Statistics Practice Problems

Part 1: ANOVA and Hypothesis Testing

Part a

Table 1: ANOVA table

From excel we find that p-value for F= 5 at 3,20 df is 0.0095. Since, p-value < a-value hence reject Null Hypothesis.

df for number of groups:

• m-1 = 3
• m = 3 + 1 = 4

Hence, number of groups = 4

Total number of observations

• n – 1 = 20
• n = 21

Hence, number of observations = 21

Table 2: Coefficients of Regression Equation

The linear trend equation

• Number of Units sold (000s) = 136.00 + 39.1818*Year (t)

The number of cars sold for t = 11

• Number of Units sold (000s) = 136.00 + 39.1818*Year (t)
• Number of Units sold (000s) = 136.00 + 39.1818*11
• Number of Units sold (000s) = 136.00 + 430.9998
• Number of Units sold (000s) = 566.9998 ≈ 567 

From the sales trend it is seen that the sales in the 11^th year would be 567 (000s)

At 0.01 level of significance the value of F = 29.6241 for df 1,5. From F-table it is seen that the F-crit value for 1,5 df at a = 0.01 is 16.26. Since F-value is more than F-crit (29.6241 > 16.26) hence we reject the Null Hypothesis.

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

At 0.01 level of significance the value of F = 29.6241 for df 1,5. The corresponding p-value is 0.0028. Since p-value is less than the level of significance hence we reject the Null Hypothesis.

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

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

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.

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.

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

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

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

At 0.05 level of significance, df (2, 12) F-crit is 3.8853. Thus if F-value is more than F-crit then we reject Null Hypothesis else we accept Null Hypothesis.

At 0.05 level of significance, F-value = 53.7111. Since F-value is more than F-crit (53.7111 > 3.8853) hence we reject Null Hypothesis. Hence, we can conclude that there are differences in the average sales of the three boxes.

Part 2: Linear Trend Expression and Sales Projection

Total mileage of all the three tyres = average mileage * number of tyres

Hence, total mileage of all the three tyres = 37*10 + 38*10 + 33*10

= 370 + 380 +330

= 1080

The total number of tyres tested = 30

Hence average mileage of all the 30 tyres = 1080/30 = 36

Thus SS (between) = Number of Tyres * (Average mileage of the brand – Average mileage of all 30 tyres)²

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

Similarly, SS (error) = (number of tyres-1)*variance of the brand

SS (error) = 9*3 + 9*4 + 9*2 = 81

df (Total) = total number of tyres tested – 1 = 30 – 1 = 29

df (Between) = types of tyres tested – 1 = 3 – 1 = 2

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

F-value for df 2, 27 is 23.333. F-crit for df 2, 27 at 0.05 level of significance is 3.3541.

Since F-value is more than F-crit (23.333 > 3.3541) hence Null Hypothesis is rejected.

Thus, we find that there is a statistically significant differences in the mileage of the tyres.

The mean square error (MSE) of the amount of tip at the car park is 0.86

The mean absolute deviation (MAD) of the amount of tip at the car park is 0.00

The coefficient of determination is the ratio of Mean Square (Regression) to Sum of Squares (Total)

The coefficient of determination R² = 70985.15/905675.74 = 0.0784

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

At df 4, 18 for 0.05 level of significance F-crit = 2.9277.

Since, F-value < F-crit (2.06 < 2.9277) hence we do not reject Null Hypothesis.

 Thus, it can be inferred that the model is not statistically significant.

For regression the sample size is given as: Total (ANOVA table) + 1 = 22 + 1 = 23

From the regression output the price of the stock can be predicted as:

y = 118.5059 – 0.0163*x1 – 1.5726*x2  

The coefficients of the regression suggest that:  

1. Keeping the volume of exchange (in millions) on the New York Stock Exchange constant for every 100 stocks of Rawlston Inc. sold the prices of Rawlston Inc. decreases by 0.0163
2. Keeping the number of stocks sold of Rawlston Inc. constant for every one-million exchange in the New York Stock Exchange the prices of Rawlston Inc. decreases by 1.5726

At 95% confidence level

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

Hence, stock prices of Rawlston Inc = 77.9408

Order an Essay Now & Get These Features For Free:

Turnitin Report

Formatting

Title Page

Citation

Outline

Place an Order