Question

Traces of illegal drugs were found on 8 of a random sample of 60 bills. A 95% confidence interval for the population proportion of tainted bills is which one of the following? (A) 10/64 (1.960)V(10/64)(1-10/64)/64 (B) 8/60± (1.645)V(8/60)(1-8/60)/60 (C) 8/60 t (1.960)(8/60)(1 - 8/60)/60 (D) 10/64± (1.645) V (10/64)(1-10764)/ 64

Answer is A !Note: Use the Agresti-Coull “plus four” method of forming a CI for p. Please show the work

Answer 1

At 95% confidence interval the critical value is z_0.025 = 1.96

$\dpi{100} \widehat p$ = 8/60

The 95% confidence interval for population proportion p is

$\widehat p$ +/- z_0.025 * $\dpi{100} \sqrt{\widehat p(1 - \widehat p)/n}$

= 8/60 +/- 1.96 * $\dpi{100} \sqrt{(8/60)(1 - 8/60)/60}$

Option - C is correct.