Statistical Analysis Results

ANOVA

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:

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.

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.  

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.

Table 5: ANOVA Table

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.

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.  

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

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.

Table 10: Moving Average Table

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.

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.

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

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