Question

You have been given data on n = 20 male customers. You are to use the data to determine the best regression model for predicting the amount spent in the current year using some or all of the other variables as predictors.

This question centers on the response variable. Question: Using the program Minitab, provide the mean, standard deviation, and five number summary, as well as a minimum of two graphs. Then use the statistics and graphs to address the shape of the distribution, as part of the answer to this question.

Data on n = 20 male customers.

Row	Customer	Age	Salary	Children	PrevSpent	Emails	AmountSpent
1	718	29	$20,100	1	$2	12	$271
2	452	40	$52,400	0	$0	18	$857
3	673	44	$53,300	0	$1,148	18	$1,720
4	399	26	$12,000	0	$313	24	$344
5	197	68	$20,800	0	$0	6	$260
6	367	53	$55,800	1	$0	6	$821
7	339	36	$74,600	2	$846	18	$705
8	1000	55	$102,500	1	$1,615	24	$2,464
9	548	73	$43,300	0	$0	12	$1,584
10	484	29	$17,300	1	$677	18	$381
11	853	25	$18,400	1	$974	6	$303
12	7	33	$34,700	0	$0	18	$1,615
13	848	68	$78,900	0	$2,422	18	$2,233
14	146	32	$73,300	0	$0	12	$1,694
15	614	70	$74,900	0	$0	12	$1,804
16	986	55	$123,600	1	$0	12	$1,892
17	361	33	$122,100	0	$2,963	12	$2,408
18	170	44	$84,800	1	$0	12	$1,826
19	683	49	$89,500	0	$2,333	24	$5,209
20	630	43	$48,700	2	$356	12	$528

Answer 1

summary statistic

	Customer	Age	Salary	Children	PrevSpent	Emails	AmountSpent
mean	523.75	45.25	60050	0.55	682.45	14.7	1445.95
Standard deviation	281.7741	15.539	34271.05	0.686333	947.2159	5.667079	1169.089186
CV	53.79934	34.34033	57.07085	124.7878	138.7964	38.55156	80.85267025
median	516	43.5	54550	0	157.5	12	1599.5
third quartile	709.25	55	83325	1	1104.5	18	1875.5
first quartile	364.75	22.75	59050	1	1104.5	6	1457.75
min	7	25	12000	0	0	6	260
max	1000	73	123600	2	2963	24	5209
range	993	48	111600	2	2963	18	4949

The regression equation is

AmountSpent = - 753 - 0.089 Customer + 10.1 Age + 0.0172 Salary - 437 Children
+ 0.279 PrevSpent + 54.6 Emails

Here from summary statistics

mean > median

Also from below graph i.e from given histogram we can conclude that the given data is positively skewed.

Also the normal probability plot tell us residuals follows normal distribution.