Download Inference for the Mean of a Population.

6.1 Inference for the Mean of a Population
Note. Some assumptions for inference about a mean are:
• Our data are a simple random sample (SRS) of size n from the
population.
• Observations from the population have a normal distribution with
mean µ and standard deviation σ. Both µ and σ are unknown
parameters.
Definition. When the standard deviation of a statistic is estimated
from the data, the result is called the standard error of the statistic.
√
The standard error of the sample mean x is s/ n.
The t Distribution
Definition. Draw an SRS of size n from a population that has the
normal distribution with mean µ and standard deviation σ. The onesample t statistic
x−µ
√
s/ n
has the t distribution with n − 1 degrees of freedom.
t=
Note. There is a different t distribution for each sample size. We
specify a particular t distribution by giving its degrees of freedom. The
degrees of freedom for the one-sample t statistic come from the sample
1
standard deviation s in the denominator of t. We will write the t
distribution with k degrees of freedom as t(k) for short.
Note. Figure 6.1 (and TM-97) compares the density curves of the standard normal distribution and the t distributions with 2 and 9 degrees
of freedom. The figure illustrates these facts about the t distribution:
• The density curves of the t distributions are similar in shape to
the standard normal curve. They are symmetric about zero and
are bell-shaped.
• The spread of the t distributions is a bit greater than that of the
standard normal distribution. The t distributions in Figure 6.1
(TM-101) have more probability in the tails and less in the center
than does the standard normal. This is true because substituting
the estimate s for the fixed parameter σ introduces more variation
into the statistic.
• As the degrees of freedom k increase, the t(k) density curve approaches the N (0, 1) curve ever more closely. This happens because s estimates σ more accurately as the sample size increases.
So using s in place of σ causes little extra variation when the
sample is large.
2
The t Confidence Intervals and Tests
Note. The one-sample t procedure is as follows: Draw an SRS of size
n from a population having unkown mean µ. A level of confidence
interval for µ is
s
x ± t∗ √
n
where t∗ is the upper (1−C)/2 critical value for the t(n−1) distribution.
This interval is exact when the population distribution is normal and is
approximately correct for large n in other cases. To test the hypothesis
H0 : µ = µ0 based on an SRS of size n, compute the one-sample t
statistic
t=
x − µ0
√ .
s/ n
In terms of a variable T having the t(n − 1) distribution, the P −value
for a test of H0 against
Ha : µ > µ0 is P (T ≥ t)
Ha : µ < µ0 is P (T ≤ t)
Ha : µ = µ0 is P (T ≥ |t|).
These P −values are exact if the population distribution is normal and
are approximately correct for large n in other cases.
Example 6.1. To study the metabolism of insects, researchers fed
cockroaches measured amounts of sugar solution. After 2, 5, and 10
hours, they dissected some of the cockroaches and measured the amount
of sugar in various tissues. Five roaches fed the sugar D-glucose and
3
dissected after 10 hours had the following amounts (in micrograms) of
D-glucose in their hindguts:
55.95 68.24 52.73 21.50 23.78.
The researchers gave a 95% confidence interval for the mean amount of
D-glucose in cockroach hindguts under these conditions. First calculate
that x = 44.44 and s = 20.741. The degrees of freedom are n − 1 = 4.
From Table C (and TM-142) we find that for 95% confidence t∗ = 2.776.
The confidence interval is
s
20.741
x ± t∗ √ = 44.44 ± 2.776 √
n
5
= 44.44 ± 25.75 = (18.69, 70, 19).
Comparing this estimate with those for other body tissues and diferent
times before dissection led to new insight into cockroach metabolism
and to new ways of eliminating roaches from homes and restaurants.
The large margin of error is due to the small sample size and the rather
large variation among the cockroaches, reflected in the large value of s.
Matched Pairs t Procedures
Note. One common design to compare two treatments makes use of
one-sample procedures. In a matched pairs design, subjects are matched
in pairs and each treatment is given to one subject in each pair.
Note. To compare the responses to the two treatments in a matched
pairs design, apply the one-sample t procedures to the observed differences.
4
Example 6.3. The National Endowment for the Humanities sponsors summer institutes to improve the skills of high school language
teachers. One institute hosted 20 French teachers for four weeks. At
the beginning of the period, the teachers took the Modern Language
Association’s listening test of understanding of spoken French. After
four weeks of immersion in French in and out of class, they took the
listening test again. (The actual spoken French in the two tests was
different, so that simply taking the first test should not improve the
score on the second test.) Table 6.1 (and TM-101) gives the pretest
and posttest scores. The maximum possible score on the test is 36. To
analyze these data, subtract the pretest score from the posttest score
to obtain the improvement for each teacher. These 20 differences form
a single sample. They appear in the “Gain” column in Table 6.1 (TM101). The first teacher, for example, improved from 32 to 34, so the
gain is 34 − 32 = 2.
Step 1: Hypothesis. To assess whether the institute significantly
improved the teacher’s comprehension of spoken French, we test
H0 : µ = 0
Ha : µ > 0.
Here µ is the mean improvement that would be achieved if the entire
population of French teachers attended a summer institute. The null
hypothesis says that no improvement occurs, and Ha says that posttest
scores are higher on the average.
Step 2: Test Statistic. The 20 differences have x = 2.5 and s =
5
2.893. The one-sample t statistic is therefore
t=
x−0
2.5 − 0
√ = 3.86.
√ =
s/ n 2.893/ 20
Step 3: P −Value. Find the P −value from the t(19) distribution.
Table C (TM-142) shows that 3.86 lies between the upper 0.001 and
0.0005 critical values of the t(10) distribution. The P −value therefore
lies between these values. A computer statistical package gives the
value P = .00053. The improvement in listening scores is very likely
to be due to chance alone. We have strong evidence that the institute
was effective in raising scores. In scholarly publications, the details
of routine statistical procedures are usually omitted. This test would
be reported in the form “The improvement in scores was significant
(t = 3.86, df = 19, P = .00053).” A 90% confidence interval for the
mean improvement in the entire population requires the critical value
t∗ = 1.729 from Table C (TM-142). The confidence interval is
s
2.8393
x ± t∗ √ = 2.5 ± 1.729 √
n
20
= 2.5 ± 1.12 = (1.38, 3.62)
The estimated average improvement is 2.5 points, with margin of error
1.12 for 90% confidence. Though statistically significant, the effect of
attending the institute was rather small.
Robustness of t Procedures
Definition. A confidence interval or significance test is called robust if
6
the confidence level or P −value does not change very much when the
assumptions of the procedure are violated.
Note. Use the t procedures when:
• Except in the case of small samples, the assumption that the data
are an SRS from the population of interest is more important than
the assumption that the population distribution is normal.
• Sample size less than 15. Use t procedures if the data are close to
normal. If the data are clearly nonnormal or if outliers are present,
do not use t.
• Sample size at least 15. The t procedures can be used except in
the presence of outliers or strong skewness.
• Large Samples. The t procedures can be used even for clearly
skewed distributions when the sample is large, roughly n ≥ 40.
Example 6.4. Consider several of the data sets we graphed in Chapter
1. Figure 6.6 (and TM-103) shows the histograms.
• Figure 6.6(a) is a histogram of the percent of each state’s residents
who are over 65 years of age. We have data on the entire population
of 50 states, so formal inference makes no sense.
• Figure 6.6(b) shows the time of the first lightning strike each day
in a mountain region in Colorado. The data contain more than 70
observations that have a symmetric distribution. You can use the
t procedures to draw conclusions about the mean time of a day’s
7
first lightning strike with high confidence.
• Figure 6.6(c) shows that the distribution of word lengths in Shakespeare’s plays is skewed to the right. We aren’t told how large the
sample is. You can use the t procedures for a distribution like this
if the sample size is roughly 40 or larger.
8