Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Round to three decimal places When 310 college students are randomly selected and surveyed, it is found that 112 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car. O A. 0.291 <p<0.432 OB. 0.298 <p<0.425 OC. 0.316<p<0.406 OD. 0.308<p <0.415

Answer 1

Solution:

Given,

n = 310 ....... Sample size

x = 112 .......no. of successes in the sample

Let $\hat p$ denotes the sample proportion.

  $\therefore$  $\hat p$ = x/n   = 112/310 = 0.3613

Our aim is to construct 99% confidence interval.

$\therefore$ c = 0.99

$\therefore$$\alpha$ = 1 - c = 1- 0.99 = 0.01

$\therefore$  $\alpha$/2 = 0.005 and 1- $\alpha$ /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

$\therefore$ = 2.576   

2012

Now , the margin of error is given by

E = $url=https://latex.codecogs.com/gif.latex?Z&t=1596156601&ymreqid=f4764063-0d40-e2b0-1cde-f00024012f00&sig=hgA_glkaYlI0EZqsFYtMdQ--~D$ $\alpha$/2 *  $url=https://latex.codecogs.com/gif.latex?%5Csqrt%7B%5Chat%20p%281-%5Chat%20p%29/n%7D&t=1596156601&ymreqid=f4764063-0d40-e2b0-1cde-f00024012f00&sig=OgQCglPdXIhHL8c_kWCRuA--~D$

= 2.576 * $url=https://latex.codecogs.com/gif.latex?%5Csqrt%7B%7D&t=1596156601&ymreqid=f4764063-0d40-e2b0-1cde-f00024012f00&sig=pxcgzcyCx.UZUhNvsIAj0g--~D$ [0.3613*(1 - 0.3613)/310]

= 0.070

Now the confidence interval is given by

($\hat p$ - E)   ($\hat p$ + E)

(0.3613 - 0.070)   (0.3613 + 0.070)

0.291   0.432