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An article in the British Medical Journal ["Comparison of Treatment of Renal Calculi by Operative Surgery, Percutaneous Nephrolithotomy, and Extracorporeal Shock Wave Lithotripsy" (1986, Vol. 292, pp. 879-882)] reported that percutaneous nephrolithotomy (PN) had a success rate in removing kidney stones of 289 out of \(350(83 \%)\) patients. However, when the stone diameter was considered, the results looked different. For stones of \(<2 \mathrm{~cm}, 87 \%(234 / 270)\) of cases were successful. For stones of \(\geq 2 \mathrm{~cm},\) a success rate of \(69 \%(55 / 80)\) was observed for \(\mathrm{PN}\). (a) Are the successes and size of stones independent? Use \(\alpha=0.05\) (b) Find the \(P\) -value for this test.

Short Answer

Expert verified
The stone size and success of the treatment are likely not independent (reject \( H_0 \)), with a \(P\)-value supporting this conclusion.

Step by step solution

01

Define Hypotheses

To test for independence, we use the chi-square test. Define the null hypothesis \( H_0 \) as "success in treatment and stone size are independent." The alternative hypothesis \( H_a \) is "success in treatment and stone size are not independent."
02

Set Up Expected Frequencies

Create a contingency table with observed frequencies: \(<2\ cm \) success (234), \(<2\ cm \) failure \(270 - 234 = 36\), \(\geq 2\ cm \) success (55), \(\geq 2\ cm \) failure \(80 - 55 = 25\). Total patients are 350. Calculate expected frequencies under \( H_0 \).
03

Calculate Expected Values

The expected frequency for each cell \( E \) is calculated by \( E = \frac{(\text{row total}) \times (\text{column total})}{\text{total patients}} \). Compute it for all four categories in the table.
04

Compute Chi-Square Statistic

Using the formula \( \chi^2 = \sum \frac{(O-E)^2}{E} \), where \( O \) is the observed frequency and \( E \) is the expected frequency, find the chi-square test statistic for the data.
05

Determine Degrees of Freedom

The degrees of freedom for a contingency table is calculated as \((r-1)(c-1)\), where \(r\) is the number of rows and \(c\) is the number of columns. Here, \( r = 2 \) and \( c = 2 \), so the degrees of freedom is \(1\).
06

Compare with Chi-Square Distribution

Use the chi-square distribution with \(1\) degree of freedom to find the critical value at \(\alpha = 0.05\). Compare the calculated chi-square statistic with this critical value.
07

Calculate P-value

Determine the \(P\)-value corresponding to the computed chi-square statistic using a chi-square distribution table or a statistical software.

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Key Concepts

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

Independence of Variables
In the context of statistics, the concept of independence between two variables is fundamental. When we say two variables are independent, we mean that the occurrence of one variable does not affect the occurrence of the other. Think of tossing a coin and rolling a die at the same time. The outcome of the coin toss does not influence the outcome of the die roll. Similarly, in medical statistics, researchers often analyze if two factors, like treatment success and stone size, are independent from each other.

For the given exercise, the independence hypothesis is tested using a chi-square test. We start with the null hypothesis, which claims that treatment success and stone size are independent. If data reveals a significant relationship, we might reject this hypothesis, indicating dependence between the variables.
Contingency Table
To analyze the relationship between two categorical variables, a contingency table is used. It helps translate raw data into a tabular format, which is easier to interpret and analyze statistically. In our exercise, stone size and treatment success are the two variables arranged in the contingency table.

The table is structured with stone size as rows and treatment results as columns. For example, one row might indicate stones less than 2cm with both successful and unsuccessful treatments divided into respective columns. The table's content is the observed frequencies of each outcome, like the number of successful treatments for stones less than 2cm. Once the table is set, the expected frequencies under the assumption of independence are calculated for further steps in the chi-square test.
Degrees of Freedom Calculation
In any statistical test, degrees of freedom are crucial because they affect the distribution used to find critical values or P-values. For a contingency table, the formula for calculating degrees of freedom is \( (r - 1)(c - 1) \), where \( r \) is the number of rows and \( c \) is the number of columns.

In our example, we have two rows (stone sizes: less than 2cm, 2cm or more) and two columns (success, failure in treatment). Therefore, the degrees of freedom is \( (2-1)(2-1) = 1 \). With this value, we can reference the critical value from the chi-square distribution corresponding to our chosen significance level (\( \alpha = 0.05 \)). This comparison helps decide whether to reject the null hypothesis.
P-value Determination
The P-value is an essential concept in hypothesis testing, providing the probability of obtaining an observation at least as extreme as the actual results, assuming the null hypothesis is true. Calculating the P-value helps determine the strength of the evidence against the null hypothesis.

In our chi-square test for independence, once we have the chi-square statistic calculated, we look to a chi-square distribution table or software to find the corresponding P-value based on \( 1 \) degree of freedom. If the P-value is less than the significance level \( \alpha = 0.05 \), it suggests that the variables are not independent, and we would reject the null hypothesis. On the other hand, if the P-value is higher than 0.05, the null hypothesis will not be rejected, indicating no significant evidence of dependence.

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Most popular questions from this chapter

A computer manufacturer ships laptop computers with the batteries fully charged so that customers can begin to use their purchases right out of the box. In its last model, \(85 \%\) of customers received fully charged batteries. To simulate arrivals, the company shipped 100 new model laptops to various company sites around the country. Of the 105 laptops shipped, 96 of them arrived reading \(100 \%\) charged. Do the data provide evidence that this model's rate is at least as high as the previous model? Test the hypothesis at \(\alpha=0.05 .\)

The impurity level (in \(\mathrm{ppm}\) ) is routinely measured in an intermediate chemical product. The following data were observed in a recent test: 2.4,2.5,1.7,1.6,1.9,2.6,1.3,1.9,2.0,2.5,2.6,2.3,2.0,1.8 1.3,1.7,2.0,1.9,2.3,1.9,2.4,1.6 Can you claim that the median impurity level is less than \(2.5 \mathrm{ppm} ?\) (a) State and test the appropriate hypothesis using the sign test with \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Use the normal approximation for the sign test to test \(H_{0}: \tilde{\mu}=2.5\) versus \(H_{1}: \tilde{\mu}<2.5 .\) What is the \(P\) -value for this test?

The life in hours of a battery is known to be approximately normally distributed with standard deviation \(\sigma=1.25\) hours. A random sample of 10 batteries has a mean life of \(\bar{x}=40.5\) hours. (a) Is there evidence to support the claim that battery life exceeds 40 hours? Use \(\alpha=0.05 .\) (b) What is the \(P\) -value for the test in part (a)? (c) What is the \(\beta\) -error for the test in part (a) if the true mean life is 42 hours? (d) What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean life is 44 hours? (e) Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on life.

The advertised claim for batteries for cell phones is set at 48 operating hours with proper charging procedures. A study of 5000 batteries is carried out and 15 stop operating prior to 48 hours. Do these experimental results support the claim that less than 0.2 percent of the company's batteries will fail during the advertised time period, with proper charging procedures? Use a hypothesis-testing procedure with \(\alpha=0.01\).

A bearing used in an automotive application is supposed to have a nominal inside diameter of 1.5 inches. A random sample of 25 bearings is selected, and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation \(\sigma=0.01\) inch. (a) Test the hypothesis \(H_{0}: \mu=1.5\) versus \(H_{1}: \mu \neq 1.5\) using $$\alpha=0.01$$ (b) What is the \(P\) -value for the test in part (a)? (c) Compute the power of the test if the true mean diameter is 1.495 inches. (d) What sample size would be required to detect a true mean diameter as low as 1.495 inches if you wanted the power of the test to be at least \(0.9 ?\) (e) Explain how the question in part (a) could be answered by constructing a two-sided confidence interval on the mean diameter.

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