Question

In 2006, the General Social Survey asked, "Do you see yourself as someone who has an active imagination?" For this question, 448 people said that they definitely did out of 1506 randomly selected people. What is the 95% confidence interval for the proportion of all Americans who believe that believe that they have an active imagination?

Answer 1

Answer:

(0.2744; 0.3206)

Step-by-step explanation:

The z-score for a 95% confidence interval is Z = 1.960.

The proportion of Americans in the sample who believe that they have an active imagination is given by the number of people who answered that "they definitely did", divided by the total sample of n =1506.

$p = \frac{448}{1506}\\ p=0.297476$

The confidence interval for a proportion 'p' is:

$p = \pm Z*\sqrt{\frac{p*(1-p)}{n}} \\0.297476 = \pm 1.960*\sqrt{\frac{0.297476*(1-0.297476)}{1506}}\\L = 0.297476 - 0.023088= 0.2744 \\U = 0.297476 + 0.023088 = 0.3206$

The 95% confidence interval is: I = (0.2744; 0.3206)