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 [tex]\hat p = \dfrac{320}{400}[/tex]
= 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:
[tex]=\hat p \pm Z_{\alpha/2} \sqrt{\dfrac{ p(1-p)}{n } }[/tex]
where;
[tex]Z_{0.05/2}[/tex] = [tex]Z_{0.025}[/tex] = 1.960
SO;
[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(1-0.8)}{400 } }[/tex]
[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.8(0.2)}{400 } }[/tex]
[tex]=0.8 \pm 1.960 \sqrt{\dfrac{ 0.16}{400 } }[/tex]
[tex]=0.8 \pm 1.960 \sqrt{4 \times 10^{-4}}[/tex]
[tex]=0.8 \pm 1.960 \times 0.02}[/tex]
[tex]=0.8 \pm 0.0392[/tex]
= 0.8 - 0.0392 OR 0.8 + 0.0392
= 0.7608 OR 0.8392
Thus; C.I = 0.7608 ≤ p ≤ 0.8392
