A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of μ = 50 and a standard deviation of σ = 10. The researcher expects a +5-point treatment effect and plans to use a two-tailed hypothesis test with α = .05.   Standard Normal Distribution Mean = 0.0 Standard Deviation = 1.0         -2.0-1.00.01.02.0z.5000.50000.00     Compute the power of the test if the researcher uses a sample of n = 4 individuals. The power for the test is the probability of obtaining a z-score    than   , which corresponds to p =   . Thus, the power of the test is    %.   Compute the power of the test if the researcher uses a sample of n = 25 individuals. The power for the test is the probability of obtaining a z-score    than   , which corresponds to p =   . Thus, the power of the test is

Given Information:

Population mean $\mu =50$

Population standard deviation $\sigma =10$

The researcher expects a +5-point treatment effect.

Significance level $\alpha =0.05$ (two tailed test)

To compute the power of the test if the researcher uses a sample of n = 4 individuals:

Calculate the parameters of null hypothesis $H_0$ is true, sampling distribution:

Mean of sampling distribution: ${\mu }_M=\mu =\mathbf{50}$

Standard deviation of sampling distribution ${\sigma }_M=\frac{\sigma }{\sqrt{n}}=\frac{10}{\sqrt{4}}=\mathbf{5}$

Now, calculate the parameter of $H_0$ is false:

$\mu =50+5=\mathbf{55}$

${\sigma }_M=5$

Calculate the critical z-score: 

Using standard normal table, critical values for a two tailed test at 0.05 significance level are: -1.96 and +1.96

Raw score (M) = $\mu +z\sigma =50+1.96\left(5\right)=59.8$

Power = area to the right of the critical value z under $H_0$ is false distribution. 

Location of critical z in $H_0$ is false distribution = $z_{mean of treatments}-z_{critical}$

$\begin{array}{rcl}z& =& \frac{M-\mu }{{\sigma }_M}\\ & =& \frac{59.8-55}{5}\\ & =& 0.96\end{array}$

Using standard normal table, value corresponding to 0.96 is 0.83147.

$\begin{array}{rcl}P\left(z>0.96\right)& =& 1-P\left(z\le 0.96\right)\\ & =& 1-0.83147\\ & =& 0.16853\end{array}$

Thus, with a sample of 4 people and 5 points treatment effect, 16.853% of the time the hypothesis test will conclude that there is a significant effect.

Power of the hypothesis is 0.1685