Test Statistics. In Exercises 13–16, refer to the exercise identified and find the value of the test statistic. (Refer to Table 8-2 on page 362 to select the correct expression for evaluating the test statistic.)

Exercise 5 “Online Data”

Out of 565 randomly selected adults, 59% would erase all of their personal information online if they could.

It is claimed that most adults would erase all of their personal information online if they could.

Corresponding to the given claim, the following hypotheses are set up:

Null hypothesis: The proportion of adults who would erase all of their personal information online if they could is equal to 0.5.

$H_0:p=0.5$

Alternative hypothesis: The proportion of adults who would erase all of their personal information online if they could is greater than 0.5.

$H_1:p>0.5$

Since the claim involves testing the equality of the sample proportion with a hypothesized value, the test statistic used will be the z-score.

The value of the sample proportion is computed below:

$$\begin{gathered} \hat{p}=59\% \\ =\frac{59}{100} \\ =0.59 \end{gathered}$$

The given value of the proportion ofadults who would erase all of their personal information online if they could is supposed to be equal to 0.5.

Thus, p=0.5.

$$\begin{gathered} q=1-p \\ =1-0.5 \\ =0.5 \end{gathered}$$

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.59-0.50}{\sqrt{\frac{0.5\left(1-0.5\right)}{565}}} \\ =4.2785 \\ \approx 4.28 \end{gathered}$$

Thus, the test statistic is equal to 4.28.