Math Problem Statement

Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using a visual analog scale. Use a 0.01 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. Construct a confidence interval appropriate for the hypothesis test.

Solution

The problem requires statistical analysis to test the hypothesis that magnets are effective in reducing pain levels compared to a sham treatment. Let's analyze it step by step based on the provided image:


Part (a): Null and Alternative Hypotheses

The hypothesis test is to determine if magnets lead to greater pain reduction compared to sham treatment.

  • Null hypothesis (H0H_0): μ1=μ2\mu_1 = \mu_2 (No difference in mean pain reduction between magnet and sham treatment)
  • Alternative hypothesis (H1H_1): μ1>μ2\mu_1 > \mu_2 (Magnet treatment results in greater mean pain reduction)

Selected Option: D


Test Statistic and P-value

From the image:

  • Test statistic (tt): 0.230.23
  • P-value: 0.4110.411

Since the P-value (0.4110.411) is greater than the significance level (α=0.01\alpha = 0.01), we fail to reject the null hypothesis. There is insufficient evidence to support the claim that magnets result in a greater mean reduction in pain than sham treatment.


Part (b): Confidence Interval

You are required to construct a confidence interval for the difference in means (μ1μ2\mu_1 - \mu_2) using the data. Let me compute this for you based on the information provided:

  1. Sample Sizes:

    • n1=18n_1 = 18, n2=18n_2 = 18
  2. Means:

    • xˉ1=0.56\bar{x}_1 = 0.56, xˉ2=0.47\bar{x}_2 = 0.47
  3. Standard Deviations:

    • s1=0.98s_1 = 0.98, s2=1.38s_2 = 1.38
  4. Standard Error (SESE) for the difference in means: SE=s12n1+s22n2SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}

    Substituting values: SE=0.98218+1.38218=0.960418+1.904418=0.053356+0.1058=0.1591560.399SE = \sqrt{\frac{0.98^2}{18} + \frac{1.38^2}{18}} = \sqrt{\frac{0.9604}{18} + \frac{1.9044}{18}} = \sqrt{0.053356 + 0.1058} = \sqrt{0.159156} \approx 0.399

  5. Margin of Error (ME): At a 99% confidence level, the critical tt-value (tt^*) for df=34df = 34 (approximation for two samples of size 18) is approximately 2.7282.728.

    ME=t×SE=2.728×0.3991.089ME = t^* \times SE = 2.728 \times 0.399 \approx 1.089

  6. Confidence Interval: (xˉ1xˉ2)±ME=(0.560.47)±1.089=0.09±1.089(\bar{x}_1 - \bar{x}_2) \pm ME = (0.56 - 0.47) \pm 1.089 = 0.09 \pm 1.089

    This gives: (1.00,1.18)(-1.00, 1.18)

The 99% confidence interval for μ1μ2\mu_1 - \mu_2 is approximately (1.00,1.18)(-1.00, 1.18).


Conclusion

Since the confidence interval includes 0, it supports the earlier result of failing to reject the null hypothesis. There is insufficient evidence to claim that magnets reduce pain more than the sham treatment.


Next Steps

If you'd like, I can provide:

  1. Detailed calculations of the standard error and confidence interval formula.
  2. Explanation of how to interpret confidence intervals in hypothesis testing.
  3. An overview of when to use one-tailed vs. two-tailed tests.
  4. Insights on improving experimental design for such studies.
  5. A guide on using software like Excel or statistical tools for such tests.

Tip: Always ensure your sample sizes are large enough to improve the reliability of your statistical conclusions.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Confidence Intervals
Statistical Inference

Formulas

t-test statistic formula: t = (x̄1 - x̄2) / SE
Standard Error formula: SE = sqrt(s1^2 / n1 + s2^2 / n2)
Confidence Interval formula: (x̄1 - x̄2) ± (t* × SE)

Theorems

Central Limit Theorem
Properties of t-distribution

Suitable Grade Level

Undergraduate (Statistics or Data Analysis)