Question

16. A study examined the effect of lumina treatments: a control with no jumping a lowm o condition (height 30 cm) and jumping on the bone density of growing rats. There were condition (00 cm). After 8 weeks of 10 mes per day 5 days per week the bon the rats (expressed in mg/cm') was measured. Here are the resuns ohta 30 cm) and a high jump week, the bone density of Flats Boneers density 0011 Control 10 Mohjup 27 33001 8387 10 18.503506 6125 10 19.329023 617.43333 Low jump Column Summary 26 270061 Test of als Bone Denety Response attribute(numeric) density Grouping attribute (categorica: group Source DF SS MSF Groups 27433.87 3716.93 Error 27 12579.5 465.91 Total 29 20013.4 S1 = mean() S2 = count ) S3 std Dev () 0.0019 a. The ANOVA F test requires that all of the populations have the same standard deviation. Do the sample standard deviations suggest that this is appropriate? Explain your answer. (2 points) b. What are the null and alternative hypotheses for the ANOVA F test? c. Use the output to carry out the ANOVA. Let a = 0.05. Calculate the F statistic, report its P-value and state your conclusion in the context of the problem. (Assume there is no strong skewness or extreme outliers.) (6 points)

Answer 1

Result:

a).

Hartley’s Fmax test used

F max= larger variance/ smaller variance = 27.3636^2/16.5935^2 =2.72

Df1 = number of groups

Df2 = 10-1= 9( for conservative test, smallest sample size used)

Critical F value (Hartley’s F_max Table) at(3,9) df is 5.34

Calculated f value 2.72 < critical Fvalue.

Ho is not rejected.

So our population have equal standard deviations. This suggests ANOVA is appropriate.

b).

Ho: µ₁= µ₂= µ₃

H1: At least one of the mean is different from the others

c).

F =MSGroups/ MSerror = 3716.93/465.91

=7.98

P value =0.0019

Critical F at 0.05 level of significance =3.35

Calculated F value 7.98 > critical F value.

Ho is rejected.

There is significant effect of jumping on the bone density of growing rats.