$f(x;\theta)=\theta e^{-\theta x},\quad x \ge 0$
then the value of the level of significance is
Level of significance:
$\alpha = P(X \ge 1 \mid \theta=2)$
$= \int_{1}^{\infty} 2e^{-2x} dx$
$= \left[-e^{-2x}\right]_{1}^{\infty}$
$= e^{-2}$
$= \frac{1}{e^2}$
Total items = 100
Probability both defective:
$=\frac{20}{100}\times\frac{19}{99}$
$=\frac{380}{9900}=\frac{19}{495}$
$t=\frac{\bar{d}}{s_d/\sqrt{n}}$
$=\frac{15}{5/\sqrt{9}}=\frac{15}{5/3}=\frac{15\times3}{5}=9$
Standard error of sample proportion:
$SE=\sqrt{\frac{PQ}{n}}$
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