Many studies have suggested that there is a link between exercise and healthy bo

Question

Many studies have suggested that there is a link between exercise and healthy bones. Exercise stresses the bones and this causes them to get stronger. One study examined the effect of jumping on the bone density of growing rats. There were three treatments: a control with no jumping, a low-jump condition (the jump height was 30 centimeters), and a high-jump condition (60 centimeters). After 8 weeks of 10 jumps per day, 5 days per week, the bone density of the rats (expressed in mg/cm) was measured. Here are the data. data242.dat (a) Make a table giving the sample size, mean, and standard deviation for each group of rats. Consider whether or not it is reasonable to pool the variances. (Round your answers for x bar, s, and to one decimal place.) (b) Run the analysis of variance. Report the F statistic with its degrees of freedom and P-value. What do you conclude? (Round your test statistic to two decimal places and your P-value to three decimal places.) F = P = Conclusion: There is statistically significant difference between the three treatment means at the alpha = 05 level. A survey looked at the value of recreational sports on college campuses. One of the questions asked each student to rate the importance of recreational sports to college satisfaction and success. Responses were on a 10-point scale with 1 indicating total lack of importance and 10 indicating very high importance. The following table summarizes these results: (a) To compare the mean scores across classes, what are the degrees of

Explanation / Answer

Solution

Part (a)

Group

n

xbar

s

Control

10

614.5

28.3

Low Jump

10

633.5

12.5

High Jump

10

602.7

16.7

DONE

Part (b)

Back-up Theory

Suppose we have data of a 1-way classification ANOVA, with r rows and n observations per cell.

Let xij represent the jth observation (density) in the ith row (treatment), k = 1,2,…,10; i = 1,2,3

Then the ANOVA model is: xij = µ + i + ij, where µ = common effect, i = effect of ith row, and ij is the error component which is assumed to be Normally Distributed with mean 0 and variance 2.

Terminology:

Row total = xi.= sum over j of xij

Grand total = G = sum over i of xi.

Correction Factor = C = G2/N, where N = total number of observations = r x n =

Total Sum of Squares: SST = (sum over i,j of xij2) – C

Row Sum of Squares: SSR = {(sum over i of xi.2)/(n)} – C

Error Sum of Squares: SSE = SST – SSR

Mean Sum of Squares = Sum of squares/Degrees of Freedom

Degrees of Freedom:

Total: N (i.e., rn) – 1;

Error: DF for Total – DF for Row;

Rows: (r - 1);

Fobs: MSSR/MSSE;

Fcrit: upper % point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for Row and n2 is the DF for Error and p-value = P(Fn1,n2 > Fcal).

Significance: Fobs is significant if Fobs > Fcrit or if p-value < level of significance.

Excel calculations summary is given below:

Grand Total G

66163

Correction Factor C

7E+07

SST

2E+06

Raw SS (Row)

7E+08

SSR

738692

SSE

868306

ANOVA TABLE

0.05

Source

DF

SS

MS

F

Fcrit

p-value

Row

2

738692

4E+05

11.485

3.354

0.00025

Error

27

868306

32159

Total

29

2E+06

55414

So, F-statistic = 11.49; DF = (2, 27); p-value = 0.00025 ANSWER 1

Conclusion

Since p-value < 0.05 (given level of significance), null hypothesis is rejected.

=> There is evidence of statistically significant difference between the three treatment means at 5% level of significance. ANSWER 2

Group

n

xbar

s

Control

10

614.5

28.3

Low Jump

10

633.5

12.5

High Jump

10

602.7

16.7

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