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Model Paper 7 — Applied Maths STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsModel Paper 73 Marks
Question
Consider the following hypothesis test:
$\begin{array}{l} H _0: p \geq 0.75 \\ H _{ a }: p <0.75\end{array}$
A sample of 300 provided a sample proportion of 0.68.
i. Compute the value of the test statistic.
ii. What is the p-value?
iii. At $\alpha=0.05$, what is your conclusion?
iv. What is the rejection rule using critical value? What is your conclusion?
✓

Answer

Given $p _0=0.75, n =300, \bar{p}=0.68$
i. $Z =\frac{\bar{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}=\frac{0.68-0.75}{\sqrt{\frac{0.75 \times 0.25}{300}}}$
$=\frac{-0.07 \times 10}{\sqrt{0.25 \times 0.25}}=\frac{-0.07}{0.25}=-2.8$
ii. $\because Z=-2.8<0$
So, p-value of -2.8 = area under the standard normal curve to the left of Z
= 0.0026
$\therefore$ p-value $=0.0026$
iii. Given $\alpha=0.05$
$\because$ p-value $<0.05$
So, reject $H _0$.
iv. Rejection rule using critical value
Reject $H _0$ if $Z \leq-Z_\alpha$
Here, $\alpha=0.05$. So $Z_\alpha= Z _{0.05}=1.645$
$\begin{array}{l}\Rightarrow-Z_\alpha=-1.645 \\ \because-2.8<-1.645\end{array}$
$\Rightarrow Z <-Z_\alpha$
So, reject $H _0$

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