MAT119 Quantitative Business Analysis

  • Subject Code :  

    MAT119

  • Country :  

    US

  • University :  

    Thomas Edison State University

Answer:-

Problem Statement

Application of one way Anova in MINITAB to solve a given task in managing Ashland MultiComm services 

Statistical Results 

Analysis of Variance 

Source

DF

Adj SS

Adj MS

F-Value

P-Value

Factor

2

173.3

86.67

5.14

0.015

Error

21

354.4

16.88

 

 

Total

23

527.7

 

 

 

Means

Factor

N

Mean

StDev

95% Upper
Bound

system 1

8

41.46

4.32

43.96

System 2

8

36.00

3.82

38.50

System 3

8

35.55

4.17

38.05

Pooled StDev = 4.10806

We can observe that the mean of system 1 is the largest while the mean of system 3 is the smallest.

The standard deviation of system 1 is also the largest and system 2 has the smallest Standard Dev.

The factor p-value is statistically significant 0.015 less than 0.05 hence we reject the null hypothesis.

Conclusion

We can see that means of different systems have significant differences, system 1 has the largest mean while system 3 has the lowest mean. On the other hand, system 1 has the largest standard deviation while system 2 has the smallest standard deviation. The p- value of is statistically significant since it is less than 0.05 significance level.

Problem Statement

In this task you are required to use one way Anova test in MINITAB worksheet to come up with the solution for managing Ashland MultiComm services

Statistical Results

Null hypothesis

All means are equal

Alternative hypothesis

Not all means are equal

Significance level

α = 0.05

 

Source

DF

Adj SS

Adj MS

F-Value

P-Value

Factor

2

173.3

86.67

5.14

0.015

Error

21

354.4

16.88

 

 

Total

23

527.7

 

 

 

 
1. From the tables above we found out that X2STAT=17.80 is bigger than 5.9915.
2. the p-value = 0.000 which is less than the significance 0.05. we can therefore reject the null     hypothesis and come to conclusion that there exist four discounts that are significantly different utilizing subscriptions after the trial session.

3. A Chi-square test is also done to provide accurate outcome especially when handling 2xc a contingency table whereby all expected frequencies have to be large. If

the combining categories is undesirable; it is advised to use the available alternative procedures.

Conclusion

Based on the analysis provided above we can see that the p-value is 0.000 which is less than the level of significance 0.05 leading to rejection of null hypothesis. Also a Chi-square test is done in order to come up with accurate results.

Problem Statement

The task here is to use MINITAB to recreate the regression equation for the managing Ashland MultiComm Services.

Statistical Results

Regression Equation

New subscriptions

=

-482 + 4.483 Hours

Model Summary

S

R-sq

R-sq(adj)

R-sq(pred)

412.438

86.36%

85.71%

84.47%

1. From the regression equation y= -482 +4.483(hours) we can calculate the number of new subscribers for a month when the value of hours is provided. For instance, given the 1000 hours we can compute it as follows y= -482 + 4.483*1000= (4001). Therefore, the more the number hours the more the subscribers.

2. the above prediction are based on the following assumptions: normality, linearity and independence of errors.
3. the p. value of independent variable (hours) is statistically significant 0.00< 0.05 while p-value of the dependent variable is not statistically significant 0.283 >0.05

Conclusion

Using regression equation, we can compute the number of new subscriptions when the value of hours is provided. We can also note that when the number of hours is bigger, the number of subscribers tend to be bigger too. However, the prediction of new subscriptions with respect to number of hours is guided by some assumptions such as linearity and independence of errors.

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