A researcher predicts that listening to music while solving math problems will make a
particular brain area more active. To test this, a research participant has her brain scanned
while listening to music and solving math problems, and the brain area of interest has a
percentage signal change of 58. From many previous studies with this same math problems
procedure (but not listening to music), it is known that the signal change in this brain area
is normally distributed with a mean of 35 and a standard deviation of 10. (a) Using the .01
level, what should the researcher conclude? Solve this problem explicitly using all five steps
of hypothesis testing, and illustrate your answer with a sketch showing the comparison
distribution, the cutoff (or cutoffs), and the score of the sample on this distribution. (b)
Then explain your answer to someone who has never had a course in statistics (but who is
familiar with mean, standard deviation, and Z scores).
(a)
Assumptions:
The population we are studying is all people who do math problems. We are interested in
testing  = the mean signal change of all people who do math problems while listening to
music. It is given that the population we are sampling from is normal, and the population
standard deviation  = 10. Therefore, we can use the z test value for the population mean  .
1. The researcher’s claim is that the mean signal change of people who do math while
listening to music is greater than 35. That is,   35 . Because there is no equality in the
claim, it must go in the alternative hypothesis. The correct hypothesis is,
H 0 :   35
H1 :   35(claim)
This is a right-tailed test.
2. The test value is,
x   58  35
z

 2.3 .
 / n 10 / 1
y
3. The cutoff value for a right-tailed test with   0.01 is z  2.3263. (The P-value is
0.0107.)
Test Value
Cutoff
Region
= 2.3
x
-2.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
2.5
Cutoff Value
= 2.3263
4. Because the test value is just below the cutoff region, (to the left of the critical value in a
right-tailed test,) we Do Not Reject H 0 . (The P-value is greater than 0.01 which would
also lead us to Do Not Reject H 0 .)
5. Because we Do Not Reject H 0 , we are not finding evidence in support of the claim.
Therefore, the researcher does not appear to be correct. Listening to music while doing
math does not appear to make this particular brain area more active.
(b)
The mean signal change for this brain area is known to be 35 with a standard deviation of
10. Based on a sample of 1 (the researcher,) the study showed an average signal change of 58.
The value of 58 has a z-score of about 2.3, that is, 2.3 standard deviations above the mean.
Although this is “far” above the mean, our chosen level of significance 0.01 yields a cutoff value
of about 2.33 and any test value above this cutoff we would consider unusual. Therefore, even
though the value of 58 is more than the mean signal change of 35, it is not significant enough of
a difference to conclude that listening to music while doing math makes this particular brain area
more active. A larger sample size may be necessary to draw any conclusions.