |
956 |
255 |
456 |
|
VIC |
95 |
381.4 |
312 |
198.8 |
102 |
937 |
254 |
582 |
|
SALES Price |
ACT |
161 |
1311.4 |
1456 |
517.7 |
258 |
2365 |
1023 |
1674.5 |
NSW |
387 |
1358.6 |
1456 |
531.3 |
258 |
2365 |
1023 |
1789 |
|
QLD |
237 |
1361.9 |
1569 |
546.1 |
258 |
2365 |
1023 |
1789 |
|
SA |
120 |
1356.4 |
1456 |
514.9 |
258 |
2365 |
1023 |
1789 |
|
VIC |
95 |
1368.4 |
1550 |
517.9 |
258 |
2365 |
1065 |
1695 |
|
Number of Customers |
ACT |
161 |
18.199 |
17 |
6.432 |
9 |
40 |
15 |
19 |
NSW |
387 |
17.214 |
17 |
4.719 |
9 |
43 |
16 |
19 |
|
QLD |
237 |
17.515 |
18 |
6.013 |
9 |
43 |
15 |
19 |
|
SA |
120 |
18.033 |
18 |
6.099 |
9 |
43 |
15.25 |
19 |
|
VIC |
95 |
17.853 |
18 |
6.459 |
9 |
43 |
15 |
20 |
|
PROFIT |
ACT |
161 |
919.3 |
1000 |
558.3 |
-531 |
1946 |
689 |
1337.5 |
NSW |
387 |
973.3 |
1000 |
554.8 |
-531 |
2153 |
625 |
1420 |
|
QLD |
237 |
982.1 |
1154 |
589.9 |
-420 |
2145 |
617 |
1408 |
|
SA |
120 |
988.2 |
1023 |
532.3 |
-198 |
2151 |
654 |
1349 |
|
VIC |
95 |
987 |
1000 |
517.3 |
-96 |
2155 |
696 |
1345 |
|
Rating |
ACT |
161 |
2.917 |
2.73 |
1.43 |
0.02 |
5 |
2.245 |
4.42 |
NSW |
387 |
2.8197 |
2.75 |
1.5273 |
0 |
5 |
2.12 |
4.33 |
|
QLD |
237 |
2.84 |
2.73 |
1.55 |
0 |
5 |
2.11 |
4.345 |
|
SA |
120 |
2.8 |
2.6 |
1.514 |
0.05 |
5 |
2.065 |
4.248 |
|
VIC |
95 |
2.801 |
2.67 |
1.451 |
0.01 |
5 |
2.13 |
4.16 |
Table 8a shows the output of the one way ANOVA for testing whether the mean product price for different region is significantly different from each other or not. As the P-value = 0.853 > 0.05, we fail to reject null hypothesis that there is no significant difference between the mean product price for different region. Output also gives the 95% confidence interval for mean price of different region.
Table 8a: Output of the one way ANOVA for testing whether the mean product price for different region
One-way ANOVA: PRICE versus REGION
Analysis of Variance for PRICE
Source DF SS MS F P
REGION 4 44614 11154 0.34 0.853
Error 995 32923389 33089
Total 999 32968004
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev ——+———+———+———+
ACT 161 392.1 184.7 (———-*———-)
NSW 387 385.3 178.2 (——*——)
QLD 237 379.7 185.7 (——–*——–)
SA 120 368.2 168.0 (————*————)
VIC 95 381.4 198.8 (————–*————-)
——+———+———+———+
Pooled StDev = 181.9 350 375 400 425
Table 8b represents the output of the one way ANOVA for testing whether the mean sale price for different region is significantly different from each other or not. As the P-value = 0.88 > 0.05, we fail to reject null hypothesis that there is no significant difference between the mean sale price for different region. Output also gives the 95% confidence interval for mean sale price of different region.
Table 8b: Output of the one way ANOVA for testing whether the mean sale price for different region
Analysis of Variance for SALES
Source DF SS MS F P
REGION 4 333084 83271 0.30 0.880
Error 995 278994796 280397
Total 999 279327880
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev —–+———+———+———+-
ACT 161 1311.4 517.7 (———-*———–)
NSW 387 1358.6 531.3 (——*——-)
QLD 237 1361.9 546.1 (———*——–)
SA 120 1356.4 514.9 (————-*————)
VIC 95 1368.4 517.9 (————–*—————)
—–+———+———+———+-
Pooled StDev = 529.5 1260 1330 1400 1470
Table 8c demonstrates the output of the one way ANOVA for testing whether the mean number of customers for different region is significantly different from each other or not. As the P-value = 0.342 > 0.05, we fail to reject null hypothesis that there is no significant difference between the mean number of customers for different region. We can look at the 95% confidence interval for mean number of customers of different region.
Table 8c: Output of the one way ANOVA for testing whether the mean number of customers for different region
Analysis of Variance for NO_OF_CU
Source DF SS MS F P
REGION 4 145.6 36.4 1.13 0.342
Error 995 32095.8 32.3
Total 999 32241.4
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev —+———+———+———+—
ACT 161 18.199 6.432 (————*————)
NSW 387 17.214 4.719 (——-*——-)
QLD 237 17.515 6.013 (———*———-)
SA 120 18.033 6.099 (————–*————-)
VIC 95 17.853 6.459 (—————*—————)
—+———+———+———+—
Pooled StDev = 5.680 16.80 17.50 18.20 18.9
Table 8d reveals the output of the one way ANOVA for testing whether the mean profit for different region is significantly different from each other or not. As the P-value = 0.795 > 0.05, we fail to reject null hypothesis that there is no significant difference between the mean profit for different region. We can observe the 95% confidence interval for mean profit of different region.
Table 8d: Output of the one way ANOVA for testing whether the mean profit for different region
Analysis of Variance for PROFIT
Source DF SS MS F P
REGION 4 520860 130215 0.42 0.795
Error 995 309678180 311234
Total 999 310199040
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev ——+———+———+———+
ACT 161 919.3 558.3 (———-*———-)
NSW 387 973.3 554.8 (——*——)
QLD 237 982.1 589.9 (——–*——–)
SA 120 988.2 532.3 (————*———–)
VIC 95 987.0 517.3 (————-*————-)
——+———+———+———+
Pooled StDev = 557.9 880 960 1040 1120
Table 8e shows the output of the one way ANOVA for testing whether the mean rating for different region is significantly different from each other or not. As the P-value = 0.959 > 0.05, we fail to reject null hypothesis that there is no significant difference between the mean rating for different region. We can observe the 95% confidence interval for mean rating of different region.
Table 8e: Output of the one way ANOVA for testing whether the mean rating for different region
One-way ANOVA: Average Rating versus REGION
Analysis of Variance for Average Rating
Source DF SS MS F P
REGION 4 1.44 0.36 0.16 0.959
Error 995 2265.06 2.28
Total 999 2266.50
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev ——+———+———+———+
ACT 161 2.917 1.430 (———–*———–)
NSW 387 2.820 1.527 (——-*——-)
QLD 237 2.840 1.550 (———*———)
SA 120 2.800 1.514 (————*————-)
VIC 95 2.801 1.451 (————–*————–)
——+———+———+———+
Pooled StDev = 1.509 2.60 2.80 3.00 3.20
Correlation Analysis:
In the following Table 9, we represent the Pearson’s correlation coefficient and P-value of its significance for Product Price, Sale Price, Number of customers, Profit and Rating.
Table 9: Pearson’s correlation coefficient and P-value of its significance for Product Price, Sale Price, Number of customers, Profit and Rating
Product Price |
Sale Price |
Number of customers |
Profit |
|
Sale Price |
0.011 |
|||
0.73 |
||||
Number of customers |
0.058 |
0.079 |
||
0.066 |
0.012 |
|||
Profit |
-0.316 |
0.945 |
0.056 |
|
0 |
0 |
0.075 |
||
Rating |
0.158 |
-0.31 |
0.016 |
-0.346 |
0 |
0 |
0.618 |
0 |
From Table 9, we can say that following pair has significant correlation
We considered the monthly profit as a response variable and monthly sale as a predictor variable. Table 10 displays the output of simple liner regression on monthly profit by monthly sales. We observed that R2 =89% which pretty high suggest that model fitting is good. We can predict the monthly profit by observing the monthly sales using regression equation.
Table 10: Regression analysis of monthly profit by monthly sales
The regression equation is
Monthly_Profit = – 3625 + 0.868 Monthly_Sale
Predictor Coef SE Coef T P
Constant -3624.6 263.6 -13.75 0.000
Monthly_ 0.868094 0.009670 89.77 0.000
S = 3928 R-Sq = 89.0% R-Sq(adj) = 89.0
Analysis of Variance
Source DF SS MS F P
Regression 1 1.24337E+11 1.24337E+11 8058.92 0.000
Residual Error 998 15397640608 15428498
Total 999 1.39735E+1
Conclusions
From the profit analysis, we can say that shipping type and customer type does not show any significant changes in the profit whereas different region have different have different profit. From two sample t-test for shipping type and customer type, average rating shows significant difference between the groups whereas other variables as product price, sale price, number of customers and profit does not show any significance difference between the groups. There is no any significant difference between the different region for product price, sale price, number of customers, profit and rating. Correlation analysis suggest that there is significant correlation between Product price and Profit, Product price and Rating, Sale price and Number of customers, Sale price and Profit, Sale price and Rating, Profit and Rating. Regression analysis shows that we can predict the monthly profit by observing the monthly sales. They are positively correlates
List of References
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