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# ANOVA Excel Worksheet

The following table contains a random sample of 40 women partitioned into three groups:
Group 1: ages below 20
Group 2: ages 20 through 40
Group 3: ages over 40
the values in the table are systolic blood pressure levels
The Hypothesis test:
H0: μ1=μ2=μ3
H1: at least one of the treatment means is different
use the Excel Analysis Toolpak to create an Anova- Single factor table.
is there sufficient evidence to support the claim that women in different age categories have different mean blood pressure levels? Give for your decision.

Group 1  Group 2  Group 3         [Place your Anova table here]
104        97           123
106        116         107            Anova: Single Factor
104        98           127
92          95           133            SUMMARY
112        108         114           Groups      Count      Sum      Average      Variance
107        106         93             Group 1
102         113            Group 2
108         114            Group 3
94           116
100         155
101         105          ANOVA
119                       Source of Variation      SS      df      MS      F      P-value      F crit
89                         Between Groups
113                       Within Groups
99
93                         Total
107
125
104
110
124
181
118

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Ryan S. | Mathematics and StatisticsMathematics and Statistics
4.8 4.8 (10 lesson ratings) (10)
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Here is the output from the Excel Add-in with the alpha set to .05

SUMMARY
Groups Count Sum Average Variance
Group 1 6 625 104.1666667 44.16666667
Group 2 23 2507 109 344.9090909
Group 3 11 1300 118.1818182 267.5636364

ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 937.930303 2 468.9651515 1.654991727 0.204926644 3.251923846
Within Groups 10484.4697 37 283.3640459

Total 11422.4 39

Notice the F value, P-value, and F critical

F is the observed value of the F statistic.

F critical is the value of the F statistic from the F distribution with numerator df=2 and denominator df=39 where Pr(F>=F critical) = .05 = alpha.

P-value is Pr(F>= F observed ).

So the probability of observing our F statistic or greater (20.5%) is much higher than than our significance level (5%). Equivalently our F observed (1.65) is much less than F critical (3.25).

Essentially the data shows that there is too high a probability that the difference between the groups is due to variability rather than treatment differences to reject the null hypothesis.

The conclusion is not to reject H0 that the group means are equal.