and plans to administers the test to a sample of n = 25 of today's high school students. Ten ago, the scores on this IQ test produced a standard- ized distribution with a mean of u standard deviation o = years 100 and a 15. If there actually has been a 7-point increase in the average IQ during the past 10 years, then find the power of the hypothesis test for each of the following. a. The researcher uses a two-tailed hypothesis test with a = .05 to determine if the data indicate a significant change in IQ over the past 10 years. b. The researcher uses a one-tailed hypothesis with a = .05 to determine if the data indicate a significant increase in IQ over the past 10 yeals test

Given
a)
sample size $n=25$
Ten years ago
The mean IQ score $\mu =100$
The standard deviation $\sigma =15$
level of significance $\alpha =0.05$

Null hypothesis :$H_0: \mu =100$
Alternate hypothesis : $H_a:\mu \ne 100$

Type II error occurs when a false null hypothesis is failed to reject.
The power of a hypothesis test is not committing type II error $Power=1-\beta$

The Z critical value for two tailed test with 5% level of significance is $Z=\pm 1.96$

As it is given that IQ score increasing by 7 points in a decade, let us consider right side critical value 1.96.
We reject the null hypothesis if test statistic $Z>1.96$
$$\begin{gathered} Z>1.96\to \frac{X-\mu }{{\displaystyle \raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}}>1.96 \\ X>1.96*\frac{15}{\sqrt{25}}+100 \\ X>105.88 \end{gathered}$$

So we reject the null hypothesis if IQ score is greater than 105.88

The power of test is probability of rejecting null hypothesis given that alternate hypothesis is correct.
If alternate hypothesis is correct then the IQ score increase by 7 points, so mean IQ score is 107.

Power of test is calculated as shown below
$$\begin{gathered} P(X>105.88|\mu =107) \\ =P\left(\frac{X-\mu }{{\displaystyle \raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}}>\frac{105.88-\mu }{{\displaystyle \raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$\sqrt{n}$}\right.}}|\mu =107\right) \\ =P\left(Z>\frac{105.88-107}{{\displaystyle \raisebox{1ex}{$15$}\!\left/ \!\raisebox{-1ex}{$\sqrt{25}$}\right.}}\right) \\ =P\left(Z>-0.37\right) \end{gathered}$$

With respect to Z table we get the following
 $$\begin{gathered} P(Z<-0.37)=0.3557 \\ P(Z>-0.37)=1-P(Z<-0.37)=1-0.3557=0.6443 \end{gathered}$$

Thus the power of the test is 64.43%