Question

PowerFor a hypothesis test with a specified significance level , the probability of a type I error is, whereas the probability of a type II error depends on the particular value ofpthat is used as an alternative to the null hypothesis.

a.Using an alternative hypothesis ofp< 0.4, using a sample size ofn= 50, and assumingthat the true value ofpis 0.25, find the power of the test. See Exercise 34 “Calculating Power”in Section 8-1. [Hint:Use the valuesp= 0.25 andpq/n= (0.25)(0.75)/50.]

b.Find the value of , the probability of making a type II error.

c.Given the conditions cited in part (a), find the power of the test. What does the power tell us about the effectiveness of the test?

Short answer

a. The power of the test is 0.7224

b. The probability of making a type II error or the value of $\beta$is 0.2776

c. The probability of making the correct conclusion of rejecting the null hypothesis when the true population proportion is equal to 0.25 is equal to 0.7224 or 72.24%.

As the power is high, the test is effective in giving accurate result.

Step-by-step solution

Given information

It is claimed that the value of the population proportion is less than 0.4. The sample size (n) is equal to 50, and the level of significance is equal to 0.05.

The true population proportion is given to be equal to 0.25.

Hypotheses

The following hypotheses are set up:

Null Hypothesis: The population proportion is equal to 0.4.

Symbolically,$H_0:p=0.4$

Alternative Hypothesis: The population proportion is less than 0.4.

Symbolically,$H_0:p<0.4$

The test is left-tailed.

Conversion of z-score to sample proportion 

The level of significance is given to be equal to 0.05.

Thus, the z-score for a left-tailed test with $\alpha =0.05$is equal to -1.645.

The value of the z-score is converted to the sample proportion as follows:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ \hat{p}=p+z\sqrt{\frac{pq}{n}} \\ =0.4+\left(-1.645\right)\sqrt{\frac{\left(0.4\right)\left(1-0.4\right)}{50}} \\ =0.2860 \end{gathered}$$

Thus, the value of the sample proportion is equal to 0.2860.

Test Statistic

The test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.286-0.25}{\sqrt{\frac{\left(0.25\right)\left(0.75\right)}{50}}} \\ =0.59 \end{gathered}$$

Therefore, the required value of z-score is equal to 0.59.

Value of βand Power of the Test

The value of $\beta$is the area on the graph to the left of the computed z-score.

Thus,

$$\begin{gathered} \beta =P\left(z<0.59\right) \\ =0.7224 \end{gathered}$$

Thus, $\beta =0.7224$.

a.

Now, the value of the power of the test is to be determined.

$$\begin{gathered} \text{Power}\text{of}\text{the}\text{Test}=1-\beta \\ =1-0.7224 \\ =0.2776 \end{gathered}$$

Thus, the power of the test is equal to 0.7224 when the true proportion is equal to 0.25

b.

The value of $\beta =0.2776$.

c.

The power equal to 0.7224 or 72.24% is high. It means that there is a 72.24% probability of making the correct conclusion of rejecting the null hypothesis when the true population proportion is equal to 0.25.