Q.25 - The Practice of Statistics for AP

Question

Power A drug manufacturer claims that fewer than 10% of patients who take its new drug for treating Alzheimer's disease will experience nausea. To test this claim, a significance test is carried out of

$\begin{array}{r}{H}_0:p=0.10\\ H_a:p<0.10\end{array}$

You learn that the power of this test at the 5 % significance level against the alternative p=0.08 is 0.64.

(a) Explain in simple language what "power =0.64" means in this setting.

(b) You could get higher power against the same alternative with the same $\alpha$ by changing the number of measurements you make. Should you make more measurements or fewer to increase power? Explain.

(c) If you decide to use $\alpha =0.01$ in place of $\alpha =0.05$, with no other changes in the test, will the power increase or decrease? Justify your answer.

(d) If you shift your interest to the alternative p= 0.07 with no other changes, will the power increase or decrease? Justify your answer.

Step-by-Step Solution

Answer

a. It means rejecting the null hypothesis if it's false

b. make more measurements

c. decrease

d. increase

Step 1: Given Information

Significance test

$\begin{array}{r}{H}_0:p=0.10\\ H_a:p<0.10\end{array}$

Significance level = $5\%$ and power =  $0.64$

Step 2: Explanation Part (a)

Given,

$\begin{array}{r}{H}_0:p=0.10\\ H_a:p<0.10\end{array}$

power = $0.64=64\%$

alternative p = $0.08$

Hence the "power =$0.64$" means rejecting the null hypothesis if it's false.

Step 3: Explanation Part (b)

Given,

$\begin{array}{r}{H}_0:p=0.10\\ H_a:p<0.10\end{array}$

power = $0.64=64\mathrm{\%}$

alternative p = $0.08$

Hence to build the power, then you require more data about the information, which can be gotten by expanding the quantity of estimation. Accordingly, we make more measurements to increase the power.

Step 4: Explanation Part (c)

Given,

If we decide to use$\alpha =0.01$in place of $\alpha =0.05$, with no other changes in the test then the power will decrease as the probability of type I error decreases and the probability of type II error increases.

Now,

$\text{ Power }=1-\beta$(probability of type II error)

Hence power decreases as $\beta$ increases.

Step 5: Explanation Part (d)

If we shift your interest to the alternative p= $0.07$ with no other changes then the power will increase as it is further from the null hypothesis p =$0.10$

Hence the power increases.