Problem 10.084 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, 14 of 17 repairs were successful while for shorter tears, 22 of 31 repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use alpha equals 0.05 . What is the Upper P -value?

The p-value = 0.181186

The p-value is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis & say that there isn't enough evidence to suggest that the success rate of repair of longer tears is greater than the success rate of repair of shorter tears.

Step-by-step explanation:

This is a two-sample hypothesis test as we want to test the whether the success rate for one group of tears is more than the other

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

In hypothesis testing, especially one comparing two sets of data, the null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test. It usually maintains that, with random chance responsible for the outcome or results of any experimental study/hypothesis testing, its statement is true.

The alternative hypothesis usually confirms the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test. It usually maintains that significant factors other than random chance, affect the outcome or results of the experimental study/hypothesis testing and result in its own statement.

For this question where we are testing if the success rate of longer tears is greater than the success rate of shorter tears,

The null hypothesis is that there is no significant difference in the success rate for the two types of tears. That is, there isn't enough evidence to suggest that the success rate of longer tears is greater than the success rate of shorter tears.

The alternative hypothesis is that there is enough evidence to suggest that the success rate of longer tears is greater than the success rate of shorter tears.

Mathematically, if the success rate of longer tears is p₁, success rate of repair of shorter tears is p₂. And the difference between the two success rates is μ = p₁ - p₂,

The null hypothesis is represented as

H₀: μ = 0 or p₁ = p₂

The alternative hypothesis is represented as

Hₐ: μ > 0 or p₁ > p₂

We then calculate the test statistic

The test statistic for two samples test is given as

t = (p₁ - p₂)/σ

σ = √{[p₁(1-p₁)/n₁] + [p₂(1-p₂)/n₂]}

p₁ = success rate of repair of longer tears = (14/17) = 0.8235

n₁ = 17

p₂ = success rate of repair of shorter tears = (22/31) = 0.7097

n₂ = 31

σ =

√{[0.8235(1-0.8235)/17] + [0.7097(1-0.7097)/31]}

σ = √[(0.8235×0.1765/17) + (0.7097×0.2903/31)]

= √(0.0085498676 + 0.0066459971)

= √0.0151958647 = 0.1232715081 = 0.1233

t = (p₁ - p₂)/σ

t = (0.8235 - 0.7097)/0.1233

t = 0.92

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n₁ + n₂ - 2 = 17 + 31 - 2 = 46

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for t = 0.92, at 0.05 significance level, df = 46, with a one tailed condition) = 0.181186

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.181186

0.181186 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there isn't enough evidence to suggest that the success rate of repair of longer tears is greater than the success rate of repair of shorter tears.

Hope this Helps!!!