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. data266.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 significant difference between the three treatment means at the α = .05 level.

obs     group   g       density
1       Control 1       602
2       Control 1       543
3       Control 1       596
4       Control 1       542
5       Control 1       650
6       Control 1       574
7       Control 1       594
8       Control 1       613
9       Control 1       573
10      Control 1       616
11      Lowjump 2       621
12      Lowjump 2       659
13      Lowjump 2       627
14      Lowjump 2       648
15      Lowjump 2       629
16      Lowjump 2       639
17      Lowjump 2       632
18      Lowjump 2       645
19      Lowjump 2       631
20      Lowjump 2       638
21      Highjump        3       605
22      Highjump        3       606
23      Highjump        3       603
24      Highjump        3       598
25      Highjump        3       634
26      Highjump        3       600
27      Highjump        3       639
28      Highjump        3       594
29      Highjump        3       606
30      Highjump        3       617

Answer 1

R-commands and outputs:

d=scan("clipboard")
d

#y1-Control, y2-Lowjump, y3-Highjump
y1=c(602,543,596,542,650,574,594,613,573,616)
y2=c(621,659,627,648,629,639,632,645,631,638)
y3=c(605,606,603,598,634,600,639,594,606,617)
y=c(y1,y2,y3)

#1
> length(y1)
[1] 10
> mean(y1)
[1] 590.3
> sd(y1)
[1] 33.43003
#2
> length(y2)
[1] 10
> mean(y2)
[1] 636.9
> sd(y2)
[1] 11.32794
#3
> length(y3)
[1] 10
> mean(y3)
[1] 610.2
> sd(y3)
[1] 15.17161

	Control 1	Lowjump 2	Highjump 3
sample size (n)	10	10	10
mean (xbar)	590.3	636.9	610.2
standard deviation (s)	33.4	11.3	15.2

#treat1-Control, treat2-Lowjump, treat3-Highjump
treat=rep(1:3,each=10)
treatments=factor(treat)
treatments
[1] 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3
Levels: 1 2 3

# Before RUNNING Analysis of Variance, we state the null hypothesis

# Null Hypothesis-H0: There is NO statistically significant difference between the three treatment means at the α = 0.05 level.
# mu1=mu2=mu3

ANOVA=aov(y~treatments)
ANOVA
Call:
aov(formula = y ~ treatments)
Terms:
treatments Residuals
Sum of Squares 10934.87 13284.60
Deg. of Freedom 2 27
Residual standard error: 22.18157
Estimated effects may be unbalanced

summary(ANOVA)
Df Sum Sq Mean Sq F value Pr(>F)
treatments 2 10935 5467 11.11 0.000301 ***
Residuals 27 13285 492   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

# From the ANOVA and summary output we get, F-statistic value=11.11 with degrees of freedom (n1=2,n2=27) and the p-value=0.000.

# Thus,p-value is less than alpha, we Reject H0.
# Conclusion: There is statistically significant difference between the three treatment means at the α = .05 level.