Question

A random sample of 915 adults showed that 34% of them are smokers. Based on this sample, the 90% confidence interval for the proportion of all adults who are smokers, rounded to four decimal places, is: the lower endpoint is i the upper endpoint is

Answer 1

Solution :

Given that,

n = 915

Point estimate = sample proportion = $\hat p$ = 0.340

1 - $\hat p$ = 1-0.34 = 0.660

At 90% confidence level

$\alpha$ = 1-0.90% =1-0.90 =0.10

$\alpha$/2 =0.10/ 2= 0.05

Z$\alpha$/2 = Z_{0.05 = 1.645}

Z$\alpha$/2 = 1.645  

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.645 * ($\sqrt$(0.340*(0.660) /915 )

= 0.0258

A 90% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.340 - 0.0258 < p < 0.340+ 0.0258

( 3142 ,0.3658)

The lower endpoint is = 0.3142

The upper endpoint is = 0.3658