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/cm3 ) was measured. Here are the data. data190.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 s_(x^^\_) to one decimal place.)

Group n x^^\_ s s_(x^^\_)

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 no/a significant difference between the three treatment means at the α = .05 level.

obs     group   g       density
1       Control 1       616
2       Control 1       613
3       Control 1       609
4       Control 1       619
5       Control 1       664
6       Control 1       602
7       Control 1       571
8       Control 1       585
9       Control 1       600
10      Control 1       609
11      Lowjump 2       623
12      Lowjump 2       620
13      Lowjump 2       622
14      Lowjump 2       653
15      Lowjump 2       622
16      Lowjump 2       634
17      Lowjump 2       647
18      Lowjump 2       636
19      Lowjump 2       642
20      Lowjump 2       660
21      Highjump        3       639
22      Highjump        3       611
23      Highjump        3       586
24      Highjump        3       622
25      Highjump        3       610
26      Highjump        3       605
27      Highjump        3       626
28      Highjump        3       630
29      Highjump        3       605
30      Highjump        3       640

Answer 1

Following is the output of ANOVA analysis:

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Control	10	6088	608.8	593.2889
Lowjump	10	6359	635.9	204.7667
Highjump	10	6174	617.4	291.1556
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	3835.4	2	1917.7	5.281896	0.011588	3.354131
Within Groups	9802.9	27	363.07037
Total	13638.3	29

(a)

Following table shows the mean, sd and se:

Groups	Count	Average	Sd	se
Control	10	608.8	24.4	7.7
Low jump	10	635.9	14.3	4.5
High jump	10	617.4	17.1	5.4

Se is calculated as follows:

(b)

F test statistics :

F = 5.28

P-value of the test is 0.012

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 alpha = .05 level.