Chapter 12: Inference about One Population
In this chapter, we presented the statistical inference methods used when the problem objective is to
describe a single population. Sections 12.2 and 12.3 addressed the problem of drawing inferences about
a single population when the data are quantitative. In Section 12.2, we introduced the interval estimator
and the test statistic for drawing inferences about a population mean. This topic was covered earl ier in
Sections 10.3 and 11.3. However, i n Chapters 10 and 11 we operat ed under the unrealistic assumption
that the population standard deviation was known. In this chapter, the population standard deviation was
assumed to be unknown. Section 12.3 presented the methods used to make inferences about a population
variance. Finally, in Section 12.4 we addressed t he problem of describing a single population when the
data are qualitative.
At the completion of this chapter, you are expected to know the following:
1. How to apply the concepts and techniques of estimation and hypothesis testing introduced in
Chapters 9 and 10, including:
a) setting up the null and alternative hypotheses
b) calculating the test statistic (by hand or using a computer)
c) setting up the rejection region
d) interpreting statistical results
2. How to recognize when the parameter of interest is a population mean.
3. How to recognize when to use the Student t distribution to estimate and test a population mean
and when to use the standard normal distribution.
4. How to recognize when the parameter of interest is a population variance.
5. How to recognize when the parameter of interest is a population proportion.
12.2 Inference About Population Mean When the Population Standard
Deviation Is Unknown
We presented the statistical techniques used when the parameter to be tested or estimated is the
population mean under the more realistic assumption that the population standard deviation is unknown.
Thus, the only difference between this section and Sections 10.3 and 11.3 is that when the population
standard deviation is known we use t he z-statistic as the basis for the inference, whereas when is
unknown we use the t-statistic.
The formula for t he i nterval est imator of t he popul ation m ean when t he popul ation st andard
deviation is unknown is
x t /2 s
137 The test statistic is
s / n
The number of degrees of freedom of this Student t distribution is n 1.
Question: How do I determine /2 ?
Answer: The degrees of freedom are n 1, where n is the sample size. Turn to Table 4
in Appendix B and locate the degrees of freedom in the left column. Pick one
of t.100, t.050, .025, t.010or t .005depending on the confidence level 1 .
The value of t is at the intersection of the row of the degrees of freedom
and the column of the value of /2.
Question: Which interval estim ator do I use when the degrees of freedom are greater
Answer: If 2 is unknown, the interval estimator of is
x t /2
regardless of the sam ple size. The fact that t /2 is approximately equal to
/2 for d.f. > 200 does not change the interval estimator.
The mean monthly sales of insurance agents in a large company is $72,000. In an at tempt to im-
prove sales, a new training program has been devised. Ten agents are randomly selected to participate in
the program. At its co mpletion, the sales of the agents in the next m onth are recorded as follows (in
63, 87, 95, 75, 83, 78, 69, 79, 103, 98
a) Do these data provide sufficient evidence at the 10% significance level to indicate that the pro-
gram is successful?
b) Estimate with 95% confidence the mean monthly sales for those agents who have taken the new
The problem objective is to describe the population of monthly sales, which is a quantitative vari-
able. Thus, the parameter of interest is . We have no knowledge of the population standard deviation.
As a result we must calculate s (as well as ) from the data. Hence, the basis of the statistical inference
is the t-statistic.
a) To establish that the program is successful, we must show that there is enough evidence to con-
clude that is greater than $72 (thousand). Thus, the alternative hypothesis is
H 1 > 72
and the null hypothesis is
H 0 = 72
The test statistic is
t = x
s / n
which is Student t distributed with n 1 degrees of freedom. The rejection region is
t > t = t = 1.383
, n1 .10, 9
From the data, we compute
x = 83.0
s = 12.85
The value of the test statistic is
t = = 83.0 72 = 2.71
s / n 12.85 / 10
Since 2.71 > 1.383, we reject H 0and concl ude that there is enough evi dence to show t hat
> 72 (thousand) and hence that the program is successful.
b) Once weve determined that the parameter to be estimated is and that the population standard
deviation is unknown, we identify the interval estimator as
x t s / n
The specified confidence level is .95. Thus,
1 = .95