written 2.6 years ago by | • modified 2.6 years ago |
Solution:
$ n_{1}=500, n_{2}=800\\ $
$ p_{1}=$ proportion of defective from first factory $=2 \%=0.02\\ $
$ p_{2}=$ proportion of defective from second factory $=1.5 \%=0.015\\ $
$\mathrm{H}_{0}$ : There is no significant difference between the two products $i$.e., the products do not differ in quality. $ \begin{aligned} &\mathrm{H}_{1}: p_{1}<p_{2} \text { (one tailed test) }\\ &\text { Under } \mathrm{H}_{0}, \quad z=\frac{p_{1}-p_{2}}{\sqrt{\mathrm{PQ}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}}\\ &\quad \mathrm{P}=\frac{n_{1} p_{1}+n_{2} p_{2}}{n_{1}+n_{2}}=\frac{0.02(500)+(0.015)(800)}{500+800}=0.01692;\mathrm{Q}=1-\mathrm{P}=0.9830 \end{aligned}\\ $ $$ z=\frac{0.02-0.015}{\sqrt{0.01692 \times 0.983\left(\frac{1}{500}+\frac{1}{800}\right)}}=0.68 $$</p>
Conclusion:
As, $|z|\lt1.645$, the significant value of $z$ at $5 \%$ level of significance. $\mathrm{H}_{0}$ is accepted. i.e., the products do not differ in quality.