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

Answer:

[0.4979, 0.5479]

Step-by-step explanation:

-We first determine the sample proportion:

$\hat p=\frac{x}{n}\\\\=\frac{528}{1008}\\\\=0.5238$

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

$CI=\hat p\pm z\sqrt{\frac{\hat p(1-\hat p}{n}}$

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

$CI=\hat p\pm z\sqrt{\frac{\hat p(1-\hat p}{n}}\\\\=0.5238\pm 1.645\times \sqrt{\frac{0.5238\times0.4762}{1008}}\\\\=0.5238\pm0.0259\\\\=[0.4979,0.5497]$

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