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Chapter 10: Problem 84

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 18 \((78 \%)\) repairs were successful, but for shorter tears, 22 of 30 \((73 \%)\) repairs were successful. (a) Is there evidence that the success rate is greater for longer tears? Use \(\alpha=0.05 .\) What is the \(P\) -value? (b) Calculate a one-sided \(95 \%\) confidence bound on the difference in proportions that can be used to answer the question

Short Answer

Expert verified
No significant evidence for greater success in longer tears at \(\alpha=0.05\). P-value is greater than 0.05.

Step by step solution

01

Define Hypotheses

We want to test if the success rate for longer tears is greater than that for shorter tears. Let \(p_1\) be the proportion of successful repairs for longer tears, and \(p_2\) be the proportion of successful repairs for shorter tears. Thus, the null hypothesis \(H_0\) is \(p_1 \leq p_2\), and the alternative hypothesis \(H_a\) is \(p_1 > p_2\).
02

Compute Sample Proportions

Calculate the sample proportions for both groups. \(\hat{p}_1\) for longer tears is \(\frac{14}{18}\) and \(\hat{p}_2\) for shorter tears is \(\frac{22}{30}\). Thus, \(\hat{p}_1 = 0.7778\) and \(\hat{p}_2 = 0.7333\).
03

Find Standard Error of the Difference

The standard error \(SE\) for the difference between two proportions is given by:\[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]Where \(n_1 = 18\) and \(n_2 = 30\). Calculate \(SE\) for these values.
04

Calculate Test Statistic

The test statistic \(z\) is given by:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \]Substitute the values of \(\hat{p}_1\), \(\hat{p}_2\), and \(SE\) to find \(z\).
05

Determine P-value

Using the z-value obtained, determine the p-value for the one-tailed test (since we are testing \(p_1 > p_2\)). Use a z-table or calculator to find the p-value.
06

Compare P-value to Alpha

Compare the p-value to the significance level \(\alpha = 0.05\). If the p-value is less than \(\alpha\), reject the null hypothesis; otherwise, do not reject it.
07

Calculate Confidence Bound

Calculate the 95% one-sided confidence bound for the difference in proportions:\[ \text{Lower bound} = (\hat{p}_1 - \hat{p}_2) - z_{\alpha} \times SE \]Use \(z_{\alpha}\) value for a 95% confidence bound which corresponds to a one-tailed \(z\) value.
08

Analyze Results

From the results of the hypothesis test and confidence bound, determine if there is enough evidence to conclude that the success rate for longer tears is greater than for shorter tears.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Intervals
A confidence interval is a range of values that likely contains the true difference in proportions between two groups. In this exercise, we're interested in determining whether the success rates of surgical repairs differ significantly based on the length of meniscal tears. By calculating a 95% confidence interval, we can estimate the difference in success rates between longer and shorter tears. This interval helps understand the reliability of our findings.
  • Confidence intervals provide insight into the precision of our estimate.
  • A one-sided confidence interval considers only one direction of effect - here we are interested in whether repairs are more successful for longer tears.
With a 95% confidence level, we are 95% confident that the true difference in proportions falls within the calculated range. This creates a scientific basis for decision-making, supporting the rejection or acceptance of the initial hypothesis.
Z-test
The Z-test is used to determine if there are significant differences between two observed samples. It's essential in hypothesis testing, especially when comparing proportions like in our exercise. The Z-test involves calculating a test statistic based on the standard error and the observed difference in proportions.
  • This helps us quantify the evidence against the null hypothesis, which assumes no difference between the two success rates.
  • The Z-value communicates how many standard deviations our observation is from the mean under the null hypothesis.
By using the Z-test, we gain a statistical framework to assess whether the success rate is truly higher for longer tears or if our finding could be due to random chance.
Proportions
In hypothesis testing, proportions help us understand the part of a study's sample that meets certain criteria. Here, proportions of successful repairs for meniscal tears provide insight into surgical outcomes.
  • We calculate these by dividing the number of successes by the total attempts for each group.
  • The two ratios obtained allow us to compare success rates between groups with different tear lengths.
Evaluating these proportions delivers a foundational understanding of our dataset by highlighting disparities or similarities in repair outcomes, laying the groundwork for a statistical test like the Z-test to further analyze these differences.
Significance Level
The significance level, denoted by \( \alpha \), defines the threshold for deciding whether an observed effect is statistically significant. In this analysis, \( \alpha = 0.05 \) signals the level of risk we are willing to accept for committing a Type I error (falsely rejecting a true null hypothesis).
  • If our p-value is less than the significance level, we have enough evidence to reject the null hypothesis.
  • This means there's a strong enough indication to conclude that the success rate is indeed higher for longer tears.
The choice of the 5% level enables us to make informed decisions while monitoring the balance between sensitivity and specificity, ensuring reliable conclusions about the data's true nature.

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Most popular questions from this chapter

A study was performed to determine whether men and women differ in repeatability in assembling components on printed circuit boards. Random samples of 25 men and 21 women were selected, and each subject assembled the units. The two sample standard deviations of assembly time were \(s_{\text {men }}=0.98\) minutes and \(s_{\text {wamen }}=1.02\) minutes. (a) Is there evidence to support the claim that men and women differ in repeatability for this assembly task? Use \(\alpha=0.02\) and state any necessary assumptions about the underlying distribution of the data. (b) Find a \(98 \%\) confidence interval on the ratio of the two variances. Provide an interpretation of the interval.

The New England Journal of Medicine reported an experiment to judge the efficacy of surgery on men diagnosed with prostate cancer. The randomly assigned half of 695 (347) men in the study had surgery, and 18 of them eventually died of prostate cancer compared with 31 of the 348 who did not have surgery. Is there any evidence to suggest that the surgery lowered the proportion of those who died of prostate cancer?

Consider the hypothesis test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(H_{1}: \sigma_{1}^{2}>\sigma_{2}^{2},\) Suppose that the sample sizes are \(n_{1}=20\) and \(n_{2}=8,\) and that \(s_{1}^{2}=4.5\) and \(s_{2}^{2}=2.3 .\) Use \(\alpha=0.01 .\) Test the hypothesis and explain how the test could be conducted with a confidence interval on \(\sigma_{1} / \sigma_{2}\),

A paper in Quality Engineering [2013, Vol. 25(1)] presented data on cycles to failure of solder joints at different temperatures for different types of printed circuit boards (PCB). Failure data for two temperatures ( 20 and \(60^{\circ} \mathrm{C}\) ) for a copper-nickel-gold PCB follow. $$\begin{array}{ll}20^{\circ} \mathrm{C} & 218,265,279,282,336,469,496,507,685,685 \\\\\hline 60^{\circ} \mathrm{C} &185,242,254,280,305,353,381,504,556,697\end{array}$$ (a) Test the null hypothesis at \(\alpha=0.05\) that the cycles to failure are the same at both temperatures. Is the alternative one or two sided? (b) Find a \(95 \%\) confidence interval for the difference in the mean cycles to failure for the two temperatures. (c) Is the value zero contained in the \(95 \%\) confidence interval? Explain the connection with the conclusion you reached in part (a). (d) Do normal probability plots of part cycles to failure indicate any violations of the assumptions for the tests and confidence interval that you performed?

An article in Industrial Engineer (September 2012\()\) reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks. lifting and using ultrasound. Assume that both sample sizes are \(6,\) the sample mean force for lifting was 6.0 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.2 pounds with standard deviation 0.3 pounds (data read from graphs in the article). Assume that the standard deviations are known. Is there evidence to conclude that the two activities result in significantly different forces on the hands?
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