Question

In a random sample of 400 residents of Boston, 320 residents indicated that they voted for Obama in the last presidential election. Develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.

Answer 1

Answer:

C.I =  0.7608   ≤ p ≤   0.8392

Step-by-step explanation:

Given that:

Let consider a  random sample n = 400 candidates where  320 residents indicated that they voted for Obama

probability $\hat p = \dfrac{320}{400}$

= 0.8

Level of significance ∝ = 100 -95%

= 5%

= 0.05

The objective is to  develop a 95% confidence interval estimate for the proportion of all Boston residents who voted for Obama.

The confidence internal can be computed as:

$=\hat p \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }$

where;

$Z_{0.05/2}$ = $Z_{0.025}$ = 1.960

SO;

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }$

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }$

$=0.8 \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }$

$=0.8 \pm 1.960 \sqrt{4 \times 10^{-4}}$

$=0.8 \pm 1.960 \times 0.02}$

$=0.8 \pm 0.0392$

= 0.8 - 0.0392     OR   0.8 + 0.0392  

= 0.7608    OR    0.8392

Thus; C.I =  0.7608   ≤ p ≤   0.8392