Question

A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice? Select one:a. ± .0736b. ± .0447c. ± .0876d. ± .0894

Answer 1

± 0.0736

Explanation:

Data provided in the question:

randomly chosen graduates of California medical schools last year intended to specialize in family practice, p =
`(48)/(120)` = 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

= ±
`z*\sqrt(p*(1-p))/(n)`

= ±
`1.645*\sqrt(0.4*(1-0.4))/(120)`

= ± 0.07356 ≈ ± 0.0736

Hence,

The correct answer is option ± 0.0736