Math 311, Spring 2007, Lab 6
The Tools
In this section you’ll learn the mechanics of 1-sample and matched pairs t-tests
Inferences With Minitab:
Have Minitab compute two columns of 10,000 rows of data. Store this data in columns C1 and
C2. Choose the first (C1) from a normally distributed population (use N(0,1)) and the second
(C2) from a uniformly distributed population, distributed between 0 and 1. Recall one does this
by selecting Calc>Random Data> …. We will use these data in our tests below.
1-sample t-test: View the data in column C1 as a sample of size 10,000. We may use Minitab
to compute both a confidence interval for the (true) mean of the population and perform a
hypothesis test for the mean of the population at one time!
a. Select Stat>Basic Statistics>1-Sample t…
b. Enter C1 in the Samples in columns box
c. Select Perform hypothesis test
d. Enter the value of the population mean from the Null Hypothesis as the
Hypothesized mean: (this is for the hypothesis test – the Alterative Hypothesis is
entered below). This time, use 0 (this is, in fact, the true mean – recall that the data
came from a N(0,1) distribution).
e. To set the confidence level for a Confidence Interval select Options. 95% is the
default setting.
f. Notice that while you’re in the Options menu, you can also select equal to, greater
than, or less than for your Alternative Hypothesis. Since this is just practice, pick
whichever floats your boat.
g. Finally, select OK (twice) and get something like this:
One-Sample T: C1
Test of mu = 0 vs not = 0
Variable
C1
N
10000
Variable
C1
P
0.970
Mean
0.000375
StDev
1.003442
SE Mean
0.010034
95% CI
(-0.019294, 0.020044)
T
0.04
h. Note: the hypotheses tested (mu = 0 vs not = 0) are listed at the top of the output
(I’ve highlight these for emphasis.)
i. Note also: T = one sample t-statistic and P = p-value. The p-value is high because
our null hypothesis was correct – mu is 0.
Matched pair t-test: We can also do a matched pair t-test by viewing the data in columns C1
and C2 as two data points collected from 10,000 individuals and by following the directions
below:
a. Select Stat>Basic Statistics>Paired t…
b. Select Samples in Columns
c. Enter C1 in First: and C2 in Second:
d. Select Options to set the following:
i.
the confidence level (α)
ii. the value of the mean in the null hypothesis
iii. and the alternative hypothesis (µ ≠, >, or <).
e. Finally, select OK (twice) and get something like this:
Paired T-Test and CI: C1, C2
Paired T for C1 - C2
C1
C2
Difference
N
10000
10000
10000
Mean
0.000375
0.501160
-0.500785
StDev
SE Mean
1.003442 0.010034
0.290175 0.002902
1.044458 0.010445
95% CI for mean difference: (-0.521258, -0.480311)
T-Test of mean difference = 0 (vs not = 0): T-Value = -47.95
P-Value = 0.000
f. Notice that the top shaded region (I added the shading) tells us what the mean
difference was (this is x1  x2 ) – it makes sense that this number should be near -0.5
as the normally distributed data in C1 has a mean around 0 and the uniformly
distributed data in C2 has a mean around .5
g. Meanwhile the lower shaded region tells the reader both the null hypothesis
(difference = 0) and the alternative hypothesis (not = 0).
The Questions
In this section you’ll apply the techniques learned above
!!!BE CERTAIN
TO READ EACH SITUATION CAREFULLY – THEY CONTAIN USEFUL CLUES!!!
Twins (or Nature vs. Nurture):
How much of our personality (or lack of personality), our likes, our individuality is
predetermined by our genes? The classical method of studying this phenomenon is the study of
identical twins separated early in life and reared apart. Identical twins, because they share an
identical genotype, make ideal subjects for investigation of the degree to which various
environmental conditions may instigate change. Again: with identical twins we have two
expressions of a single genotype.
Over the past twenty years, several studies of identical twins have been conducted. The most
publicized study was begun in 1979 by University of Minnesota psychologist Thomas Bouchard
and continues today. Bouchard and his colleagues at the Minnesota Center for Twin and
Adoption Research have published over 129 scientific papers on the subject. In this lab, we
consider a similar study carried out by psychologist Susan Farber and published in her book,
Identical Twins Reared Apart (Basic Books, 1981).
Farber chronicles and analyzes data for 95 pairs of identical twins reared apart. Much of her
discussion focuses on a comparison of IQ scores. The question of concern is, “Are there
significant differences between the IQ scores of identical twins, where one member of the pair is
reared by the natural parents and the other member is not?” IQ scores for 32 of Farber’s 95
twins are stored in Twins.mtw.
(Get the file now from the course webpage
http://www.cwu.edu/~englundt/). One member (A) from each of the sets was reared by a natural
parent, whereas the other member (B) was reared by a relative or some other person.
1. Which test would be most appropriate: a one-sample t-test or a matched pair test? Why?
2. Construct a Null Hypothesis and an Alternative Hypothesis. Be certain to define your
parameter.
3. Perform the appropriate test and report the p-value you found (you can do this by just
cutting and pasting the Minitab analysis into Word)
4. State your conclusions.
Note: we are now well past the halfway point of the course. Thus, I expect a complete
and correctly worded explanation and summary of your results. If it helps, imagine that
you are Dr. Farber and write accordingly.
Football (or Winning for Dummies):
Data for all NFL games played over 3 seasons is contained in the file football.mtw. Get this file
now. An explanation of the column headings follows:
 Home/Away: Favored team is at home (1) or away (0).
 Favorite Points: Points scored during the game by the favored team.
 Underdog Points: Points scored during the game by the underdog team.
 Pointspread: Oddsmaker’s points (determined before the game) to handicap the favored
team.
 Favorite Name: Code for favored team’s name.
 Underdog Name: Code for the underdog team’s name
 Year: 89, 90, or 91 season (determined by beginning of season).
 Week: week of the season the game was played.
 Special: Monday night (M), Sat (S), Thur (H), Sunday night (N), overtime game (ot).
Let’s investigate these data:
1. Can the oddsmakers pick the right team? Consider
Points Scored by the Favorite – Points Scored by the Underdog.
If the oddsmakers were just randomly guessing, then it’s reasonable to expect that the mean
of this difference would be zero. Is there evidence significant at the α = 0.05 level that this is
not so? Perform the appropriate test; report the test used and corresponding t-score & Pvalue (as above, you can do this by cutting and pasting from Minitab into Word). Then, give
your conclusion.
2. Bonus Question (i.e. extra credit): We’ll now investigate how often is the “spread”
covered? That is, how often are the points scored by the favored team minus the points
scored by the underdog team at least as great as the spread?
Does the point differential (i.e. Points Scored by the Favorite – Points Scored by the
Underdog) usually exceed the spread? Perform the appropriate test; report the test used
and corresponding t-score & P-value. List your conclusion. Note: you will need to
manipulate the data using either the calculator function or by exporting the data to Excel
and manipulating it there.