# QNT/351

Jul 3rd, 2013
KateS
Category:
Price: \$10 USD

Question description

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

(Top Tutor) Daniel C.
(997)
School: Rice University

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## Review from our student for this Answer

Sigchi4life
Jul 3rd, 2013
"Excellent!"

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