A ) G i v e n t h a t t h e s a m p l e s i z e i s n = 200 S a m p l e p r o p o r t i o n p ′ = 140 200 = 0.7 ⇒ q ′ = 1 − p ′ = 1 − 0.7 = 0.3 99 % c o n f i d e n c e i n t e r v a l f o r p o p u l a t i o n p r o p o r t i o n i s p ′ − z α 2 p ′ q ′ n < p < p ′ + z α 2 p ′ q ′ n ⇒ 0.7 − 2.58 ( 0.7 × 0.3 200 ) < p < 0.7 + 2.58 ( 0.7 × 0.3 200 ) 0.7 − 0.0836 < p < 0.7 + 0.0836 0.6164 < p < 0.7836 T h e r e f o r e , t h e 99 % c o n f i d e n c e i n t e r v a l f o r p o p u l a t i o n p r o p o r t i o n i s 0.6164 < p < 0.7836. B ) L e t t h e n u l l h y p o t h e s i s H 0 : P r o p o r t i o n o f s c h o o l s t u d e n t s o w n l a p t o p p = 0.75 A l t e r n a t i v e h y p o t h e s i s H 1 : P r o p o r t i o n o f s c h o o l s t u d e n t s o w n l a p t o p i s l e s s t h a n 0.75 ; ( p < 0.75 ) T h e s i g n i f i c a n c e l e v e l α = 0.05 T h e c r i t i c a l v a l u e i s z α = 1.65 T h e t e s t s t a t i s t i c i s z = p − p 0 p q = 0.7 − 0.75 0.7 × 0.3 = − 0.1091 S i n c e , t h e t e s t s t a t i s t i c z > − c r i t i c a l v a l u e , w e d o n o t h a v e a r e a s o n t o r e j e c t t h e n u l l h y p o t h e s i s . T h e r e f o r e t h e p r o p o r t i o n o f s t u d e n t s o w n l a p t o p i s 0.75. A) Given \ that \ the \ sample \ size \ is \ n=200\\
Sample\ proportion \ p'= \frac{140}{200}=0.7\\
\Rightarrow \ q'=1-p'=1-0.7=0.3\\
99\% confidence \ interval \ for\ population \ proportion \ is \ p'-z_\frac{\alpha}{2}\sqrt{\frac{p'q'}{n}}\lt p \lt p'+z_\frac{\alpha}{2}\sqrt{\frac{p'q'}{n}}\\
\Rightarrow 0.7-2.58(\sqrt{\frac{0.7×0.3}{200}})\lt p\lt 0.7+2.58(\sqrt{\frac{0.7×0.3}{200}})\\
0.7-0.0836\lt p\lt 0.7+0.0836\\
0.6164\lt p \lt 0.7836\\
Therefore, \ the \ 99\% confidence \ interval \ for \ population \ proportion \ is \ 0.6164\lt p \lt 0.7836.\\
B) Let \ the \ null \ hypothesis \ H_{0}:Proportion \ of \ school \ students \ own \ laptop \ p = 0.75\\
Alternative \ hypothesis \ H_{1}: Proportion \ of \ school \ students \ own \ laptop\ is \ less \ than \ 0.75 ; (p\lt 0.75)\\
The \ significance \ level \ \alpha =0.05\\
The\ critical \ value \ is \ z_{\alpha}=1.65\\
The \ test \ statistic \ is \ z=\frac{p-p_{0}}{\sqrt{pq}}=\frac{0.7-0.75}{\sqrt{0.7×0.3}}=-0.1091\\
Since,\ the \ test \ statistic \ z\gt -{critical value},\ we\ do \ not \ have \ a \ reason \ to \ reject \ the \ null \ hypothesis.\\
Therefore \ the \ proportion \ of \ students \ own \ laptop \ is \ 0.75. A ) G i v e n t ha t t h e s am pl e s i ze i s n = 200 S am pl e p ro p or t i o n p ′ = 200 140 = 0.7 ⇒ q ′ = 1 − p ′ = 1 − 0.7 = 0.3 99% co n f i d e n ce in t er v a l f or p o p u l a t i o n p ro p or t i o n i s p ′ − z 2 α n p ′ q ′ < p < p ′ + z 2 α n p ′ q ′ ⇒ 0.7 − 2.58 ( 200 0.7 × 0.3 ) < p < 0.7 + 2.58 ( 200 0.7 × 0.3 ) 0.7 − 0.0836 < p < 0.7 + 0.0836 0.6164 < p < 0.7836 T h ere f ore , t h e 99% co n f i d e n ce in t er v a l f or p o p u l a t i o n p ro p or t i o n i s 0.6164 < p < 0.7836. B ) L e t t h e n u ll h y p o t h es i s H 0 : P ro p or t i o n o f sc h oo l s t u d e n t s o w n l a pt o p p = 0.75 A lt er na t i v e h y p o t h es i s H 1 : P ro p or t i o n o f sc h oo l s t u d e n t s o w n l a pt o p i s l ess t han 0.75 ; ( p < 0.75 ) T h e s i g ni f i c an ce l e v e l α = 0.05 T h e cr i t i c a l v a l u e i s z α = 1.65 T h e t es t s t a t i s t i c i s z = pq p − p 0 = 0.7 × 0.3 0.7 − 0.75 = − 0.1091 S in ce , t h e t es t s t a t i s t i c z > − cr i t i c a l v a l u e , w e d o n o t ha v e a re a so n t o re j ec t t h e n u ll h y p o t h es i s . T h ere f ore t h e p ro p or t i o n o f s t u d e n t s o w n l a pt o p i s 0.75.
#332078 on Oct 2022
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