Compute The Coefficient Of Determination And Fully Interpret Its Meaning

ANOVA table

Analysis of variation (ANOVA) is a test for differences between three of more groups. The F-value found through ANOVA is used to test for differences.

Source	df	SS	MS Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	F
Treatments	k-1	SST
Error	n-k	SSE
Total	n-1	Total SS

SST is used to measure the variation between means of k samples.

SSE is used to measure the pooled variation between means of k samples.

SSE =  where  denotes the number of sample in the group and  is the standard deviation of the samples in the group.

F value is compared with F-crit values at the desired level of significance. If F-value > F-crit value, then we reject Null Hypothesis else accept it (Mendenhall, Beaver and Beaver 2012).

Table 1: ANOVA table

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Between treatments	90	3
Within Treatments (Error)	120	20
Total

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value	F-crit
Between Treatments	90	3	30	5	0.0095	4.9382
Within Treatments (Error)	120	20	6
Total	210	23

MS-EXCEL is used to find p-value’s and F-crit values. Since F-value > F-crit value at 0.01 level of significance hence we reject Null Hypothesis.

df between groups = 3

• m – 1 = 3
• m = 4

Thus, number of groups = 4

Total number of observations = df (Treatment) + df (Error) + 1= 3 + 20 + 1 = 24

Total number of observations = 24

Table 2: Coefficients of Regression Equation

	Coefficients	Standard Error	t Stat	P-value
Intercept	136.0000	13.7631	9.8815	0.0000
Year (t)	39.1818	2.2181	17.6643	0.0000

 

• Number of Cars sold (in 1000s of Units) = 136.0000 + 39.1818*Year (t)

The number of Car units sold for t = 11 can be calculated as

• Number of Cars sold (in 1000s of Units) = 136.0000 + 39.1818*Year (t)
• Number of Cars sold (in 1000s of Units) = 136.0000 + 39.1818*11 = 567

ANOVA
	df	SS	MS	F	Significance F
Regression	1	59.89	59.89	29.6241	0.0028
Residual	5	10.11	2.02
Total	6	70

At a = 0.01, the p-value for the test is 0.0028. Since p-value < a-value (0.0028 < 0.01) Null hypothesis is rejected (Courtinhas and Black 2012).

Thus, there is a statistically significant relationship between price and number of flash drives.

	Coefficients	Standard Error	t Stat	P-value
Intercept	40.0329	1.0695	37.4309	0.0000
Units sold (y)	-1.1743	0.2158	-5.4428	0.0028

The coefficient for the number of units sold = -1.1743. The standard Error of the coefficient = 0.2158.

Thus the t-stat

At a = 0.01, the p-value for the t-statistics is 0.0028. Since p-value < a-value (0.0028 < 0.01)

Null hypothesis is rejected.

Thus, there is a statistically significant relationship between price and number of flash drives.

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Between treatments		4	800
Within Treatments (Error)		65
Total	10600	69

df (Total) = 5*14 – 1 = 70 – 1 = 69  

df (Treatment) = 5 – 1 = 4

df (Error) = 70 – 5 = 65

Since there are more than 2 groups hence ANOVA is used to check for differences in the three groups.

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	324	2	162	40.5000	0.0000	4.2565
Within Groups	36	9	4
Total	360	11

From ANOVA table it is found that p-value < 0.0000. Since, p-value < a-value  (0.0000 < 0.05) hence we reject Null Hypothesis.

Thus there are statistically significant differences in the average sales of the three stores.  

To test for differences in the sales of the three stores a hypothesis was developed.

Null Hypothesis: The average sales of the three stores are equal

Alternate Hypothesis: The average sales of at least one of the stores is different

Linear trend equation

For the testing of the hypothesis 5% level of significance is used.

At 0.05 level of significance, df (2,9) F-crit is 4.256. Thus if F-value is more than F-crit then we reject Null Hypothesis else we accept Null Hypothesis.

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	324	2	162	40.5	3.16E-05	4.256
Within Groups	36	9	4
Total	360	11

At 0.05 level of significance, F-value = 40.5. Since F-value is more than F-crit (40.5 > 4.256) hence we reject Null Hypothesis. Hence, we can conclude that there are differences in the average sales of the stores.

Null Hypothesis: There are no differences in the average sales of the three boxes

Alternate Hypothesis: There are differences in the average sales of the three boxes

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Box1	5	1000	200	400
Box2	5	948	189.6	133.3
Box3	5	1400	280	150
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	24467.2	2	12233.6000	53.7111	0.0000	3.8853
Within Groups	2733.2	12	227.7667
Total	27200.4	14

For the testing of the hypothesis a = 0.05.

At 0.05 level of significance, F-value = 53.7111.

p-value for the ANOVA = 0.0000.

Since, p-value < a-value (0.0000 < 0.05) hence we reject Null Hypothesis.

Thus, there are statistically significant differences in the average sales of the three boxes.

	Brand A	Brand B	Brand C
Average Mileage	37	38	33
Sample Variance	3	4	2
Number of tyres	10	10	10

Total number of tyres = 30

Hence df (total) = 30 – 1 = 29

df (Between) = 3 (Brands) – 1 = 2

df(Error) = df(total) – df(Between) = 29 – 2 = 27 

Total mileage of all 10 tyres of Brand A = 37*10 = 370

Total mileage of all 10 tyres of Brand B = 38*10 = 380

Total mileage of all 10 tyres of Brand C = 33*10 = 330

Thus total mileage of all three brands = 370 + 380 + 330 = 1080

Hence Average mileage of all 3 brands = 1080/30 = 36

SS(between) =

SS (between) = 10*(37 – 36)²+10*(38 – 36)²+10*(33 – 36)²  = 140

SS (error) =  

SS (error) = 9*3 + 9*4 + 9*2 = 81

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between	140	2	70	23.3333	0.0000	3.3541
Error	81	27	3.000
Total	221	29

At 0.05 level of significance, F-value = 23.333.

p-value for the ANOVA = 0.0000.

Since, p-value < a-value (0.0000 < 0.05) hence we reject Null Hypothesis.

Thus, there are statistically significant differences in the average mileage of the three brands of tyres.

Day	Tips	Simple Moving average
1	18
2	22		19
3	17		19
4	18		21
5	28		22
6	20		20
7	12

Day	Tips	Deviation	Deviation2	Absolute Deviation
1	18	-1.2857	1.6531	1.2857
2	22	2.7143	7.3673	2.7143
3	17	-2.2857	5.2245	2.2857
4	18	-1.2857	1.6531	1.2857
5	28	8.7143	75.9388	8.7143
6	20	0.7143	0.5102	0.7143
7	12	-7.2857	53.0816	7.2857
AVERAGE	19.2857		20.7755	3.4694

The mean square error (MSE) of the amount of tip at the car park is 20.7755

The mean absolute deviation (MAD) of the amount of tip at the car park is 3.4694

Source of Variation	Degrees of Freedom	Sum of Squares	Mean Square	F
Regression	4	283940.60
Error	18	621735.14
Total	22	905675.74

The value of coefficient of determination indicates that 7.84% of the variations in sales of Very Fresh Juice Company (dependent variable) can be predicted from the four variables of Price per unit, Competitor’s price, advertising and type of container (independent variables).

Source of Variation	Degrees of Freedom	Sum of Squares	Mean Square	F	Significance	F-crit
Regression	4	283940.60	70985.15	2.06	0.13	2.9277
Error	18	621735.14	34540.84
Total	22	905675.74

At a = 0.05 Significance = 0.13

Since, p-value > a-value (0.13 > 0.05), the Null Hypothesis is accepted.  

Hence, the model is statistically not significant.

Sample Size = df(Total) + 1 = 22 + 1 = 23

ANOVA
	df	SS	MS	F	Significance F
Regression	2	118.8474	59.4237	40.9216	0.0000
Residual	9	13.0692	1.4521
Total	11	131.9167
	Coefficients	Standard Error	t Stat	P-value
Intercept	118.5059	33.5753	3.5296	0.0064
x1	-0.0163	0.0315	-0.5171	0.6176
x2	-1.5726	0.3590	-4.3807	0.0018

The equation to predict the prices of the stock

• y= 118.5059 – 0.0163*x1 – 1.5726*x2  

From the coefficients of the variables it can be interpreted:

1. For every 100 stocks of Rawlston Inc. sold the price of stocks of Rawlston Inc. decreases by 0.0163, when the volume of exchange (in millions) for the New York Stock Exchange remains constant
2. For every one-million exchange in the New York Stock Exchange the price of stocks of Rawlston Inc. decreases by 1.5726, when there is no sale of the stocks of Rawlston Inc. constant

Part c

At 95% confidence level

1. The p-value for the coefficient of Rawlston Inc (x1) is 0.6176. Since p-value > confidence level (0.6176 > 0.05) hence, the coefficient of Rawlston Inc is statistically not significant.
2. The p-value for coefficient of New York Stock Exchange (x2) is 0.0018. Since p-value < confidence (0.0018 < 0.05) level hence, the coefficient is statistically significant.

The number of stocks of Rawlston Inc. sold = 94500 = 945 (100s)

The volume of exchange at the New York Stock Exchange = 16 million

• y = 118.5059 – 0.0163*x1 – 1.5726*x2  

  = 118.5059 – 0.0163*945 – 1.5726*16

  = 118.5059 – 15.4035 – 25.1616 = 77.9408

Thus, the stock prices of Rawlston Inc = 77.9408

Conclusion

The following assignment was on the learning on the basics of ANOVA, Regression. We learnt the how to find the degrees of freedom, sum of squares and the F-value. We learnt to evaluate the F-value on the basis p-values (significance levels). MS Excel was used to find the p-values. Thus, we also learnt the use of MS-EXCEL in doing ANOVA and Regression.

Regression analysis was used to find the relationship between variables, and forecasting.

References

Cortinhas, C. and Black, K., 2014. Statistics for business and economics. Wiley Global Education.

Mendenhall, W., Beaver, R.J. and Beaver, B.M., 2012. Introduction to probability and statistics. Cengage Learning.