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Carpal tunnel The painful wrist condition called carpal tunnel syndrome can be treated with surgery or less invasive wrist splints. In September \(2002,\) Time magazine reported on a study of 176 patients. Among the half that had surgery, \(80 \%\) showed improvement after three months, but only \(54 \%\) 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
The confidence interval is [0.1267, 0.3933]; surgery is more effective.

Step by step solution

01

Define the populations and sample sizes

We have two groups: those who had surgery and those who used wrist splints. Both groups comprise half of the total sample size which is 176 patients. Therefore, each group has a sample size of \( n_1 = n_2 = \frac{176}{2} = 88 \) patients.
02

Define the sample proportions

For the surgery group, the proportion of improved patients is \( p_1 = 0.8 \). For the wrist splint group, the proportion of improved patients is \( p_2 = 0.54 \).
03

Calculate the standard error of the difference in proportions

The standard error of the difference in two independent proportions is given by:\[ SE = \sqrt{\frac{p_1 (1-p_1)}{n_1} + \frac{p_2 (1-p_2)}{n_2}} \]Plugging in the values:\[ SE = \sqrt{\frac{0.8 \times 0.2}{88} + \frac{0.54 \times 0.46}{88}} = \sqrt{\frac{0.16}{88} + \frac{0.2484}{88}} \approx \sqrt{0.001818 + 0.002823} = \sqrt{0.004641} \approx 0.0681 \]
04

Determine the critical value for a 95% confidence interval

To find the 95% confidence interval, we use the critical value for a standard normal distribution, which is approximately 1.96 for 95% confidence.
05

Calculate the confidence interval for the difference in proportions

The difference in proportions is \( \hat{p_1} - \hat{p_2} = 0.8 - 0.54 = 0.26 \).The confidence interval is calculated using the formula:\[ CI = (\hat{p_1} - \hat{p_2}) \pm Z \times SE \]Substituting the values:\[ CI = 0.26 \pm 1.96 \times 0.0681 \]\[ CI \approx 0.26 \pm 0.1333 \]This gives us a confidence interval of \([0.1267, 0.3933]\).
06

State the conclusion based on the confidence interval

Since the 95% confidence interval for the difference in proportions \([0.1267, 0.3933]\) does not include 0, we conclude that there is a statistically significant difference between the effectiveness of surgery and wrist splints in treating carpal tunnel syndrome.

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Key Concepts

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

Understanding Standard Error
The standard error is a critical concept when dealing with the differences in proportions. It helps us understand the variability in the estimated statistic between two samples. In the context of comparing two treatments, like surgery and wrist splints for carpal tunnel syndrome, it acts as a measure of accuracy for the sample's representation.
Put simply, the standard error of the difference in proportions quantifies how much the observed sample proportions can be expected to vary from the true population proportions. This calculation takes into account that both groups (surgery and wrist splints) may have different rates of improvement, and we need to know how reliable our observed difference is.
The formula for calculating the standard error of the difference in proportions is:
  • Let's define the proportions first: for the surgery group as \( p_1 \) and for the splint group as \( p_2 \).
  • The sample sizes for both groups are \( n_1 \) and \( n_2 \) respectively.
  • The standard error (SE) formula is: \[ SE = \sqrt{\frac{p_1 (1-p_1)}{n_1} + \frac{p_2 (1-p_2)}{n_2}} \]
This formula allows us to quantify the level of uncertainty in the difference in observed proportions. Smaller standard errors imply more precise estimates, while larger ones suggest greater variability.
Difference in Proportions
The difference in proportions is a fundamental concept for comparing two groups. In research and statistical analysis, we often want to know if there's a meaningful difference between the effects of two treatments or interventions.
In the example of carpal tunnel syndrome treatment, we are comparing two forms of treatment: surgery and wrist splints. The improvement rates, represented by proportions, for these treatments are compared to understand which method is more effective.
The difference in proportions, denoted as \( \hat{p_1} - \hat{p_2} \), highlights the contrast between the success rates of the two groups. Here’s how it works:
  • First, we calculate the improvement rate (proportion) for each group. This is done using the formula \( \hat{p} = \frac{\text{number of successful outcomes}}{\text{total number in the group}} \).
  • Then, we determine the difference by subtracting the proportion of the wrist splints group from that of the surgery group.
This calculated difference tells us how much more or less effective one treatment is compared to the other. If the difference is large, it might indicate a substantial preference for one treatment over the other.
Critical Value in Confidence Intervals
A critical value plays an essential role when calculating confidence intervals. It helps define the range within which the true difference in proportions is likely to lie with a specified level of confidence, commonly 95%.
For the difference in proportions of carpal tunnel treatments, we first calculate the observed difference and the standard error, and then use the critical value to construct our interval:
  • The critical value for a given confidence level is derived from the standard normal distribution. For a 95% confidence interval, the critical value is typically 1.96.
  • This critical value represents how many standard errors we need to extend in either direction from our sample difference to capture the true difference.
The formula for the confidence interval is:
  • \[ CI = (\hat{p_1} - \hat{p_2}) \pm Z \times SE \]
  • \( Z \) here is the critical value.
Understanding and calculating this confidence interval allows us to infer whether the observed difference between the two treatment groups might just be due to variability in samples or truly indicates a consistent difference. If the interval does not include zero, as in this example, it suggests a significant difference in treatment effectiveness.

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Most popular questions from this chapter

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