QNT/351

Question 1

Last week we discussed the one sample hypothesis test. This week we will discuss ANOVA. The ANOVA allows us to test the mean between  two or more groups.

 Teachers Example, I have five pizza delivery areas and I want to know if the average sales per run is different in one area than any of the other areas.

H0: There is no significant difference in the sales per run between the delivery areas

H1: There is a significant difference in the sales per run between the delivery areas

Notice how the hypothesis is very simple. There is a difference or there is not a difference. Again, we will use MegaStat. You cannot calculate ANOVA by hand.

Just put your own sales for each area in a column and click the Megastat menu, select analysis of variance, and highlight your data and click enter. Check your p-value. Interpret.

Here is the teachers example :

      Area 1    Area 2     Area 3   Area 4    Area 5

 $  45.00  $  13.00  $  12.00  $  34.00  $  37.00

 $  34.00  $  24.00  $  37.00  $  25.00  $  22.00

 $  36.00  $  34.00  $  27.00  $  23.00  $  32.00

 $  45.00  $  23.00  $  17.00  $  36.00  $  42.00

 $  53.00  $  32.00  $  28.00  $  23.00  $  35.00

 $   43.00  $  21.00  $  39.00  $  35.00  $  37.00

 $  45.00  $  23.00  $  35.00  $  45.00  $  28.00

 $  65.00  $  24.00  $  26.00  $  34.00  $  13.00

 $  32.00  $  37.00  $  45.00  $  23.00  $  24.00

 $  30.00  $  45.00  $  27.00  $  53.00  $  15.00

 $  40.00  $  32.00  $  34.00  $  29.00  $  63.00

 $  23.00  $  34.00  $  16.00  $  28.00  $  35.00

             

One factor ANOVA           

             

  Mean  n  Std. Dev       

32.45  40.917  12  11.1719  Area 1     

32.45  28.500  12  8.6707  Area 2     

32.45  28.583  12  9.9951  Area 3     

32.45  32.333  12  9.3355  Area 4     

32.45  31.917  12  13.3448  Area 5     

  32.450  60  11.2347  Total     

             

ANOVA table             

Source  SS    df  MS  F    p-value 

Treatment  1,230.4333  4  307.60833  2.72  .0386 

Error  6,216.4167  55  113.02576       

Total  7,446.8500  59         

             

The p-value is less than .05 so I do not reject the null

 

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