Answer:
The right answer is:
(a) 0.01252
(b) 0.8882
(c) 0.9544
Step-by-step explanation:
Given:
Population proportion,
P = 0.17
Sample size,
n = 900
(a)
The standard error of proportion will be:
⇒ [tex]SE_p=\sqrt{\frac{P(1-P)}{n} }[/tex]
By putting the values, we get
[tex]=\sqrt{\frac{0.17(1-0.17)}{900} }[/tex]
[tex]=\sqrt{0.000158}[/tex]
[tex]=0.01252[/tex]
(b)
We know that,
[tex]\hat{P}-P=\pm 0.02[/tex]
Now,
⇒ [tex]z=\frac{\hat{P}-P}{\sqrt{\frac{P(1-P)}{n} } }[/tex]
By substituting the values, we get
[tex]=\frac{\pm 0.02}{\sqrt{\frac{0.17(1-0.17)}{900} } }[/tex]
[tex]=\frac{\pm 0.02}{0.01252}[/tex]
[tex]=\pm 1.59[/tex]
⇒ [tex]P(I\hat{P}-PI<0,02)=P(-1.59<z<1.59)[/tex]
[tex]=1-2P(z<-1.59)[/tex]
[tex]=1-2(0.0559)[/tex]
[tex]=0.8882[/tex]
(c)
We know that,
[tex]\hat{P}-P=\pm 0.02[/tex]
[tex]n = 1400[/tex]
Now,
⇒ [tex]z=\frac{\hat{P}-P}{\sqrt{\frac{P(1-P)}{n} } }[/tex]
By substituting the values, we get
[tex]=\frac{\pm 0.02}{\sqrt{\frac{0.17(1-0.17)}{1400} } }[/tex]
[tex]=\frac{\pm 0.02}{0.01}[/tex]
[tex]=\pm 2[/tex]
⇒ [tex]P(I \hat P-PI<0.02)=P(-2<z<2)[/tex]
[tex]=1-2P(z<-2)[/tex]
[tex]=1-2(0.0228)[/tex]
[tex]=0.9544[/tex]
