Part a

Table 1: ANOVA table

The p-value for the F-value at 3,20 degrees of freedom is 0.0095. At 0.01 level of significance since p-value is less hence we reject the NULL hypothesis.

Part b

There are 4 groups in the above problem.

Number of Groups = degrees of freedom (between treatment) + 1 = 3+1 = 4

Part c

There are 21 observations in the above problem.

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

Answer 2

Table 2: Coefficients of Regression Equation

From table 2 the trend of the number of units sold by the auto manufacturer can be written as:

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

Thus, in the 11^th year the number of units sold = 136+39.18*11 = 566.98 ≈ 567 (000s)

Trend of the number of units sold by major auto manufacturer

The trend shows that the number of units that can be sold by the auto manufacturer = 567000

Answer 3

Part a

At 0.01 level of significance there is statistically significant relationship between price and the number of flash drives sold, p-value =0.0028.

Part b

At 0.01 level of significance there is statistically significant relationship between price and the number of flash drives sold, p-value =0.0028.

Answer 4

Answer 5

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

At 0.05 level of significance we reject the Null Hypothesis, p-value (0.000). Thus, the sales of the three stores are not similar, there are significant differences in the average sales of the three stores.

Answer 6

Part a

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

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

Part b

Part c

At 0.05 level of significance we reject the Null Hypothesis, p-value (0.000). Thus, the sales of the three boxes are not similar.  

Answer 7

The total average = 37*10+38*10+33*10 = 370+380+330 = 1080

The total average mileage =

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

SS_Error = 10*3+10*4+10*2 = 90

F-crit from f-table with 2, 29 at 0.05 level of significance = 3.33

Since, F-value > F-crit, hence at 0.05 level of significance there is sufficient evidence to reject Null Hypothesis. Thus, there is statistically significant differences in the mean mileage of the tyres.

Answer 8

Part a

Part b

The mean square error of the forecast is 0.86

Part c

The mean absolute deviation of the forecast is 0.00

Answer 9

Part a

The coefficient of determination =

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

Part b

The corresponding p-value for F is 0.129. At 0.05 level of significance we do not have sufficient evidence to reject Null Hypothesis. Hence, the model is statistically not significant.

Significance value can be found in MS-EXCEL by =F.DIST.RT(2.055,4,18)

Part c

The total sample size = 23 (Total +1)

Answer 10

Part a

The price of the stock can be predicted from the equation

• y = 118.5059 – 0.0163*x₁– 1.5726*x₂

Part b

From the 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
2. For every million increase in exchange of the New York Stock Exchange the price of Rawlon Inc. Stock would decrease by 1.5726.

Part c

At 95% confidence level the volume of exchange of New York stock exchange is statistically significant (p-value = 0.0018).

At 95% confidence level the number of shares sold by Rawlon Inc. is statistically not significant (p-value = 0.6176).

Part d

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

• y = 118.5059 – 0.0163*x₁– 1.5726*x₂
• y =118.5059 – 0.0163*945 – 1.5726*16
• y = 77.95