EDF 6486
Advanced Analysis in Educational Research
Assignment due 1-23-12
Solutions
3. The null hypothesis for this exercise is H0: μ = 100; the alternative hypothesis is
Ha: μ=105. For a sample size greater than 120, the normal distribution can be used as
the sampling distribution. The means for both sampling distributions is defined in the
null and alternative hypotheses; the standard error of the mean for both distributions
is
sX  s
n  24 128  24 11.31  2.12
To determine the power of this test of H0 against Ha, we will assume a one-tailed
test and α = .05. Note that in the illustration, this critical value is 1.645 above the
mean. You can determine this by looking at the table of areas under the standard
normal curve for values of z in the back of your textbook. We need to find a
value that has .05 (5%) of the area under the curve beyond it. A z-score of 1.64
has .0505 of the area above it while a z-score of 1.65 has .0495 of the area beyond
it. Therefore, the value that cuts off exactly .05 of the area of the curve is right in
the middle of these two values. That is, a z-score of 1.645. In reality, you could
have used either 1.64 or 1.65 and gotten the same answer since we are rounding to
two decimal places. Assuming the null hypothesis (μ = 100) is true, the raw score
corresponding to a z-score of 1.645 is equal to the mean plus 1.645 standard
errors of the mean.
X    1.645s X  100  1.6452.12  100  3.49  103.49
2
Now, we assume that the alternative hypothesis (μ = 105) is true and determine
the z-score that corresponds to a raw score of 103.49 in this new distribution (see
the bottom figure).
z
X   103.49  105  1.51


 .71
sX
2.12
2.12
Now, since we have assumed that the alternative hypothesis is true (i.e. the null
hypothesis is false), the power of the statistical test (probably a one-tailed t-test) is
the probability of rejecting this null hypothesis. We can find this by determining
what proportion of the area under the curve is beyond the critical value in this
new distribution. In this case, what proportion of the area under the curve in the
distribution that assumes the null is true is above (beyond) a z-score of -.71? This
is a little tricky because the z-score is negative (i.e. below the mean), but I’m sure
you remember how to do it from your first statistics class. Look at the lower
distribution pictured on the previous page. We know that 50% of the scores are
above the mean. So, now all we need to find out is the proportion of scores that
fall between a z-score of -.71 and the mean. Then we can add the two proportions
together. According to the table of areas under the standard normal curve for
values of z, .2611 (26.11%) of the area under the normal curve falls between a zscore of -.71 and the mean. So, the area of rejection consists of .5000 + .2611 =
.7611. Now we know that the chances of rejecting this false null hypothesis is
.7611 and this, of course, is the definition of power.
The second part of the question asks us to determine what happens to the power
when the sample size is doubled to 256. Of course, this changes the standard
error, which is now
sX  s
n  24 256  24 16  1.50
Now, we can determine the raw score that cuts off the upper 5% of the scores
when the null hypothesis is true (that is when μ = 100) by simply replacing the
value of the standard error of the mean in the first part of the problem with the
current standard error of the means of 1.50. So
X    1.645s X  100  1.6451.50  100  2.47  102.47
Now, let us determine what the z-score of 102.47 would be if the null were false
and the alternative hypothesis (μ = 105) were true.
z
X   102.47  105  2.53


 1.69
sX
1.50
1.50
3
As in the first part of this problem, we now must determine the proportion of
scores that falls above a z-score of –1.69. Because this value is negative, and
therefore below the mean, we know that .5000 of the scores fall above the mean.
Now we simply look into the table of the normal curve to find the proportion of
scores between z-score and the mean. We find that this value is .4545. By adding
these two proportions, we see that .9545 of the scores fall in the area of rejection
and that the power of this test is .9545.
By comparing these two measures of power we see that increasing the sample size
in this case considerably increases the power of the statistical test.
10. a. In order to find the appropriate sample size to give us a power of .80 for
detecting a standardized effect size of .25 using a statistical that that tests a onetailed null hypothesis H0: μ1 – μ2 = 0 at the α = .05 level of significance, we use
the following formula.
2z   z 
2 .84  1.64
2 2.48
26.15 12.30




 205 people
2
.25
.06
.06
.06
2
n
2
d2
2
For a two-tailed test,
2z   z 2 
2 .84  1.96
2 2.80
27.84 15.68




 261.33  262
2
.25
.06
.06
.06
2
n
2
d2
2
b. In order to find the appropriate sample size to give us a power of .80 for
detecting a standardized effect size of .40 using a statistical that that tests a twotailed null hypothesis H0: μ1 – μ2 = 0 at the α = .05 level of significance, we use
the following formula.
2z   z 2 
2
n
2 .84  1.96
2 2.80
27.84 15.68



 98 persons
2
.40
.16
.16
.16
2

d2
2
c. In order to find the appropriate sample size to give us a power of .95 for
detecting a standardized effect size of .25 using a statistical that that tests a
one-tailed null hypothesis H0: μ1 – μ2 = 0 at the α = .01 level of significance,
we use the following formula.
2z   z 
2
n
d
2 1.64  2.33
2 3.97
215.76 31.52



 525.33  526
2
.25
.06
.06
.06
2

2
2
For a two-tailed test:
2z   z 2 
2
n
d2
2 1.64  2.58
2 4.22
217.81 35.62



 593.67  594
.06
.06
.06
.252
2

2
people
4
d. In order to find the appropriate sample size to give us a power of .95 for
detecting a standardized effect size of .40 using a statistical that that tests a
two-tailed null hypothesis H0: μ1 – μ2 = 0 at the α = .01 level of significance,
we use the following formula.
2z   z 2 
2
n
people.
d
2
2 1.64  2.58
2 4.22
217.81 35.62



 222.62  223
.16
.16
.16
.40 2
2

2