Question

In: Statistics and Probability

Date Company 1 Company 2 Company 3 Jan-19 107 111 102 Feb-19 105 124 112 Mar-19...

Date Company 1 Company 2 Company 3 Jan-19 107 111 102 Feb-19 105 124 112 Mar-19 113 102 121 Apr-19 121 102 116 May-19 120 120 125 Jun-19 123 107 125 Jul-19 120 122 120 Aug-19 112 100 109 Sep-19 106 108 121 Oct-19 100 113 122 Nov-19 124 122 100 Dec-19 101 110 102 Jan-20 95 75 74 Feb-20 90 71 70 Mar-20 101 51 73 Apr-20 101 64 73 1)

1.Perform a regression analysis and determine what is the expected ouput for Company 1 in May-20 2)?

Perform an ANOVA to determine if the means are significantly different from each other or not. Set up Hypothesis and state whether you accept or reject it.

Solutions

Expert Solution

Solution :

a)

the expected output for company 1 in may-20 = 98.4

Explanation:

The regression equation is defined as,

where the dependent variable Y = expected output for company 1 and the independent variable, X = Date

Now, the regression analysis is done in excel by following steps

Step 1:

Write the data values in excel. The screenshot is shown below,

Step 2:

DATA > Data Analysis > Regression > OK. The screenshot is shown below,

Step 3:

Select Input Y Range: 'Company 1' column, Input X Range: 'Date' column then OK. The screenshot is shown below,

The result is obtained. The screenshot is shown below,

The regression equation is,

Prediction

For may-20, X = 17

b)

The one way ANOVA analysis is performed in excel by following these steps,

Step 1:

Write the data values in excel. The screenshot is shown below

Step 2:

DATA > Data Analysis > ANOVA: Single Factor > OK.  The screenshot is shown below,

Step 3:

Select Input Range: All the data values column. The screenshot is shown below,

The result is obtained.  The screenshot is shown below,

Hypothesis

Null Hypothesis: The mean expected outputs are equal,  

Alternative Hypothesis: At least one mean expected output differs significantly.

Let the significance level = 0.05

Test statistic and P-value

From the ANOVA table,

F-statistic = 0.8487

P-value = 0.4347

Decision

Since the p-value is greater than 0.05 at a 5% significance level, the null hypothesis is failed to reject.

Conclusion

There is not sufficient evidence to conclude that the mean expected output differs significantly.


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