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ANOVA

I have the following information that I need to draw up a conclusion from:

        Machine 1       Machine 2       Machine 3
        0.546   0.573   0.573
        0.526   0.592   0.57
        0.587   0.571   0.527
        0.563   0.556   0.572
SUM     2.222   2.292   2.242

What is the SUM of SQUARES (SS) for FACTOR?                                             0.00065
What is the SUM of SQUARES (SS) for ERROR?                                              0.00383
What is the SUM of SQUARES (SS) TOTAL?                                          0.00448
What is the Degrees of Freedom (Df) for FACTOR?                                         2
What is the Degrees of Freedom (Df) for the ERROR term?                                         9
What is the Degrees of Freedom (Df) TOTAL?                                              11
What is the MEAN SQUARED (MS) for FACTOR?                                               0.00033
What is the MEAN SQUARED (MS) for the ERROR term?                                               0.00043
What is the F Calculated value?                                         F-Calculated
What is F Critical value (from the table)?                                              4.2600
Will you reject the null? Yes or No

Base on my calculations above, should I reject the Null? If so for what reason?"
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Jamie33
Asked:
Jamie33
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1 Solution
 
TommySzalapskiCommented:
If your F Calculated (which I don't see a number for) is greater than your F Critical, then you should reject the NULL hypothesis. That is also your reason. F Calc > F Crit.
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Jamie33Author Commented:
The F Calc is 0.7629.  So How do I write the justification on why I am rejecting the Null in non-statistical terms, where anyone could understand it?
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TommySzalapskiCommented:
Well, that would depend a lot on what you were measuring and what the null hypothesis was.

You could say something like "there is a statistically significant difference in the two machines." I can't do any better than that without knowing what the machines are measuring.

But, if F Calc is .76 and F Crit is 4.2, then you can't reject the NULL hypothesis because there is no statistically significant difference.
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