Answer:
± 0.0736
Step-by-step explanation:
Data provided in the question:
randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = [tex]\frac{48}{120}[/tex] = 0.4
Confidence level = 90%
sample size, n = 120
Now,
For 90% confidence level , z-value = 1.645
Width of the confidence interval = ± Margin of error
= ± [tex]z\times\sqrt\frac{p\times(1-p)}{n}[/tex]
= ± [tex]1.645\times\sqrt\frac{0.4\times(1-0.4)}{120}[/tex]
= ± 0.07356 ≈ ± 0.0736
Hence,
The correct answer is option ± 0.0736
