Chapter 6: The One-Sample t Test and Interval Estimation
1. The t Distributions and the Test for the Mean of One Population When σ Is Not Known
Considering that the whole point of inferential statistics is to make inferences about populations,
it shouldn’t be surprising that you often don’t have the standard deviation of the population to
use in the z-score formula. But fret not, mini-statisticians, since there is a mathematical fix to this
dilemma. Instead of using a population standard deviation (σ), you just need to find the standard
deviation of your sample (s). Once you’ve done that, plug your numbers into the following
formula:
Keep in mind, the distributions that this formula yields won’t be normally distributed, yet will
somewhat resemble that shape—still bell-shaped, but it doesn’t precisely look like the Liberty
Bell anymore. (Though keep in mind, as your sample size (N) increases, the curve will look more
and more similar to the normal distribution.) And just to add another notion to blow your mind,
each sample size has its very own curve. But again, you don’t need to channel that calculus from
yesteryear; some kind mathematicians have created a handy little t distribution table that gives
you the critical values you need. As a reminder, an edited portion of the t table is shown below:
Level of Significance for a Two-Tailed Test
df
...
t.05
t.01
...
.
4
2.78 4.60
.
10
2.23 3.17
11
2.20 3.11
12
2.18 3.06
13
2.16 3.01
14
2.15 2.98
15
2.13 2.95
.
40
2.02 2.70
.
120
1.98 2.62
.
∞
1.96 2.58
You should note that the t table has a column on the left-hand side, corresponding to degrees of
freedom (df). What is the value of df for a one-sample t test, you ask? Just df = N–1. Yup, as
simple as that.
So let’s find a value for t and then look up whether it would be out in the critical range (a good
thing if you want to reject the null) for α = .05 and α = .01 (for your convenience, you can always
assume that two-tailed tests are being performed in this study guide, unless otherwise specified).
Suppose we want to find out if a subgroup of students from Terrific Tower, who are always
known for being the Debbie Downers of the dorm, vary significantly in their happiness ratings
from the rest of the dorm’s population. The statistics are as follows:
μ = 4.6 (The happiness rating of all 120 students.) Please note, we are now using this figure as
the population parameter, because we are measuring the Debbie Downers against the whole
dorm, not the whole world. Also, we are assuming someone misplaced the rest of the dorm’s
data, and thus we don’t know the standard deviation of the population anymore.
= 4.1 (The happiness rating for the 12 Debbie Downeresque students.)
s = .6
N = 12 (df = N – 1 = 11)
Now, plug in the numbers:
(4.1 – 4.6)/(.6/(√12) = -2.887
We are certainly significant at the .05 level (t.05 = ±2.201), but we do not attain significance at
the .01 level (t.01 = ±3.106). So we can reject the null at .05 but must retain it at .01. So yes, those
Debbie Downers are an anomaly, but just how dramatically different are they really? And that’s
where further research and statistics can be used to find out more specific information, which
you’ll learn in later chapters!
Now, you may be surprised that those results didn’t attain significance at .01 when they would
have easily done so in a z-score example (if you recall, z only has to beat 2.58 to be significant at
the .01 level). But don’t forget, those sneaky little degrees of freedom can make a BIG
difference. This is another reason (the size of the SEM is the first) that you must find a
considerably larger difference between a sample and the general population when you have just a
few individuals as part of the study to determine that you’ve truly got something significant
going on!
Keep in mind, the important difference between the t-distributions and the z-distribution is that
the t-distribution tails will be fatter, which implies that the critical values will be a bit higher and
harder to beat. Even if you have a large sample, it is best to report it as beating the t (rather than
the z) critical value, simply so no one thinks you’re trying to give yourself an easier value to
surpass.
Try this example:
1. You want to determine if the number of all-nighters architecture majors pull per year
differs significantly from the general college population (μ = 9). You are given the raw
scores for 15 architecture majors: 25, 42, 35, 10, 2, 40, 12, 16, 10, 4, 3, 0, 29, 14, 6.
2. Interval Estimation and Confidence Intervals versus Null Hypothesis Testing
Sometimes proving that the null should be rejected isn’t enough, as it just tells us that one
particular point is not likely to be the parameter of the data we found. So when you want
additional information, it’s time to turn to confidence intervals (CIs). These bad boys will
provide, with varying amounts of confidence (usually 95% or 99%, your choice), a range of
values within which we believe the parameter we are looking for truly exists. In other words, we
can be pretty darn sure that the actual population mean is in whatever range we determine, and
that is based on this formula:
As an example, let’s suppose that you want to know, with 95% confidence, the mean number of
beers that fraternity brothers force the pledges to drink at all American fraternities during rush
week. We’re going to use the average number of beers consumed by 24 randomly selected
pledges from the U.S. as our sample mean, = 39, s = 8. (Luckily, it’s actually the Beermaster’s
job to take note of this number for each pledge, so we have data to use!) Next, we need to know
our df; since the number of pledges sampled was 24, our df = 24 – 1 = 23. Next, we can look for
the t-value for the CI for 95%, which is 2.069 (the .05, two-tailed critical t for 23 df). Now, we
just plug in the numbers: 39 ± 2.069 (8/√24) = 39 ± 3.38. Usually this is written as 35.62 ≤ μ ≤
42.38, which is saying that we can assume, with 95% accuracy, that the population mean for
number of beers that a pledge drinks in a week during rush falls somewhere between about 35
and 42 beers. And, well, my liver is cringing just thinking about that number!
Keep in mind, the 99% CI would be very similar to calculate; however, you would use the tvalue of 2.787, which would make the range a bit larger. If you think about this for a moment (let
those wheels crank for a bit), it should be fairly intuitive that if you want to be MORE confident
that the mean truly exists within your range, you’ll need to expand it somewhat. So, the 99% CI
comes out to: 34.45 ≤ μ ≤ 43.55. As you can see, the likely beer consumption range is now
roughly 34.5 to 43.5, which means it is now about 9 beers wide, instead of just under 7 beers
wide for the 95% CI.
Now let’s try some examples . . .
2. Every summer, Anthropology 101 takes a class trip to Africa to learn firsthand about the
roots of humankind. Mary really wants to go on the trip but is worried about how much it
will cost her to go. She wants to be 95% confident that the trip will cost her no more than
$975, and she is going to base her decision on the amount of money the 28 students from the
class last year spent ( = 905, s = 145, N = 28). Can she be 95% sure? Could she also be
99% sure?
3) Jake is a biochemistry major, but he really wants to take the course Human Sexuality this
semester. He wants to be 95% confident that he has enough time (only 70 free hours to offer)
to study for the course to ensure he won’t fail (and he’ll even learn a thing or two!). He is
using data from the past year the course was taught ( = 60, s = 12, N = 16) to figure out the
maximum and minimum amount of hours other students have put into the course. Can Jake
take the course?
3. The Standard Error of a Proportion
Let’s imagine that your friend is a math major, and she says that there seems to be an unusually
large number of left-handed students among her fellow math majors (they never have enough
left-handed desks, etc.). She searched for estimates of the percentage of lefties in the U.S.
population, and found them to range from 7% to 13%, so she decided that a population
proportion (π) of .1 was reasonable for purposes of comparison with the math majors. If she
surveys all of the 120 math majors at her university, and finds that the proportion of left-handed
students is .15 (i.e., 18 of the 120), can she conclude that the proportion of lefties among math
majors is actually more than the national average of .10? She can answer this question by
computing a simple z score to compare the math majors with the population – a z score for
proportions, that is. Here is what that z score formula looks like:
P 
z
 (1   )
N
Now let’s fill in the values for this example, and compute the z score.
z
.15  .1

.1(1  .1)
120
.05
.05

 1.83
.00075 .0274
You can see from Table A in your text that the area beyond a z score of 1.83 is 50 – 46.64 =
3.36, so the one-tailed p value is 3.36/100 = .0336. That makes her fellow math majors pretty
unusual as a group, and she could reject the null hypothesis that π = .1 for math majors, if she
could justify a one-tailed test. Unfortunately, her two-tailed p value is 2 * .0336 = .0672, which
is not significant at the .05 level with the more conservative two-tailed test. (And like many math
majors, she’s a bit conservative, though we do not have any data to back up that observation.)
Now, here’s an example for you to work out:
4. A friend thinks that there are an unusual number of women with blond hair among the
psychology majors at his college. He conducts a survey and finds that 36 of the 160 female
psych majors at his school are naturally blond. If the percentage of blondes in the female collegeaged population of the U.S. is 15%, is he correct that there is an unusual percentage of blondes in
his department? Can he reject the null hypothesis that the percentage of blondes in his
department is no different from the national average?
Answers to Exercises
1.
= 16.5333; s = 14.182; μ = 9; N = 15; s /√N = 3.662; t = 7.533/3.662 = 2.06 < t.05 (14) =
2.145, not sig.
2. Mary can be confident that it won’t be more than $975 because the 95% CI upper limit is
961.23: t (27) = 2.052; CI = 905 ± 2.052 (145/√28) = 905 ± 56.2298 = 848.77 ≤ μ ≤ 961.23.
However, she cannot be sure at the 99% level, as CI = 905 ± 2.771 (145/√28) = 905 ± 75.932=
829.07 ≤ μ ≤ 980.93.
3. Jake can take the course, because the 95% CI upper limit is 66.39, which is less than 70
hours: t (15) = 2.131; CI = 60 ±2.131 (12/√16) = 53.61 ≤ μ ≥ 66.39. He can even be 99%
confident, too! (51.16 ≤ μ ≤ 68.84)
4. For this example, π = 15/100 = .15; N = 160; and P = 36/160 = .225.
z
.225  .15
.075
.075


 2.68
.15(.85)
.0008 .028
160
The two-tailed p value corresponding to z = 2.68 is .0074, so p < .01. The null hypothesis can be
rejected.