Question

in a random sample of 150 voters, 60 support Trump. Set up a 96 percent confidence interval for the proportion of voters who support trump

Answer 1

Given:
n = 150, sample size
Denote the sample proportion by q (normally written as $\hat{p}$).
That is,
 q = 60/150 = 0.4, sample proportion.

At the 96% confidence level, the z* multiplier is about 2.082, and the confidence interval for the population proportion is
$q \pm z^{*}[ \frac{q(1-q)}{ \sqrt{n} } ]$

That is,
0.4 +/- 2.082* √[(0.4*0.6)/150]
= 0.4 +/- 0.0833
= (0.3167, 0.4833)
= (31.7%, 48.3%)

Answer: The 96% confidence interval is about (32% to 48%)