# QNT/351

Question 1Last 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 areasH1: There is a significant difference in the sales per run between the delivery areasNotice 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