Answer to Question #215502 in Statistics and Probability for lew

Question

Question 1

In an investigation into ownership of laptops among school students, 200 randomly chosen studens were interviewed.  It is found that 140 of them owned a laptop. 

(a)        Find the 99% confidence interval for the proportion of all school students owning a laptop.

(5 marks)

(b)        Test, at 5% level of significance, whether the proportion of school students owning a laptop is less than 75%.

(8 marks)

Answer 1

$A) Given \ that \ the \ sample \ size \ is \ n=200\\ Sample\ proportion \ p'= \frac{140}{200}=0.7\\ \Rightarrow \ q'=1-p'=1-0.7=0.3\\ 99\% confidence \ interval \ for\ population \ proportion \ is \ p'-z_\frac{\alpha}{2}\sqrt{\frac{p'q'}{n}}\lt p \lt p'+z_\frac{\alpha}{2}\sqrt{\frac{p'q'}{n}}\\ \Rightarrow 0.7-2.58(\sqrt{\frac{0.7×0.3}{200}})\lt p\lt 0.7+2.58(\sqrt{\frac{0.7×0.3}{200}})\\ 0.7-0.0836\lt p\lt 0.7+0.0836\\ 0.6164\lt p \lt 0.7836\\ Therefore, \ the \ 99\% confidence \ interval \ for \ population \ proportion \ is \ 0.6164\lt p \lt 0.7836.\\ B) Let \ the \ null \ hypothesis \ H_{0}:Proportion \ of \ school \ students \ own \ laptop \ p = 0.75\\ Alternative \ hypothesis \ H_{1}: Proportion \ of \ school \ students \ own \ laptop\ is \ less \ than \ 0.75 ; (p\lt 0.75)\\ The \ significance \ level \ \alpha =0.05\\ The\ critical \ value \ is \ z_{\alpha}=1.65\\ The \ test \ statistic \ is \ z=\frac{p-p_{0}}{\sqrt{pq}}=\frac{0.7-0.75}{\sqrt{0.7×0.3}}=-0.1091\\ Since,\ the \ test \ statistic \ z\gt -{critical value},\ we\ do \ not \ have \ a \ reason \ to \ reject \ the \ null \ hypothesis.\\ Therefore \ the \ proportion \ of \ students \ own \ laptop \ is \ 0.75.$