Question

During a certain​ week, a poll asked 1370 ​adults, aged 18 or​ over, how they rated economic conditions. Only 24​% rated the economy as​ Excellent/Good. Construct a 99​% confidence interval for the true proportion of adults who rated the economy as​ Excellent/Good. g

Answer 1

Answer:

99% Confidence interval: (0.21,0.27)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1370

Proportion of adults who rated economy as good =

$\hat{p} = 24\% = 0.24$

99% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Putting the values, we get:

$=0.24\pm 2.58(\sqrt{\dfrac{0.24(1-0.24)}{1370}}) = 0.24\pm 0.0297\\\\=(0.2103,0.2697)\\\approx (0.21,0.27)$