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Chapter 20: Problem 30

The painful wrist condition called carpal tunnel syndrome can be treated with surgery or, less invasively, with wrist splints. Recently, Time magazine reported on a study of 176 patients. Among the half that had surgery, \(80 \%\) showed improvement after three months, but only \(48 \%\) of those who used the wrist splints improved. a. What's the standard error of the difference in the two proportions? b. Construct a \(95 \%\) confidence interval for this difference. c. State an appropriate conclusion.

Short Answer

Expert verified
First, calculate the standard error of the difference between the two proportions using the proportions and sizes for each group. Then calculate the difference in proportions. Use this information to construct the 95 percent confidence interval for the difference. Draw a final conclusion based on whether or not the confidence interval contains zero.

Step by step solution

01

Identify Proportions and Sizes

Identify the proportions and sizes for each group. If we call the surgery group 'group A' and the splints group 'group B', then:Proportion of improvement in group A, \(p_A = 0.80\).Number in group A, \(n_A = 0.50 \times 176 = 88\).Proportion of improvement in group B, \(p_B = 0.48\).Number in group B, \(n_B = 0.50 \times 176 = 88\).
02

Calculate Standard Error

Calculate the standard error of the difference in the two proportions. Using the formula:\[SE = \sqrt{\frac{{p_A\times (1-p_A)}}{{n_A}} + \frac{{p_B\times (1-p_B)}}{{n_B}}}\]Substitute the identified values:\[SE = \sqrt{\frac{{0.80\times (1-0.80)}}{{88}} + \frac{{0.48\times (1-0.48)}}{{88}}}\]
03

Calculate Difference in Proportions

Calculate the difference in proportions, which is \(p_A - p_B = 0.80 - 0.48 = 0.32\). This difference is needed for constructing the confidence interval in the next step.
04

Construct 95 percent Confidence Interval

Construct the 95 percent confidence interval for the difference using the formula \[(p_A - p_B) \pm 1.96 \times SE\]Substitute the calculated difference in proportions and standard error into the formula to calculate the confidence interval.
05

Conclusion Drawing

Draw a conclusion on whether or not there is a statistically significant difference in the proportions based on the calculated confidence interval.If the confidence interval does not contain zero, it suggests there is a significant difference between the two treatment methods.

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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 gives us a range where we expect our true value to lie. It shows us how much uncertainty is around a measurement or estimate. For the case of carpal tunnel treatments, we're interested in the difference between the improvement rates of surgery and wrist splints.
The confidence level, often set at 95%, tells us how sure we are about the interval capturing the true difference. Here's why it's useful:
  • It tells us the precision of our estimate.
  • It helps us make decisions. If the interval doesn鈥檛 contain zero, we see a significant difference.
To find the confidence interval, we use the formula: \[(p_A - p_B) \pm 1.96 \times SE\] Where \(1.96\) is the z-value for 95% confidence. By computing this, we get a range for the difference in improvement rates.
Standard Error
The standard error measures the variability of a statistic. It's like a yardstick for how much variability there might be in our estimates from sample to sample. In our scenario of treating carpal tunnel syndrome, the standard error of the difference in proportions helps us understand how much the difference between the effectiveness of surgery and wrist splints could change by chance alone.
The formula for calculating the standard error (SE) of the difference between two proportions is: \[SE = \sqrt{\frac{{p_A \times (1 - p_A)}}{{n_A}} + \frac{{p_B \times (1 - p_B)}}{{n_B}}}\] This tells us how spread out the difference in proportions could be. A smaller SE indicates a more precise estimate, meaning less variation by chance.
Proportion Difference
The proportion difference is simply the difference between two proportions. In statistics, this measures how one group differs from another. For our carpal tunnel study, we are comparing two treatments: surgery and wrist splints.
This difference is calculated as:\[p_A - p_B = 0.80 - 0.48 = 0.32\] This result means that surgery leads to a higher improvement rate by 32% compared to splints. Understanding this helps in evaluating which treatment is more effective, guiding better medical decisions. Knowing the difference allows us to quantify the advantage of one treatment over another.
Statistical Significance
Statistical significance helps us decide whether an observed effect is real or could just be due to chance. In analyzing carpal tunnel treatments, determining whether the difference in improvement rates is statistically significant is crucial.
First, we look at the confidence interval. If the interval for the difference between surgery and splints does not include zero, the difference is considered statistically significant.
  • If zero is not in the interval, it implies that there is a real difference.
  • If zero is in the interval, we can't confidently say one treatment is better.
This conclusion informs doctors whether surgery or splints make a significant difference in treatment success rates. Statistical significance, therefore, boosts confidence in clinical decisions and can affect treatment recommendations.

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Most popular questions from this chapter

In Chapter 6 , Exercise 25 , we looked at collected samples of water from streams in the Adirondack Mountains to investigate the effects of acid rain. Researchers measured the pH (acidity) of the water and classified the streams with respect to the kind of substrate (type of rock over which they flow). A lower pH means the water is more acidic. Here is a boxplot of the \(\mathrm{pH}\) of the streams by substrate (limestone, mixed, or shale): Here are selected parts of a software analysis comparing the pH of streams with limestone and shale substrates: 2 -Sample \(t\) -Test of \(\mu_{1}-\mu_{2}\) Difference Between Means \(=0.735\) \(t\) -Statistic \(=16.30 \mathrm{w} / 133 \mathrm{df}\) \(\mathrm{p} \leq 0.0001\) a. State the null and alternative hypotheses for this test. b. From the information you have, do the assumptions and conditions appear to be met? c. What conclusion would you draw?

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One month before the election, a poll of 630 randomly selected voters showed \(54 \%\) planning to vote for a certain candidate. A week later, it became known that he had had an extramarital affair, and a new poll showed only \(51 \%\) of 1010 voters supporting him. Do these results indicate a decrease in voter support for his candidacy? a. Test an appropriate hypothesis and state your conclusion. b. If you concluded there was a difference, estimate that difference with a confidence interval and interpret your interval in context.

The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65 and older, which found that 411 of 1012 men and 535 of 1062 women suffered from some form of arthritis. a. Are the assumptions and conditions necessary for inference satisfied? Explain. b. Create a \(95 \%\) confidence interval for the difference in the proportions of senior men and women who have this disease. c. Interpret your interval in this context. d. Does this confidence interval suggest that arthritis is more likely to afflict women than men? Explain.
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