Calculating Power Consider a hypothesis test of the claim that the Ericsson method of gender selection is effective in increasing the likelihood of having a baby girl, so that the claim is p>0.5. Assume that a significance level of $\alpha$= 0.05 is used, and the sample is a simple random sample of size n = 64.

a. Assuming that the true population proportion is 0.65, find the power of the test, which is the probability of rejecting the null hypothesis when it is false. (Hint: With a 0.05 significance level, the critical value is z = 1.645, so any test statistic in the right tail of the accompanying top graph is in the rejection region where the claim is supported. Find the sample proportion in the top graph, and use it to find the power shown in the bottom graph.)

b. Explain why the green-shaded region of the bottom graph represents the power of the test.

It is claimed that the proportion of a baby girl being born is greater than 0.5. The sample size (n) is equal to 64, and the level of significance is equal to 0.05.

The true population proportion is equal to 0.65.

The following hypotheses are set up:

Null hypothesis: The proportion of a baby girl being born is equal to 0.5.

Symbolically, $H_0:p=0.5$.

Alternative hypothesis: The proportion of a baby girl being born is more than 0.5.

Symbolically, $H_0:p>0.5$.

The level of significance is equal to 0.05.

The test is right-tailed.

Thus, the z-score for a right-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.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{\hat{p}\hat{q}}{n}}} \\ \hat{p}=p+z\sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =0.5+\left(1.645\right)\sqrt{\frac{\left(0.5\right)\left(1-0.5\right)}{64}} \\ =0.6028 \end{gathered}$$

Thus, the value of the sample proportion $\left(\hat{p}\right)$is equal to 0.6028.

The value of the test statistic (z-score) corresponding to the given value of the true proportion of baby girls (p) equal to 0.65 is computed below.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{\hat{p}\hat{q}}{n}}} \\ =\frac{0.6028-0.65}{\sqrt{\frac{\left(0.65\right)\left(1-0.65\right)}{64}}} \\ =-0.79 \end{gathered}$$

Therefore, the required value of the z-score is equal to -0.79.

The value of $\beta$ or the value of the type II error is the area to the left of the green shaded region.

In other words, the value of $\beta$ is the area on the graph to the left of the computed z-score.

The required probability is obtained by the standard normal table.

Thus,

.$$\begin{gathered} \beta =P\left(z<-0.79\right) \\ =0.2148 \end{gathered}$$

Thus, $\beta =0.2148$.

a.

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

Thus, the power of the test that is the green shaded area (in the second graph) has the following value:

.$$\begin{gathered} \text{Power}\text{of}\text{the}\text{Test}=1-\beta \\ =1-0.2148 \\ =0.7852 \end{gathered}$$

Thus, the power of the test is equal to 0.7852 when the true proportion of baby girls is equal to 0.65.

b.

As mentioned in the second graph, the area to the left of the green shaded area represents the value of $\beta$.

It is known that

$\text{Power}\text{of}\text{the}\text{Test}=1-\beta$

This implies that the area corresponding to the power of the test is the total area of the graph minus the area that represents $\beta$.

Thus, the green shaded area (area to the right of role="math" localid="1648621451385" $\beta$ ) is the area on the graph that represents the power of the test.