A researcher studied the flexibility of 10 women in an aerob

A researcher studied the flexibility of 10 women in an aerobic exercise class, 8 women in a modern dance class, and a control group of 10 women. One measurement she made was on the degree of spinal extension. Measurements were made before and after a 16 week training program and the difference between the two was calculated. Summary statistics for the data are given in Table 1 below. Let the significance level ? alpha = 0.1.

A) Complete the ANOVA table by filling in the missing cells. Use 2-decimal precision where appropriate. MStreatment has been reported for you.

The null/alternative pair appropriate for this ANOVA setting is as follows:
Ho: ?1 = ?2 = ?3
Ha: At least one ?i differs

The p-value for this hypothesis test is: _____________


Table 2: ANOVA results for the flexibility study
Source of Variation df Sum of Squares Mean Square F
Treatment 3.68
Error Not meaningful
Total Not meaningful Not meaningful

Solution

Let there be n observations and m treatments. Here n = 10+8+10 = 28 and m = 3.

Knowing any of the one value (SSError,SSTotal,MSError or F), the rest can be calculated.

p-value can be calculated using the FDIST function in excel {=FDIST(F,2,25)}

Source of Variation Degrees of freedom Sum of Squares Mean Square F Ratio
Treatment m-1=3-1=2 MST x (m-1) = 3.68 x 2 = 7.36 3.68 MSTreatment/MSError
Error n-m=28-3=25 SSError MSError x
Total n-1=28-1=27 SSTreatment+SSError=SSTotal x x
A researcher studied the flexibility of 10 women in an aerobic exercise class, 8 women in a modern dance class, and a control group of 10 women. One measurement

Get Help Now

Submit a Take Down Notice

Tutor
Tutor: Dr Jack
Most rated tutor on our site