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Chapter 9: Problem 89

An article in the British Medical Journal [鈥淐omparison of Treatment of Renal Calculi by Operative Surgery, Percutaneous Nephrolithotomy, and Extra- Corporeal Shock Wave Lithotrips," (1986, Vol. 292, pp. \(879-882\) ) ] found that percutaneous nephrolithotomy (PN) had a success rate in removing kidney stones of 289 out of 350 patients. The traditional method was \(78 \%\) effective. (a) Is there evidence that the success rate for \(\mathrm{PN}\) is greater than the historical success rate? Find the \(P\) -value. (b) Explain how the question in part (a) could be answered with a confidence interval.

Short Answer

Expert verified
(a) P-value = 0.0187; Yes, the PN rate is greater. (b) Use a 95% confidence interval to check if 0.78 is excluded.

Step by step solution

01

Define the Hypotheses

We need to test whether the success rate of PN is greater than the traditional method. The null hypothesis \((H_0)\) is that the success rate of PN \(p\) is equal to 0.78, and the alternative hypothesis \((H_a)\) is that \(p > 0.78\).
02

Calculate Sample Proportion

The sample proportion \(\hat{p}\) is calculated by dividing the number of successful cases by the total number of cases in the sample: \(\hat{p} = \frac{289}{350} \approx 0.826\).
03

Compute the Test Statistic

The test statistic for a one-sample proportion test is calculated using the formula \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\), where \(p_0 = 0.78\) and \(n = 350\). Substituting the values, we get \(z \approx \frac{0.826 - 0.78}{\sqrt{\frac{0.78 \times 0.22}{350}}} \approx 2.082\).
04

Find the P-value

Using the standard normal distribution table, find the P-value corresponding to the test statistic \(z = 2.082\). This gives us \(P(z > 2.082) = 0.0187\).
05

Conclusion for Hypothesis Test

Since the P-value \(0.0187\) is less than the typical significance level of 0.05, we reject the null hypothesis. There is evidence to suggest that the success rate for PN is greater than 78%.
06

Explain Confidence Interval Approach

We could alternatively calculate a 95% confidence interval for the true proportion and check if 0.78 lies within this interval. If 0.78 is not contained within the interval, it confirms that the PN success rate is significantly higher.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values within which the true population parameter is expected to lie. This range is determined based on a specific level of confidence, often 95% or 99%.
A 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect 95 of those intervals to contain the actual population parameter.
In the context of our PN study, the confidence interval helps us assess whether the true success rate is significantly different from the historical rate of 78%.
  • Calculate the sample proportion (\(\hat{p}\approx0.826\)
  • Determine the margin of error using a formula that incorporates the sample standard deviation and the z-value associated with your confidence level.
  • Construct the interval by adding and subtracting the margin of error from the sample proportion.
If 0.78 is outside this interval, we have strong evidence that the PN procedure has a higher success rate.
P-Value
The P-value is a statistical measure to help you determine the significance of your test results.
A P-value tells us the probability of obtaining a test statistic (or extreme value) as observed, assuming the null hypothesis is true.
In our exercise, the P-value is 0.0187. This value is evaluated against a chosen significance level, typically 0.05, to determine the outcome of the hypothesis test.
  • If the P-value is less than the significance level, we reject the null hypothesis, implying that the results are statistically significant.
  • Conversely, if it is higher, we fail to reject the null, suggesting insufficient evidence against it.
Since the P-value here is less than 0.05, we conclude that there is strong evidence that the success rate for PN is greater than the historical rate of 78%.
Null and Alternative Hypothesis
In hypothesis testing, forming the right hypotheses is crucial.
The null hypothesis (\(H_0\)) usually represents the status quo or a position of no effect, while the alternative hypothesis (\(H_a\)) represents the claim to be tested.
For the PN success rate, our null hypothesis is that the rate is 78%, matching the traditional method.
  • \(H_0: p = 0.78\) (success rate equals 78%)
  • \(H_a: p > 0.78\) (success rate is greater than 78%)
Testing these hypotheses helps us determine if the procedural change from traditional methods to PN results in a statistically significant improvement in success rates.
Sample Proportion
In the analysis of a sample, the sample proportion (\(\hat{p}\)) represents the fraction of successful outcomes over the total sample size.
It is a crucial statistic in evaluating claims about population proportions.
In the PN study, the sample proportion is used to assess whether the observed success rate significantly differs from the known success rate of 78%.
  • Number of successful PN cases: 289
  • Total PN cases: 350
  • Sample proportion \(\hat{p} = \frac{289}{350} \approx 0.826\)
This proportion forms the basis for further statistical tests, including computations of confidence intervals and the P-value, to make informed conclusions about the procedure's effectiveness.

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Most popular questions from this chapter

Suppose we are testing \(H_{0}: p=0.5\) versus \(H_{0}: p \neq 0.5 .\) Suppose that \(p\) is the true value of the population proportion. (a) Using \(\alpha=0.05,\) find the power of the test for \(n=100\), 150, and 300 assuming that \(p=0.6\). Comment on the effect of sample size on the power of the test. (b) Using \(\alpha=0.01\), find the power of the test for \(n=100\), 150 , and 300 assuming that \(p=0.6\). Compare your answers to those from part (a) and comment on the effect of \(\alpha\) on the power of the test for different sample sizes. (c) Using \(\alpha=0.05\), find the power of the test for \(n=100\), assuming \(p=0.08\). Compare your answer to part (a) and comment on the effect of the true value of \(p\) on the power of the test for the same sample size and \(\alpha\) level. (d) Using \(\alpha=0.01\), what sample size is required if \(p=0.6\) and we want \(\beta=0.05 ?\) What sample is required if \(p=0.8\) and we want \(\beta=0.05\) ? Compare the two sample sizes and comment on the effect of the true value of \(p\) on sample size required when \(\beta\) is held approximately constant.

A hypothesis will be used to test that a population mean equals 7 against the alternative that the population mean does not equal 7 with unknown variance \(\sigma\). What are the critical values for the test statistic \(T_{0}\) for the following significance levels and sample sizes? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

A company operates four machines in three shifts each day. From production records, the following data on the number of breakdowns are collected: $$ \begin{array}{cccrc} \hline & {\text { Machines }} \\ { 2 - 5 } \text { Shift } & A & B & C & D \\ \hline 1 & 41 & 20 & 12 & 16 \\ 2 & 31 & 11 & 9 & 14 \\ 3 & 15 & 17 & 16 & 10 \\ \hline \end{array} $$ Test the hypothesis (using \(\alpha=0.05\) ) that breakdowns are independent of the shift. Find the \(P\) -value for this test.

The mean pull-off force of an adhesive used in marufacturing a connector for an automotive engine application should be at least 75 pounds. This adhesive will be used unless there is strong evidence that the pull-off force does not meet this requirement. A test of an appropriate hypothesis is to be conducted with sample size \(n=10\) and \(\alpha=0.05 .\) Assume that the pull-off force is normally distributed, and \(\sigma\) is not known. (a) If the true standard deviation is \(\sigma=1\), what is the risk that the adhesive will be judged acceptable when the true mean pull-off force is only 73 pounds? Only 72 pounds? (b) What sample size is required to give a \(90 \%\) chance of detecting that the true mean is only 72 pounds when \(\sigma=1 ?\) (c) Rework parts (a) and (b) assuming that \(\sigma=2\). How much impact does increasing the value of \(\sigma\) have on the answers you obtain?

A semiconductor manufacturer collects data from a new tool and conducts a hypothesis test with the null hypothesis that a critical dimension mean width equals \(100 \mathrm{nm}\). The conclusion is to not reject the null hypothesis. Does this result provide strong evidence that the critical dimension mean equals \(100 \mathrm{nm}\) ? Explain.
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