Question

From past knowledge A Check knows that the true proportion of ghost profiles on Facebook is 0.2. Suppose we take a sample of 200 Facebook profiles and found only 34 to be ghost profiles. Use this information to answer questions 23 to 25.

Question 23

What is the value of the population proportion and sample proportion?

(1) π = 0.2 and p = 0.34
(2) π = 0.34 and p = 0.2
(3) π = 0.17 and p = 0.2
(4) π = 0.2 and p = 0.17
(5) π = 0.34 and p = 0.17

Question 24

Keeping in mind that the true proportion is known, what is the value of the standard error of the
proportion?

(1) 0.0283
(2) 0.0008
(3) 0.0335
(4) 0.0266
(5) None of the above.

Question 25

P(p ≥ 0.17) =?

(1) 0.8554
(2) 0.5120
(3) 0.4880
(4) 0.1446
(5) None of the above.

Answer 1

Question 23.

We have population proportion $p=0.2.$

Sample proportion equals $\pi=\frac{34}{200}=0.17.$

The answer is (3).

Question 24.

$SE_p=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{(0.2)(0.8)}{200}}\approx 0.0283 (1).$

Question 25.

$np=200(0.2)=40\geq 10\\ n(1-p)=200(0.8)\geq 10$

So $p$ has normal distribution with mean 0.2 and standard deviation $\sqrt{\frac{(0.2)(0.8)}{200}}\approx 0.0283.$

$P\{p\geq 0.17\}=1-P\{p<0.17\}=1-F(0.17).\\ F(x)=\frac{1}{\sqrt{2\pi}(0.0283)^2}\int_{-\infty}^{x} e^{-\frac{(t-0.2)^2}{2(0.0283)^2}}dt\\ F(0.17)=\Phi(\frac{0.17-0.2}{0.0283})=\Phi(-1.06)=1-\Phi(1.06)\\ \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{z^2}{2}}dz\\ P\{p\geq 0.17\}=1-(1-\Phi(1.06))=\Phi(1.06)\approx 0.8554.$

The answer is (1).