Question

An article in Knee Surgery, Sports Traumatology, Arthroscopy (2005, Vol. 13, pp. 273-279), considered arthroscopic meniscal repair with an absorbable screw. Results showed that for tears greater than 25 millimeters, 13 of 18 repairs were successful while for shorter tears, 21 of 29 repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use

Answer 1

Answer:

$z=\frac{0.722-0.724}{\sqrt{0.723(1-0.723)(\frac{1}{18}+\frac{1}{29})}}=-0.015$  

$p_v =P(Z<-0.015)= 0.494$  

If we compare the p value with any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can say the the proportion of longer tears it's not significant less than the porportion for shorter tears.  

Step-by-step explanation:

1) Data given and notation  

$X_{L}=13$ represent the number of repairs that were successful for longer tears

$X_{S}=21$ represent the number of repairs that were successful for shorter tears

$n_{L}=18$ sample of longer tears selected

$n_{S}=29$ sample of shorter tears selected

$p_{L}=\frac{13}{18}=0.722$ represent the proportion of repairs that were successful for longer tears

$p_{S}=\frac{21}{29}=0.724$ represent the proportion of repairs that were successful for shorter tears

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion for longer tears it's higher than the proportion for shorter tears , the system of hypothesis would be:  

Null hypothesis:$p_{L} \geq p_{S}$  

Alternative hypothesis:$p_{L} < p_{S}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{L}-p_{S}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{L}}+\frac{1}{n_{S}})}}$   (1)

Where $\hat p=\frac{X_{L}+X_{S}}{n_{L}+n_{S}}=\frac{13+21}{18+29}=0.723$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.722-0.724}{\sqrt{0.723(1-0.723)(\frac{1}{18}+\frac{1}{29})}}=-0.015$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

Since is a one left tailed test the p value would be:  

$p_v =P(Z<-0.015)= 0.494$  

If we compare the p value with any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can say the the proportion of longer tears it's not significant less than the porportion for shorter tears.