Solve the problem. A hypothesis test is performed to test the claim that a population proportion is greater than $0.7$. Find the probability of a type II error, $\beta$, given that the true value of the population proportion is $0.72$. The sample size is 50 and the significance level is $0.05$.
A) $0.7123$
B) $0.4129$
C) $0.9706$
D) $0.5754$

Question

Quizplus · Accepted Answer

We can approach this problem by first finding the rejection region for the null hypothesis ($H_0: p \leq 0.7$) at the significance level of 0.05.

Using a z-test with the given sample size of 50, we can calculate the critical value as:

$$z_{\alpha} = 1.645$$

Next, we can calculate the probability of a type II error ($\beta$) given the true population proportion is 0.72. For this, we need to calculate the non-rejection region (i.e., acceptance region) for the null hypothesis.

Using the given population proportion of 0.72 and the sample size of 50, we can calculate the standard error as:

$$SE = \sqrt{\frac{0.7 \cdot 0.3}{50}} = 0.0606$$

The critical value for the acceptance region can be calculated as:

$$z_{\beta} = z_{\alpha} - \frac{\left|p - 0.7\right|}{SE} = 1.645 - \frac{\left|0.72 - 0.7\right|}{0.0606} = 0.5754$$

Note that we use the null value of 0.7 in the formula because that is the value we are testing against.

Therefore, the non-rejection region is $z < 0.5754$.

To find $\beta$, we need to find the area under the true sampling distribution that falls in the non-rejection region. Since we know that the true population proportion is 0.72, the sampling distribution follows a normal distribution with mean 0.72 and standard error 0.0606.

Changing the inequality to an equality ($z = 0.5754$), we can find the corresponding p-value as:

$$p = P(Z \leq 0.5754) = 0.7123$$

Finally, we can calculate $\beta$ as:

$$\beta = 1 - P(\text{reject } H_0 \mid p = 0.72) = 1 - P(Z \geq 1.645 | \mu = 0.72, SE = 0.0606) = 0.9706$$

Therefore, the answer is:

Answer: C

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