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:± 0.0736Step-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] = 0.4Confidence level = 90%sample size, n = 120Now,For 90% confidence level , z-value = 1.645Width 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 ≈ ± 0.0736Hence,The correct answer is option  ± 0.0736