Out of a sample of 30 people, 12 individuals have diabetes. Compute a 95% confidence interval for the proportion of individuals who have diabetes.

Answer:
Given:
Favorable Cases (x)=12

Sample Size (n)=30

Confidence Interval Level =95%

Solution:
The Sample Proportion (p^):
p^=1230=0.4

The Significance Level (α):
α=10.95=0.05

The critical value (zc):
zc=Zα/2=Z0.05/2=1.96

The confidence interval (CI):
CI=p^±zc×p^(1p^)n =0.4±1.96×0.4(10.4)30 =(0.225,0.575)

Final Answer:
The 95% confidence interval =(0.225,0.575)

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