Question

Significance tests A test of $H_0:p=0.65$ against $H_a:p<0.65$

based on a sample of size $400$ yields the standardized test statistic $z=-1.78$ .

a. Find and interpret the $P$-value.

b. What conclusion would you make at the $\alpha =0.10$ significance level? Would

your conclusion change if you used $\alpha =0.05$ instead? Explain your reasoning.

c. Determine the value of $\stackrel{⏜}{p}$= the sample proportion of successes.

Short answer

a. Null hypothesis is rejected.

b. $\alpha =0.1:\text{fail to reject the null hypothesis.}$and $\alpha =0.05:\text{reject the null hypothesis}$

c.$\hat{p}=0.6510$

Step-by-step solution

Part (a) Step 1: Given Information

It is given that $z=1.78$

$n=400,\alpha =0.65$

$H_0:p=0.65$

$H_a:p<0.65$

Part (a) Step 2: Explanation

Value of $z$score, $z=1.78$is:

$P(x<z)=0.96246$

$P(x>z)=1-0.96246=0.037538$

If $P\mathrm{value}>\alpha$, null hypothesis is rejected.

$P=0.037538<0.65$, null hypothesis is rejected here.

$H_0:p=0.65$

Part (b) Step 1: Given Information

It is given that$\alpha =0.1,\alpha =0.05$

Part (b) Step 2: Explanation

As studied above, for $\alpha =0.65$, hypothesis is rejected.

For $\alpha =0.1,P=0.037538<\alpha$, null hypothesis is rejected. $H_0:p=0.65$

For $\alpha =0.05,P=0.037538<\alpha$, null hypothesis is rejected. $H_0:p=0.5$

$P$value changes due to significance level changes.

Part (c) Step 1: Given Information

It is given that $n=400$

$p=0.65$

$z=1.78$

Part (c) Step 2: Explanation

Test statistic is calculated as:

$Z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$1.78=\frac{\hat{p}-0.65}{\sqrt{\frac{0.65(1-0.65)}{400}}}$

$1.78=\frac{\hat{p}-0.65}{\sqrt{\frac{0.65(0.35)}{400}}}$

$1.78=\frac{\hat{p}-0.65}{5.6875\times {10}^{-4}}$

Hence,$\hat{p}=0.6510$