Paired t-Test Math Example 4

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Example 4

hard
A paired t-test for blood pressure before and after medication: dห‰=โˆ’8\bar{d}=-8 mmHg, sd=5s_d=5 mmHg, n=16n=16. Construct a 95% CI for the true mean difference and interpret.

Solution

  1. 1
    SE=sd/n=5/16=5/4=1.25SE = s_d/\sqrt{n} = 5/\sqrt{16} = 5/4 = 1.25
  2. 2
    df=nโˆ’1=15df = n-1 = 15; t0.025,15โˆ—=2.131t^*_{0.025,15} = 2.131
  3. 3
    95% CI: dห‰ยฑtโˆ—ร—SE=โˆ’8ยฑ2.131ร—1.25=โˆ’8ยฑ2.66=(โˆ’10.66,โˆ’5.34)\bar{d} \pm t^* \times SE = -8 \pm 2.131 \times 1.25 = -8 \pm 2.66 = (-10.66, -5.34)
  4. 4
    Interpretation: we are 95% confident the medication reduces blood pressure by between 5.34 and 10.66 mmHg on average

Answer

95% CI: (โˆ’10.66,โˆ’5.34)(-10.66, -5.34) mmHg. Medication significantly reduces BP (entire CI below 0).
The CI for paired differences entirely below zero confirms a statistically significant reduction. The clinical meaning: with 95% confidence, the medication reduces BP by at least 5.34 mmHg โ€” a clinically meaningful reduction for most patients.

About Paired t-Test

A hypothesis test for the mean difference in a paired (matched) data design, where each subject provides two related measurements. The test analyzes the differences di=x1iโˆ’x2id_i = x_{1i} - x_{2i} as a single sample.

Learn more about Paired t-Test โ†’

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