Question

Significance tests A test of $H_o:p=0.5$versus $H_a:p>0.5$based on

a sample of size $200$yields the standardized test statistic $z=2.19$. Assume that the conditions for performing inference are met.

a. Find and interpret the P-value.

b. What conclusion would you make at the $\alpha =0.01$ significance level? Would

your conclusion change if you used α=0.05 instead? Explain your reasoning.

c. Determine the value of p^= the sample proportion of successes.

Short answer

a. Hypothesis is rejected.

b. $\alpha =0.01:\text{fail to reject the null hypothesis.}$and $\alpha =0.05:\text{reject the null hypothesis}$

c.$\hat{p}=0.57774$

Step-by-step solution

Part (a) Step 1: Given Information

It is given that $z=2.19$

$n=200$

$\alpha =0.5$

$H_0:p=0.5$

$H_a:p>0.5$

Part (a) Step 2: Explanation

As $z=2.19$

$P(x<z)=0.98574$

$P(x>z)=1-0.98574=0.014262$

When $P\mathrm{value}\le \alpha$, null hypothesis is rejected.

From above $P=0.014262$

$P=0.014262<\alpha$, null hypothesis is rejected. $H_0:p=0.5$

Part (b) Step 1: Given Information

It is given that$\alpha =0.01,\alpha =0.05$

Part (b) Step 2: Explanation

Higher the significance level, higher is probability to reject null hypothesis.

As $P=0.014262<0.5$, so null hypothesis is rejected.

If $\alpha =0.01$$P=0.014262>0.01$, null hypothesis is not rejected.

If $\alpha =0.05$, $P=0.014262>0.05$, null hypothesis is not rejected.

Therefore,$P$value changes with change in significance levels.

Part (c) Step 1: Given Information

It is given that $n=200$

$p=0.5$

$z=2.19$

Part (c) Step 2: Explanation

We know that $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

Hence, $2.19=\frac{\hat{p}-0.5}{\sqrt{\frac{0.5(1-0.5)}{200}}}$

$2.19=\frac{\hat{p}-0.5}{\sqrt{\frac{0.5(0.5)}{200}}}$

$\hat{p}=(2.19\times 0.03535)+0.5$

Therefore,$\hat{p}=0.57774$