Question

a. Determine the sample proportion.

b. Decide whether using the one-proportion $z-$test is appropriate.

c. If appropriate, use the one-proportion $z-$test to perform the specified hypothesis test.

$x=16$

$n=20$

$H_0:p=0.7$

$H_{a:}p\ne 0.7$

$a=0.05$

Short answer

(a) The sample proportion is $0.8$.

(b) It is appropriate to use the one proportion $z-$test.

(c) The hypothesis $H_0$is not rejected.

Step-by-step solution

Part(a) Step 1: Given information

The given data is

$x=16$

$n=20$

$H_0:p=0.7$

$H_a:p\ne 0.7$

$a=0.05$

Part(a) Step 2: Explanation

Write the expression for the sample proportion as

$\hat{p}=\frac{x}{n}$

The sample proportion is calculated as

$\hat{p}=\frac{16}{20}$

$=0.8$.

Part (b) Step 1: Given information

The given data is

$x=16$

$n=20$

$H_0:p=0.7$

$H_{a:}p\ne 0.7$

$a=0.05$

Part (b) Step 2: Explanation

Test the hypothesis as

$H_0:p=0.7$

$H_0:p\ne 0.7$

Calculate the value of$n{p}_0$

$n{p}_0=\left(20\right)(0.7)$

$=14$

Find the value of$n\left(1-p_0\right)$

$n\left(1-p_0\right)=\left(20\right)(1-0.7)$

$=20(0.3)$

$=6.$

Both $n{p}_0$and $n(1-p_0)$have a value greater than $5$. So, one proportion $z-$ test is appropriate to use. Therefore, it is appropriate to use the one proportion $z-$test.

Part (c) Step 1: Given information

The given data is

$x=16$

$n=20$

$H_0:p=0.7$

$H_{a:}p\ne 0.7$

$a=0.05$

Part (c) Step 2: Explanation

Write the expression for $z$as

$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

The value is calculated as

$z=\frac{0.8-0.7}{\sqrt{\frac{0.7(1-0.7)}{20}}}$

$=\frac{0.1}{0.1025}$

$=0.9759$

$\approx 0.98$.

And also, $\alpha =0.05$

For, $\alpha =0.025,$the critical value from the standard table is

$z_{0.025}=\pm 1.96$

The test statistic falls in the acceptance region.

So, the hypothesis$H_0$ is not rejected.