Question

Suppose we are estimating a population proportion by its sample equivalent.
(a) We have a sample of n = 10 units and we find the proportion is p = .4. If the true proportion is p= .3 find PC Ô-p|>.2)
(b) Consider the same problem as in part (a) but now, our sample size is n = 400. Find P( P-p> .001)

Answer 1

Answer:

(a) 0.16759

(b)  0.9649

Step-by-step explanation:

Given that:

n = 10 , p = 0.3 and  $\hat p = 0.4$

$P(|\hat p - p| > 0.2 ) = 1 - P ( |\hat p -p| \leq 0.2)$

= $1 - P(-0.2 \leq \hat p-p \leq 0.2)$

= $1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{pq}{n}}} \leq \dfrac{\hat p -p}{\sqrt{\dfrac{pq}{n}}} \leq \dfrac{0.2}{\sqrt{\dfrac{pq}{n}}} \end {pmatrix}$

$=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{(0.3)(1-0.3)}{10}}} \leq Z \leq \dfrac{0.2}{\sqrt{\dfrac{(0.3)(1-0.3)}{10}}} \end {pmatrix}$

$=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{\dfrac{(0.3)(0.7)}{10}}} \leq Z \leq \dfrac{0.2}{\sqrt{\dfrac{(0.3)(0.7)}{10}}} \end {pmatrix}$

$=1 - P \begin {pmatrix} \dfrac{-0.2}{\sqrt{0.021}} \leq Z \leq \dfrac{0.2}{\sqrt{0.021}} \end {pmatrix}$

$=1 - P \begin {pmatrix} -1.380 \leq Z \leq 1.380 \end {pmatrix}$

= 1 - P( Z ≤ 1.380) - P(-1.380)

= 1 - ( 0.91620 - 0.08379 )

= 1 - 0.83241

= 0.16759

b) when n = 400; p =0.3 , q = 1 - p = 1 - 0.3 = 0.7

$P( |\hat p - p | > 0.001) = 1- P ( |\hat p - p | < 0.001 )$

$= 1- P ( -0.001 < \hat p - p < 0.001 )$

$= 1- P ( \dfrac{-0.001}{\sqrt{\dfrac{pq}{n}}} < \dfrac{ \hat p - p}{\dfrac{pq}{n}} < \dfrac{0.001}{\sqrt{\dfrac{pq}{n}}} )$

$= 1- P ( \dfrac{-0.001}{\sqrt{\dfrac{0.3\times 0.7}{400}}} < Z < \dfrac{0.001}{\sqrt{\dfrac{0.3 \times 0.7}{400}}} )$

$= 1- P ( \dfrac{-0.001}{\sqrt{5.25 \times 10^{-4}}} < Z < \dfrac{0.001}{\sqrt{5.25 \times 10^{-4}}} )$

$= 1- P ( -0.0436< Z < 0.0436)$

= 1 - P ( Z < 0.0436) - P ( -0.0436)

= 1 - (0.5176 - 0.4825)

= 1 -  0.0351

= 0.9649