Statistical Analysis Results

ANOVA

Source of Variation Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	Sum of squares	Degrees of Freedom	Mean Square	F Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper
Between treatments	90	3
Within treatments (Error)	120	20
Total	= (90 + 120) = 210	= (3+20) =23

Table 1: ANOVA table

The F-statistic is calculated as 5. The p-value for the F-statistic with 3 and 20 degrees of freedom is found to be 0.00951034. At 1% level of significance (α=0.01), as calculated p-value is less than 0.01, we decide to reject the null hypothesis.

Total number of groups = (degrees of freedom of between treatments) + 1 = (3+1) = 4

Hence, there are 4 groups in the question.

Total number of observations = (degrees of freedom of within treatment) + 1= (20+1) = 21

There are 21 observations in the question.

Linear regression model:

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	136	13.76	9.881	0.000	104.2621	167.7379
Year (t)	39.18	2.22	17.664	0.000	34.0668	44.29684

Table 2: Linear Regression Model

The simple linear regression, Y = a + b*X.

Here, the intercept (a) = 136, the slope (b) = 39.18.

According to the table 2, the estimated number of units sold by the auto manufacturer is:

Number of Units sold (‘000s) = 136 + 39.18*Year

It is the linear model of this analysis.

Thus, in the 11^th (t=11) year, the number of units sold = (136+39.18*11) = 567 (‘000s). 

Figure 1: Trend of the number of units sold by major auto manufacturer

The estimation of trend indicates that the number of units that can be sold by the auto manufacturer = 567000.

ANOVA
	df	SS	MS	F	Significance F
Regression	1	59.89145	59.89145	29.62415	0.002842
Residual	5	10.10855	2.021711
Total	6	70

Table 3: Linear Regression Model calculates F-statistic

The p-value of the F-statistic is calculated as 0.002842. Therefore, at 0.01 level of significance, there is statistically significant relationship between price and the number of flash drives sold.  

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	40.033	1.070	37.4309	0.0000	37.28362	42.78217
Units sold (y)	-1.174	0.216	-5.4428	0.0028	-1.72897	-0.61971

Table 4: Linear Regression Model calculates t-statistic

The p-value is calculated as 0.002842. Therefore, at 0.01 level of significance, there is statistically significant relationship between price and the number of flash drives sold. 

In a completely randomized experimental design, 14 experimental units were utilised for each of the five levels of factor (5 treatments). We executed ANOVA table.

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Between treatments	= (4*800) = 3200	= (5 – 1) = 4	800.00
Within treatments (Error)	= (10600 – 3200) = 7400	= (14 – 1) = 13
Total	10600	17

Table 5: ANOVA Table

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	324	2	162	40.500	0.000	4.256
Within Groups	36	9	4
Total	360	11

Table 6: ANOVA Table

The hypotheses are:

Null Hypothesis (H₀): The average sales of the three stores are equal.

Alternate Hypothesis (H_A): There exists at least one inequality of average sales of the stores.

At 5% level of statistical significance, we can reject the Null Hypothesis for its p-value (0.000<0.05). Hence, the mean values of sales of the three stores are not equal. There exist significant differences in the mean sales of the three stores.

The hypotheses are-

Null Hypothesis (H₀): The mean values of  sales of the three boxes are equal.

Alternate Hypothesis (H_A): There exists at least one equality in mean sales of the boxes.

ANOVA
Source of Variation	SS	df	MS	F	P-value	F-crit
Between Groups	24467.2	2	12233.6000	53.7111	0.000	3.8853
Within Groups	2733.2	12	227.7667
Total	27200.4	14

Linear Regression

Table 7: ANOVA Table

The p-value is calculated as 0.0. Therefore, at 0.05 level of statistical significance, we can reject the Null Hypothesis of equality of mean values of sales. Therefore, the sales of the three boxes are unequal.  

	Brand A	Brand B	Brand C
Average Mileage	37	38	33
Sample Variance	3	4	2
Count	10	10	10

Table 8: Descriptive statistics table

The total average = [(37*10) + (38*10) + (33*10)] = [370 + 380 + 330] = 1080

The total average mileage =

Thus, SS_Between Groups (SSB) = [10*(37 – 36)²+10*(38 – 36)²+10*(33 – 36)²] = 140

SS_Error (SSE) = [(10*3) + (10*4) + (10*2)] = 90

ANOVA
Source of Variation	SS	df	MS	F
Between Groups	140	2
Within Groups	90	=(29 – 2) = 27
Total	210	29

Table 9: ANOVA Table

F-critical calculated from F-table with (2, 27) degrees of freedom at 5% level of significance is 0.00000313.

As, F-value (21.02)> F-critical (0.00000313), at 5% level of significance, there exists sufficient evidence to reject Null Hypothesis. Therefore, there is a statistically significant variability in the average mileage of the 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

Table 10: Moving Average Table

Days (x)	Forecast (Y)	Y’	(Y-Y’)	(Y-Y’)2
1	19	19.2	-0.2	0.04
2	19	19.7	-0.7	0.49
3	21	20.2	0.8	0.64
4	22	20.7	1.3	1.69
5	20	21.2	-1.2	1.44
			0.00	0.86

Table 11: Forecasting Table

The mean square error of the forecast is calculated as 0.86.

The mean absolute deviation of the forecast is calculated as 0.00.

Source of Variation	Degrees of Freedom	Sum of Squares (SS)	Mean Square (MSS)	F-statistic
Regression	4	283940.60	70985.15	2.055
Error	18	621735.14	34540.84
Total	22	905675.74 (TSS)

Table 12: ANOVA Table

The coefficient of determination = (

From the linear regression model, it can be inferred that 7.84% of the variability in sales of “Very Fresh Juice Company” could be explained by the independent variables – “price per unit”, “competitor’s price”, “advertising” and “type of container.”

Significance value of F-statistic can be found in MS-Excel by the excel function FDIST(2.055,4,18). According to the p-value (α=0.05), F-statistic with degrees of freedom 4 and 18 is 0.12942. As, 0.12942>0.05, at 0.05 level of significance, we do not have sufficient evidence to reject Null Hypothesis. Therefore, the model statistically is insignificant.

The total sample size = (Total degrees of freedom +1) = (22+1) = 23.

ANOVA
	df	SS	MS	F	Significance F
Regression	2	118.8474369	59.4237	40.9216	0.000
Residual	9	13.0692	1.4521
Total	11	131.9166667
	Coefficients	Standard Error	t Stat	P-value
Intercept	118.5059	33.5753	3.5296	0.0064
Number of shares sold (in ’00s) (x1)	-0.0163	0.0315	-0.5171	0.6176
New York Stock Exchange (x2)	-1.5726	0.3590	-4.3807	0.0018

Table 13: ANOVA Table of Regression Model

The price of the stock can be predicted from the linear regression provided trend equation-

y = 118.5059 – 0.0163*x₁ – 1.5726*x₂

From the linear regression equation, it can be said that:

1. For every 100 stocks of company sold the price of Rawlon Inc. stock would decrease by 0.0163 and vice versa.
2. For every million enhance in exchange of the New York Stock Exchange the price of Rawlon Inc. Stock would reduce by 1.5726 and vice versa.
3. For, zero amount of sold price of both Rawlon Inc. and New York Stock Exchange, the Inc. Stock is found to be 118.5059.

At 0.05 level of significance, the volume of exchange of New York stock exchange is statistically significant (as p-value = 0.0018< 0.05).

The p-value is calculated as 0.6176 (>0.05). Therefore, at 0.05 level of significance, the number of shares sold by Rawlon Inc. is not statistically significant.

For 94500 stocks sold and 16 million the volume of exchange on the New York Stock Exchange, the price of the stock (Inc. stock) would be:  

y = [118.5059 – (0.0163*x₁) – (1.5726*x₂)]

   = [118.5059 – (0.0163*945) – (1.5726*16)]

   = 77.95