Question

Researchers compared meniscal repair techniques using cadaveric knee specimens. One of the variables of interest was the load at failure (in newtons) for knees fixed with Method 1 and Method 2. Each technique was applied to six specimens. The sample standard deviation for Method 1 (s1) was 30.62 and the sample standard deviation for Method 2 (s2) was 11.37. Can we conclude that the variance of load at failure is higher for Method 1 (σ12) than for Method 2 (σ2)?

[a] What assumptions will you make to ensure the validity of this test?

[b] State the null and the alternative hypotheses. Then calculate the value of the observed test statistic.

[c] Calculate the p-value within table accuracy and state the strength of the evidence against the null hypothesis.

Answer 1

a)

(1) the samples are normally distributed,

(2) the samples are independent of each other.

b)

Sample Variance,    s₁² =    937.5844
Sample size,    n₁ =    6
   Sample 2:  
Sample Standard deviation,    s₂² =    129.2769
Sample size,    n₂ =    6
   α =    0.05

Null and alternative hypothesis:  
Hₒ : σ₁² = σ₂²  
H₁ : σ₁² > σ₂²  
Test statistic:  
F = s₁² / s₂² = 937.5844 / 6 =    7.2525

c)

Degree of freedom:  
df₁ = n₁-1 =    5
df₂ = n₂-1 =    5

P-value = F.DIST.RT(7.2525, 5, 5) =    0.0242

Conclusion:
As p-value < α, we reject the null hypothesis.

There is sufficient evidence that the variance of load at failure is higher for Method 1 (σ12) than for Method 2 (σ2).

Please revert in case of any doubt.

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