How do I andser this. I’m stuck
1 answer:
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Answer:
is an attachment
test statistic = 9.87
p value is less than 0.01
Not all means of the three assembly methods are equal
Step-by-step explanation:
the anova table is an attachment
total number of methods = k = 3
number of observations n = 30
for treatment, df = 3-1 = 2
for error, df = 30 -3 (n-k) = 27
sst = 10800, sstr = 4560, sse = 10800-4560 = 6240
we find the mean of squares for error
sse/df of error = 6240/27 = 231.11
mean of squares for treatment = 4560/2 = 2280
test stat
F = 2280/231.11
= 9.8654
using f distribution table;
alpha = 0.05
df = 2 , 27
we get 3.35
h0; no difference in mean
h1; there is difference
9.87 > 3.35 so we reject H0, there is difference in means. all are not equal.
pvalue calculation
using excel,
FDIST(9.8654, 2, 27) = 0.0006078
p value is less than 0.01
we conclude that a. Not all means of the three assembly methods are equal
