The distribution of total body protein in healthy adult men is normally distributed. A 95% confidence interval for the true mean total body protein of adult men with liver cirrhosis, based on 58 randomly selected adult males with liver cirrhosis, has a width of 0.052 kg. A 90% confidence interval based on the same study would have单项选择题
A
a different width, but length would depend on the sample mean.
B
a longer width.
C
a shorter width.
D
the same width.

Question

The distribution of total body protein in healthy adult men is normally distributed. A 95% confidence interval for the true mean total body protein of adult men with liver cirrhosis, based on 58 randomly selected adult males with liver cirrhosis, has a width of 0.052 kg. A 90% confidence interval based on the same study would have单项选择题

A

a different width, but length would depend on the sample mean.

B

a longer width.

C

a shorter width.

D

the same width.

BlackTom AI · Accepted Answer

Confidence interval width is proportional to the critical value (z or t) times the standard error, so lowering the confidence level reduces the multiplier and therefore narrows the interval.

Relevant concepts:
- A two-sided CI for a mean has half-width = (critical value) × (standard error). For large samples or known σ, critical values are zα/2; for n = 58 the z-approximation is appropriate.
- Changing the confidence level changes only the critical value, not the sample standard error (which depends on data and sample size).

Step-by-step reasoning and calculation:
1. The 95% CI width is given as 0.052 kg, so the half-width (margin of error) is 0.052/2 = 0.026 kg.
2. For a 95% two-sided CI, the critical z-value is z0.025 = 1.96. The implied standard error (SE) equals half-width divided by the critical value:
   SE = 0.026 / 1.96 ≈ 0.0132653 kg.
3. For a 90% CI the critical z-value is z0.05 = 1.645. The 90% CI half-width would be:
   half-width90 = 1.645 × SE ≈ 1.645 × 0.0132653 ≈ 0.021815 kg.
4. The full 90% CI width is 2 × 0.021815 ≈ 0.04363 kg, which is smaller than 0.052 kg.

Why the correct answer is correct:
- Because the critical value for 90% (1.645) is smaller than for 95% (1.96), the margin of error and thus the overall width decrease while the sample size and SE remain unchanged.

Why other options are incorrect:
- “Different width, but length would depend on the sample mean.” The sample mean shifts the center of the interval but does not affect its width.
- “A longer width.” Lowering confidence makes the interval narrower, not wider.
- “The same width.” Changing confidence level changes the critical multiplier, so width changes.