Question

: A random sample of n = 200 observations from a binomial population yield

$\hat{\mathbf{p}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{29}$

a. Test ${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathit{p}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}$ against ${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathit{p}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{35}$. Use$\mathit{a}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$.

Short answer

a. Reject the null hypothesis. So, there is enough evidence to claim that the population proportion is less than 0.35, at the $a=0.05$ significance level.

Step-by-step solution

Given Information

The random sample is taken from the Binomial population.

The sample size, n=200.

The sample proportion,$\hat{p}=0.29$.

State the large sample test of the hypothesis about a population proportion

The large sample test of the hypothesis about a population proportion is valid in the two given conditions:

⦁ A random sample is selected from a binomial population.

⦁ The sample size n is large.

Compute the hypothesis test H0:p=0.35 against  H0:p<0.35 at the significance level a=0.05.

The null and alternative hypothesis are:

$$\begin{gathered} H_0:p=0.35 \\ {H}_0:p<0.35 \end{gathered}$$

The significance level,$a=0.05$

The test statistic is,

$$\begin{gathered} Z=\frac{\hat{p}-p_0}{\sqrt{{\displaystyle \frac{p_0{q}_0}{n}}}} \\ =\frac{0.29-0.35}{\sqrt{{\displaystyle \frac{0.35(1-0.35)}{200}}}} \\ =\frac{-0.06}{0.0337} \\ =-1.7804 \end{gathered}$$

This is a one-tailed test. So, the $Z_a$ corresponding significance level $a=0.05$obtained from the standard normal table is 1.645.

Here, the rejection region is $Z_c<-Z_a\Rightarrow -1.7804<-1.645$. So, we reject the null hypothesis.

Hence, we reject the null hypothesis. So, there is enough evidence to claim that the population proportion is less than 0.35, at the $a=0.05$ significance level.