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Zorluk: ZorStatistical Inference and Hypothesis Testing

A local public health department evaluates a community-based dietary intervention aimed at reducing fasting blood glucose levels in patients with prediabetes. A preliminary study of 3636 participants reveals a mean fasting blood glucose reduction of 18.018.0 mg/dL after 6 months. The reported 95% confidence interval for this mean reduction is 12.012.0 to 24.024.0 mg/dL.

Based on these promising results, the investigators are planning a larger follow-up study using an identical intervention in the same population. They assume the population variance will remain exactly the same and that the standard normal distribution (z-score) will be used to construct the confidence intervals in both analyses.

If the follow-up study successfully enrolls and analyzes 144144 participants, what will be the total width of the 95% confidence interval for the mean reduction in fasting blood glucose in the new study?

Cevap: 6 mg/dL

Cevap

6.0
The total width of a confidence interval is defined by the formula: Width=2×z×σnWidth = 2 \times z \times \frac{\sigma}{\sqrt{n}}. When the population standard deviation (σ\sigma) and the critical value (zz) are held constant, the width of the confidence interval relies solely on the sample size (nn). Because nn is in the denominator under a square root, the width is inversely proportional to n\sqrt{n}. In this scenario, the initial sample size is 3636, resulting in a width of 24.0−12.0=12.024.0 - 12.0 = 12.0 mg/dL. The new sample size is 144144. The ratio of the new sample size to the old sample size is 14436=4\frac{144}{36} = 4. Therefore, the new standard error will be divided by 4=2\sqrt{4} = 2. Consequently, the original total width (12.012.0 mg/dL) is halved, resulting in a new total width of 6.06.0 mg/dL.

Adım Adım Çözüm

1
Calculate the width of the original confidence interval.
The width is 24.0−12.0=12.024.0 - 12.0 = 12.0 mg/dL.
The total width of a confidence interval represents twice the margin of error.
2
Identify the relationship between sample size (nn) and confidence interval width.
Width ∝1n\propto \frac{1}{\sqrt{n}}.
The width of a confidence interval is calculated as 2×z×σn2 \times z \times \frac{\sigma}{\sqrt{n}}. Assuming zz and σ\sigma are constant, the width is inversely proportional to the square root of the sample size.
3
Determine the scale factor for the sample size increase.
14436=4\frac{144}{36} = 4. The sample size is multiplied by 44.
Comparing the new sample size (144144) to the original sample size (3636) provides the proportional increase.
4
Calculate the new confidence interval width.
12.04=12.02=6.0\frac{12.0}{\sqrt{4}} = \frac{12.0}{2} = 6.0 mg/dL.
Because the sample size increased by a factor of 44, the standard error (and therefore the confidence interval width) decreases by a factor of 4\sqrt{4}, which is 22.

Anahtar Kavram

The width of a confidence interval is inversely proportional to the square root of the sample size. Increasing the sample size by a factor of XX reduces the confidence interval width by a factor of X\sqrt{X}.
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Statistical Inference and Hypothesis Testing

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