Question

obs group g density 1 Control 1 605 2 Control 1 604 3 Control 1 640 4 Control 1 602 5 Control 1 580 6 Control 1 599 7 Control 1 597 8 Control 1 617 9 Control 1 566 10 Control 1 578 11 Lowjump 2 625 12 Lowjump 2 624 13 Lowjump 2 632 14 Lowjump 2 623 15 Lowjump 2 635 16 Lowjump 2 623 17 Lowjump 2 624 18 Lowjump 2 627 19 Lowjump 2 630 20 Lowjump 2 630 21 Highjump 3 649 22 Highjump 3 630 23 Highjump 3 632 24 Highjump 3 615 25 Highjump 3 633 26 Highjump 3 625 27 Highjump 3 615 28 Highjump 3 634 29 Highjump 3 598 30 Highjump 3 619

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/cm3 ) was measured. Here are the data. data379.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, s, and to one decimal place.) Group n s Control Low jump High jump (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 ? = .05 level.

Sorry here is the Date

Answer 1

Following is the output of ANOVA analysis:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
control 1	10	5988	598.8	441.0666667
Lowjump2	10	6273	627.3	17.78888889
Highjump3	10	6250	625	195.5555556
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	5013.266667	2	2506.63333	11.49109462	0.000245135	3.354130829
Within Groups	5889.7	27	218.137037
Total	10902.96667	29

(a)

Following table shows the mean, sd and se:

Groups	Count	Average	SD	SE
control 1	10	598.8	21	6.6
Lowjump2	10	627.3	4.2	1.3
Highjump3	10	625	14	4.4

Se is calculated as follows:

(b)

F test statistics :

F = 11.49

P-value of the test is 0.0002

Since p-value is less than 0.05 so we reject the null hypothesis.

Conclusion: There is a statistically significant difference between the three treatment means at the ? = .05 level.