Many studies have suggested that there is a link between exercise and healthy bo
ID: 3269825 • Letter: M
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.
(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.)
Conclusion: There is ---Select--- no a statistically significant difference between the three treatment means at the = .05 level.
Group n s Control Low jump High jumpExplanation / Answer
using minitab
a)
Variable group N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
density Control 10 0 592.90 8.22 25.99 560.00 571.00 588.50 609.75 643.00
Highjump 10 0 607.90 7.12 22.53 574.00 583.00 610.50 625.25 641.00
Lowjump 10 0 638.20 5.26 16.62 623.00 626.00 630.50 651.00 674.00
Variable group Range
density Control 83.00
Highjump 67.00
Lowjump 51.00
b)
One-way ANOVA: density versus group
Method
Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values
group 3 Control, Highjump, Lowjump
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
group 2 10651 5325.3 10.95 0.000
Error 27 13131 486.3
Total 29 23782
Model Summary
S R-sq R-sq(adj) R-sq(pred)
22.0533 44.78% 40.69% 31.83%
Means
group N Mean StDev 95% CI
Control 10 592.90 25.99 (578.59, 607.21)
Highjump 10 607.90 22.53 (593.59, 622.21)
Lowjump 10 638.20 16.62 (623.89, 652.51)
Pooled StDev = 22.0533
f value =10.95 p value =0.0000
reject null hypothesis
p< alpha