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Section 9-3
Inferences About Two Means:
Independent Samples
Key Concept
This section presents methods for using sample data
from two independent samples to test hypotheses
made about two population means or to construct
confidence interval estimates of the difference between
two population means.
Key Concept
In Part 1 we discuss situations in which the standard
deviations of the two populations are unknown and are
not assumed to be equal. In Part 2 we discuss two
other situations: (1) The two population standard
deviations are both known; (2) the two population
standard deviations are unknown but are assumed to
be equal. Because is typically unknown in real
situations, most attention should be given to the
methods described in Part 1.
Part 1: Independent Samples with σ1
and σ2 Unknown and Not Assumed
Equal
Definitions
Two samples are independent if the sample values
selected from one population are not related to or
somehow paired or matched with the sample
values from the other population.
Two samples are dependent if the sample values
are paired. (That is, each pair of sample values
consists of two measurements from the same
subject (such as before/after data), or each pair of
sample values consists of matched pairs (such as
husband/wife data), where the matching is based
on some inherent relationship.)
Notation
1 = population mean
σ1 = population standard deviation
n1 = size of the first sample
x1 = sample mean
s1 = sample standard deviation
***When we have SAMPLE st. dev. (s)
we use Student T Distributions! (t* and t)
Corresponding notations for 2, σ2, s2, x2 , and n2 apply to
population 2.
Hypothesis Test for Two Means:
Independent Samples
Test Statistic for Two Means: Independent Samples
x1  x2    1  2 

t
2
1
2
2
s
s

n1 n2
(where 1 – 2 is often assumed to be 0)
Hypothesis Test - cont
Degrees of freedom: In this book we use this simple and conservative estimate: df = the
smaller of n1 – 1 and n2 – 1. In other words, use whichever sample is smaller, and take one
away.
Critical values: invT(area to the left, df), with degrees of freedom (df) = the smaller of n1 –
1 and n2 – 1.
*left tailed test, α is in the left tail: invT(, df)
*right tailed test, α is in the right tail: invT(1- , df)
𝛼
*two tailed test, α is divided equally between the two tails: invT( 2 , df)
P-values: Use the tcdf feature on your calculator, with degrees of freedom (df) = the smaller
of n1 – 1 and n2 – 1.
*For right-tailed tests: To find this probability in your calculator, type:
tcdf(t test statistic, 99999999, df)
*For left-tailed tests: To find this probability in your calculator, type:
tcdf(–99999999, –t test statistic, df)
***Don’t forget if your test is two-sided, double your P-value.***
Example 1: A headline in USA Today proclaimed that “Men, women are equal talkers.” That
headline referred to a study of the numbers of words that samples of men and women spoke
in a day. Given below are the results from the study. Use a 0.05 significance level to test the
claim that men and women speak the same mean number of words in a day. Does there
appear to be a difference?
a) State the null hypothesis and
the alternative hypothesis.
H0: μ1 = μ2
H1: μ1 ≠ μ2
b) What is/are the critical value(s)?
.05
= 0.025
2
The degrees of freedom is 185 (186 is the smaller sample. n – 1 = 185)
Type into your calculator: invT(0.025, 185)
z* = ±1.973
𝛼
Since this test is two-tailed, find 2 .
c) Calculate the test statistic.
Use the TMEAN program: The calculator
t = –0.676
will ask for these
6 pieces of data
d) What is the P-value?
Type into your calculator: tcdf(–99999, –0.68, 185)
Since this test is two-tailed, don’t forget to DOUBLE the p –value!
= 0.250×2 = 0.500
e) What is the conclusion?
Compare the p-value to . 0.5 > 0.05
Fail to reject H0. There is NOT enough evidence to suggest that men and
women DO NOT speak the same mean number of words in a day.
Make sure your context matches your alternative hypothesis; not the null hypothesis!
Example 2: A researcher wishes to determine whether people with high blood pressure can
reduce their blood pressure, measured in mm/Hg, by following a particular diet. Use a
significance level of 0.01 to test the claim that the treatment group is from a population with
a smaller mean than the control group.
Treatment Group
Control Group
n1 = 101
n2 = 105
x1  120.5
x2  149.3
s1 = 17.4
s2 = 30.2
a) State the null hypothesis and the alternative hypothesis.
H0: μ1 = μ2
H1: μ1 < μ2
b) What is/are the critical value(s)?
Since this test is left-tailed,  is in the left tail
The degrees of freedom is 100 (101 is the smaller sample. n – 1 = 100)
Type into your calculator: invT(0.01, 100) t* = –2.364
c) Calculate the test statistic. The calculator
Use the TMEAN program:
will ask for these
6 pieces of data
Treatment Group
Control Group
n1 = 101
n2 = 105
x1  120.5
s1 = 17.4
x2  149.3
s2 = 30.2
t = –8.426
d) What is the P-value?
Type into your calculator: tcdf = (–99999, –8.426, 100) = 0
e) What is the conclusion?
Compare the p-value to . 0 < 0.01
Reject H0. There IS enough evidence to suggest that the treatment group
is from a population with a smaller mean than the control group.
Example 3: A researcher was interested in comparing the resting pulse rates of people who
exercise regularly and of those who do not exercise regularly. Independent simple random
samples of 16 people who do not exercise regularly and 12 people who exercise regularly
were selected, and the resting pulse rates (in beats per minute) were recorded. The summary
statistics are as follows.
Do not exercise regularly
Exercise regularly
n1 = 16
n2 = 12
x1  73.4 beats/min
x2  68.1 beats/min
s1 = 10.9 beats/min
s2 = 8.2 beats/min
Use a 0.025 significance level to test the claim that the mean resting pulse rate of people who
do not exercise regularly is larger than the mean resting pulse rate of people who exercise
regularly.
a) State the null hypothesis and the alternative hypothesis.
H0: μ1 = μ2
H1: μ1 > μ2
b) What is/are the critical value(s)?
Since this test is right-tailed,  is in the right tail
The degrees of freedom is 11 (12 is the smaller sample. n – 1 = 11)
Type into your calculator: invT(1 – 0.025, 11) t* = 2.201
c) Calculate the test statistic.
Use the TMEAN
The calculator
will ask for these
program: 6 pieces of data
t = 1.468
Do not exercise regularly
Exercise regularly
n1 = 16
n2 = 12
x1  73.4 beats/min
s1 = 10.9 beats/min
s2 = 8.2 beats/min
x2  68.1 beats/min
d) What is the P-value?
Type in your caclulator: tcdf = (1.468, 99999, 11)
= 0.085
e) What is the conclusion?
Compare the p-value to . 0.085 > 0.025
Fail to reject H0. There IS NOT enough evidence to suggest that the
mean resting pulse rate of people who do not exercise regularly is larger
than the mean resting pulse rate of people who exercise regularly.