Chapter 10, Problem 1P

Consider the Physicians’ Health Study data presented in Example 10.37 (p. 411).

How many participants need to be enrolled in each group to have a 90% chance of detecting a significant difference using a two-sided test with α = .05 if compliance is perfect?

To determine

Find the number of participants need to be enrolled in each group to have a 90% chance of detecting a significant difference using a two-sided test with α = 0.05 and compliance is perfect.

Answer to Problem 1P

The number of participants need to enrol in each group to achieve 90% power is 18,469.

Explanation of Solution

From Example 10.37, it is known that $p_1=0.020,p_2=0.025$.

Where,

    $\begin{array}{c}\overline{p}=\frac{p_1+p_2}{2}\\ =\frac{0.020+0.025}{2}\\ =0.0225\\ \overline{q}=1-\overline{p}\\ =1-0.0225\\ =0.9775\\ \Delta =\left|p_1-p_2\right|=\left|0.020-0.025\right|=0.005\end{array}$   

From the given information, it is known that $\alpha =0.05,\beta =0.9$.

Step-by-step procedure to obtain critical value using Excel:

• In an empty cell of an Excel sheet, type “=NORMSINV(1-(0.05/2))”.
• Click Enter.

The output obtained is as follows:

Thus, $z_{1-\frac{\alpha }{2}}=1.96$.

Similarly, the value of $z$ for 90% confidence level is obtained below:

It can be written às follows:

$\begin{array}{c}{z}_{1-\beta }=z_{0.9}\\ =1.28\end{array}$

Using Equation 10.13, the sample size is computed as follows

$\begin{array}{l}n=\frac{{\left[\sqrt{2\overline{p}\overline{q}}{z}_{1-\alpha /2}+\sqrt{p_1{q}_1+p_2{q}_2}{z}_{1-\beta }\right]}^2}{{\Delta }^2}\\ n=\frac{{\left[\sqrt{2(0.0225)(0.9775)}1.96+\sqrt{0.020(0.980)+0.025(0.975)}1.28\right]}^2}{{(0.05)}^2}\\ n=18468.5\\ n\cong 18,469\end{array}$

Thus, the number of participants need to enrol in each group to achieve 90% power is 18469.