Question

In a survey of 1008 ​adults, a polling agency​ asked, "When you​ retire, do you think you will have enough money to live comfortably or not. Of the 1008 ​surveyed, 528 stated that they were worried about having enough money to live comfortably in retirement. Construct a 90​% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.

Answer 1

[0.4979, 0.5479]

Explanation:

-We first determine the sample proportion:

`\hat p=(x)/(n)\\\\=(528)/(1008)\\\\=0.5238`

-The confidence intervals of a sample proportion is calculated using the formula:

`CI=\hat p\pm z\sqrt{(\hat p(1-\hat p)/(n)}`

#We substitute for the sample proportion and z value to get the Confidence interval:

`CI=\hat p\pm z\sqrt{(\hat p(1-\hat p)/(n)}\\\\=0.5238\pm 1.645* \sqrt{(0.5238*0.4762)/(1008)}\\\\=0.5238\pm0.0259\\\\=[0.4979,0.5497]`

Hence, the 90% confidence intervals is [0.4979,0.5479]