,/
/
Brittany Poyer
3/23/07
ANOV A assignment
#9
1.
The null hypothesis
is exercise duration has no effect on strength test results
among the 30 subjects.
2.
The research hypothesis
for the ANOV A is the longer the subjects exercises, the
stronger they will be.
3.
The mean of all 30 scores was 84.23.
The standard deviation of all 30 scores was
10.45.
4.
The SST was 3169.37.
5.
The explained
sum of squaresbetweenas
a result of the treatment,
was 1414.47.
6.
The error sum of squares was 1754.90.
7.
The degree of freedom for exercise duration is equal to 2 because there are 3
different groups in the class, to get df you need to subtract one. (31=2)
8.
The degrees of freedom for residual is calculated
by subtracting
one from 30 to
get the degrees of freedom, then subtract the explained degree of freedom from
that, resulting in the value for residual degree of freedom.
N  ~
9.
If the null is true, then both mean square values should be equal to 64.996
10. The F value of 10.881 was obtained by dividing the mean square between from
the mean square within.
(707.233/64.996=10.881)
11. You should reject the null hypothesis
because if the null was true, the f ratio
should be one.
In this case, the f ratio was 10.881, which means we should reject
it.
12. The longer a person spends exercising,
the stronger the person will become
13. the mean for 30 min: 75.60,45
min: 84.70 and 60 min: 92.40.
14. the null hypothesis
is that there is no difference
of the mean test scores between
each of the classes.
15. the mean for all 60 scores was 26.25. The standard deviation
for all 60 scores was
12.18867.
16. The SS for between:
8375.25, the SS for within: 390.0 and SS total: 8765.25.
17. the degrees of freedom are 3 and 56 because there are 4 groups in the class (4
1=3) there is a total of 60 (df=nk) which is the total subjects and n is 4 which is
the number of groups (604=56)
18. the value of the F ratio is 400.867.
19. the significance
value for this f ratio was .000 which means it is extremely
significant
20. the MS for error (within) is so small (6.964) because there is little variation when
looking at the difference
within a class.
21. I would reject the null because when looking at results
between
the classes, there
would be large variation among the classes.
When looking at the pattern to
measure Integrative
Physiology
knowledge,
this pattern does make sense.
I
would expect to see a greater variation between the classes (comparing
f vs h vs j
vs s.
.. ) because a senior would be expected to know more than a freshman per
say, ultimately
scoring higher percentages.
When comparing
within
the classes
however,
I would expect little variation because you are comparing,
for example,
sophomores
versus sophomores
knowledgethey
have all been equally prepared,
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 Spring '08
 DECKER,MIK
 Statistics, Standard Deviation, Mean, Null hypothesis, Statistical hypothesis testing, Brittany Poyer

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