Math Problem Statement

Construct a confidence interval appropriate for the hypothesis test in part (a). according to the picture

Construct a confidence interval appropriate for the hypothesis test in part (a).

Solution

To construct the confidence interval for the hypothesis test, let's summarize the information given:

  • Magnet Treatment (μ1\mu_1):

    • Sample size (n1n_1) = 17
    • Mean (xˉ1\bar{x}_1) = 0.53
    • Standard deviation (s1s_1) = 1.02
  • Sham Treatment (μ2\mu_2):

    • Sample size (n2n_2) = 17
    • Mean (xˉ2\bar{x}_2) = 0.45
    • Standard deviation (s2s_2) = 1.38

Given this, we can calculate the confidence interval for μ1μ2\mu_1 - \mu_2 using the formula for the confidence interval of the difference in means:

(xˉ1xˉ2)±tα/2×(s12n1)+(s22n2)(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \times \sqrt{\left( \frac{s_1^2}{n_1} \right) + \left( \frac{s_2^2}{n_2} \right)}

From the image, it seems the test statistic tt is calculated as 0.190.19 and the P-value is 0.425. However, we need to calculate the critical value tα/2t_{\alpha/2} for constructing the confidence interval. For a typical 95% confidence level, α=0.05\alpha = 0.05, and the degrees of freedom for this two-sample t-test can be approximated using the following formula:

df(s12n1+s22n2)2(s12n1)2n11+(s22n2)2n21df \approx \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}}

Let's calculate this step-by-step.The 95% confidence interval for the difference in means (μ1μ2\mu_1 - \mu_2) is:

0.77<μ1μ2<0.93-0.77 < \mu_1 - \mu_2 < 0.93

This means we are 95% confident that the true difference in the reduction of pain levels between the magnet treatment and sham treatment lies between 0.77-0.77 and 0.930.93.

Do you want further details or have any questions?

Here are 5 related questions to expand this topic:

  1. What does it mean if 0 is included in the confidence interval?
  2. How does the choice of confidence level affect the width of the confidence interval?
  3. What are the assumptions of the two-sample t-test used here?
  4. How would a larger sample size affect the confidence interval?
  5. Can a non-overlapping confidence interval with zero suggest a significant difference between treatments?

Tip: Always check whether the confidence interval includes 0 to infer if there is a significant difference between groups!

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Interval
Two-Sample t-test

Formulas

Confidence Interval for Difference of Means: (x̄1 - x̄2) ± t(α/2) × √((s1²/n1) + (s2²/n2))
Degrees of Freedom for t-test: df ≈ ((s1²/n1 + s2²/n2)²) / ((s1²/n1)² / (n1-1) + (s2²/n2)² / (n2-1))

Theorems

Two-Sample t-Test

Suitable Grade Level

Grades 11-12, College Level