CHAPTER 7, Sections 7.1 & 7.2
Revised Jan 27, 2011
Inference for the Mean of a Population when sigma is unknown, Sec 7.1
Previously we made the assumption that we knew the population standard
deviation, σ. We then developed a confidence interval and used tests of
significance to evaluate evidence for or against a hypothesis, all with a known σ.
In many situations, σ is unknown. In this section, we will continue doing inference
for the population mean, but we will use the sample standard deviation , s, as a
substitute for the unknown population standard deviation. The procedures are
called t procedures.
Confidence Interval for a Mean
First, Assumptions for Inference about the population mean:
 Our data are a simple random sample (SRS) of size n taken from a normally
distributed population with mean µ and standard deviation σ, both of which
are unknown.
 Unless a small sample is used, the assumption that the data comes from a SRS
is more important than the assumption that the population distribution is
normal.
Because we do not know the population sigma, σ, we make two changes in our
procedure:
1. The sample standard deviation , s, is used in place of the unknown σ to
estimate the standard deviation of the sample mean. The result is called the
standard error of the mean, SEM.
SEM = SEx 
s
n
Where s is the sample standard deviation and n is the sample size.
2. We calculate a different test statistic, t instead of z, and our P Value comes
from the t distribution instead of the Normal distribution.
The t-distributions:
 The t-distribution is used when we do not know the population sigma, σ. The
t-distributions have density curves similar in shape to the standard normal
curve, but with more spread.
Lecture 9, Section 7.1 & 7.2
Page 1
 The t-distributions have more probability in the tails and less in the center
when compared with the standard normal distribution. This is because
substituting the estimate s for the unknown value of σ introduces more
uncertainty.
 For sample sizes less than 30, s tends to underestimate σ. As the sample size
increases, the t-density curve approaches the N(0,1) distribution. This is
because s estimates σ with less bias as the sample size increases.
The t Distributions
Suppose that an SRS of size n is drawn from a N ( , ) population, but the values
of µ and σ are unknown. Then the one-sample t statistic
t
x
s/ n
has the t distribution with n-1 degrees of freedom. There is a separate distribution
for every sample size. For any given sample size the degrees of freedom = n-1.
Each line on the t table is for one specific degree of freedom.
For any line on the t table there are columns which show the value of t which
coincides with certain specific probabilities in the right tail of the t distribution.
The One-Sample t Confidence Interval
Suppose that a SRS of size n is drawn from a normal population having unknown
mean µ and σ. A level C confidence interval for µ is
xt* s
n
where t* is the value for the t(n-1) density curve with area C between –t* and t*.
Xbar is the point estimate, and t* (s)/√(n) is the margin of error.
This interval is exact when the population distribution is normal and is
approximately correct for large n in other cases.
Examples :
1. Suppose X, Bob’s golf scores, are approximately normally distributed with
unknown mean and standard deviation. A SRS of n = 16 scores is selected and a
sample mean of x = 77 and a sample standard deviation, s, = 3 is calculated.
Calculate a 90% confidence interval for  .
Lecture 9, Section 7.1 & 7.2
Page 2
2. (Example 7.1 in Textbook) Corn soy blend, CSB is highly nutritious, low-cost
fortified food that can be incorporated into different food preparations
worldwide. As part of a study to evaluate appropriate vitamin C levels in this
commodity, measurements were taken on samples of CSB produced in a factory.
The following data are the amounts of vitamin C, measured in milligrams per
100 grams of blend, for a random sample of size 8 from a production run.
Compute a 95% confidence interval for  where  is the population mean
vitamin C content of the CSB.
26 31 23 22 11 22 14 31
By hand:
x = 22.5
Lecture 9, Section 7.1 & 7.2
Page 3
s = 7.191
n = 8 df = 7
Using SPSS:
analyze > descriptive statistics > explore
Move “vitaminC” to “dependent list”.
Click “statistics” and select “descriptives” and change/keep a
confidence interval.
Click “continue” followed by “OK”.
95%
Descriptives
Vitamin C
Statistic
22.50
Mean
95% Confidence
Interval for Mean
Lower Bound
Upper Bound
5% Trimmed Mean
Median
Variance
Std. Deviation
Std. Error
2.542
16.49
28.51
22.67
22.50
51.714
7.191
Minimum
11
Maximum
31
Range
20
Interquartile Range
14
Skewness
-.443
.752
Kurtosis
-.631
1.481
The One-Sample t test:
1. State the Null and Alternative hypothesis.
2. Find the test statistic:
Suppose that a SRS of size n is drawn from a normal population having
unknown mean µand unknown σ. To test the hypothesis H 0 :   0 based
on a SRS of size n, compute the one-sample t statistic
t  x  0
s/ n
3. Calculate the p-value.
In terms of a random variable T having the t(n-1) distribution, the Pvalue for a test of H 0 :   0 against
Ha :   0 is P(T  t ) , one side right
Ha :   0 is P(T  t ) , one side left
Ha :   0 is 2 P(T | t |) , two side
These P-values are exact if the population distribution is normal and
are approximately correct for large n in other cases.
Lecture 9, Section 7.1 & 7.2
Page 4
4. State the conclusions in terms of the problem. Use the given α or if the
value is not specified use α = 0.05 as the default. Then compare the P-value
to the α level.
If P-value  α, then reject H0 . We have sufficient evidence to reject
Ho.
If P-value > α, then we fail to reject H0 . We lack sufficient evidence
to reject H0 .
Either way, we should use the words of the story in the conclusion.
Examples:
1. Experiments on learning in animals sometimes measure how long it takes mice
to find their way through a maze. Suppose the population mean time is 18
seconds for one particular maze. A researcher thinks that loud noise will
decrease the time it takes a mouse to complete the maze. She measures how long
each of 30 mice take to complete the maze with loud noise stimulus. She gets a
sample mean, Xbar = 16 seconds and a sample standard deviation, s = 3
seconds. Do a one sided hypothesis test to test the researchers assertions with α
= 0.1.
2. (Example 7.2 in Textbook) Suppose that we know that sufficient vitamin C was
added to the CSB mixture to produce a mean vitamin C content in the final
product of 40 mg/100 g. It is suspected that some of the vitamin C is lost in the
production process. To test this hypothesis we can conduct a one-sided test to
determine if there is sufficient evidence to conclude that vitamin C is lost. A
sample of 8 batches of CSB was tested for vitamin C. The sample mean = 22.50
and the sample standard deviation = 7.191. Use α = 0.05 level.
By hand: Ho: µ = 40
Lecture 9, Section 7.1 & 7.2
Page 5
Ha: µ < 40
one side left test
Using SPSS:
analyze > compare means > One sample T test
Move “vitaminc” into the “test variable box” and type in 40 for the
test value.
To change the confidence interval, Click “options” and
change confidence interval from 95% to whatever. I did
not do this as I will keep the 95% default.
Click “continue”. Lastly click “OK”.
One-Sample Statistics
N
Vitamin C
Mean
8
Std. Deviation
22.50
Std. Error
Mean
7.191
2.542
One-Sample Test
Test Value = 40
95% Confidence Interval
of the Difference
vitamin C
t
-6.883
df
7
Sig. (2-tailed)
.000
Mean
Difference
-17.500
Lower
-23.51
Upper
-11.49
Matched Pairs Design:
A common design to compare two treatments is the matched pairs design. One type
of matched pair design has 2 subjects who are similar in important aspects matched
in pairs and each treatment is given to one of the subjects in each pair.
A 2nd common design does not use matched subjects. Instead each subject is given
2 treatments in random order. Ex: Each subject does a left hand test and a right
hand test.
Lecture 9, Section 7.1 & 7.2
Page 6
Another common design uses before-treatment and after-treatment observations on
each of the subjects in the experiment. Each subject thereby provides data on the
difference, or improvement, or reduction associated with the treatment.
Assumptions for the matched pair t test:
1. The data values are paired and we analyze the line-by-line differences.
2. The line-by-line differences are independent of each other.
3. The line-by-line differences are normally distributed with unknown
population mean and unknown population standard deviation.
Paired t Procedures:
To compare the responses to the two treatments in a matched pairs design,
determine the difference between the two treatments for each subject and analyze
the observed differences.
Example: (Problem 7.31 is done by hand and using SPSS):
The researchers studying vitamin C in CSB in example 7.1 were also interested in a
similar commodity called wheat soy blend (WSB). Both these commodities are
mixed with other ingredients and cooked. Loss of vitamin C as a result of cooking
was a concern of the researchers. One preparation used in Haiti called gruel can be
made from WSB, salt, sugar, milk, banana, and other optional items to improve the
taste. Five samples of gruel prepared in Haitian households were obtained. The
vitamin C content of these 5 samples was measured before and after cooking.
Set up appropriate hypotheses and carry out a significance test for these data.
Hypotheses: Ho: µ of differences (before – after) =0; µ = pop mean of differences
Ha: µ of differences (before – after) >0
Here are the data:
Sample
1
Before
73
After
70
Difference 3
2
79
77
2
3
86
79
7
4
88
86
2
5
78
67
11
Xbar
80.8
75.8
5.0
s
6.140
7.530
3.937
BY HAND:
Xbar of differences =
Test statistic, t = 2.840
Lecture 9, Section 7.1 & 7.2
Page 7
5.0
Std Dev of differences =s = 3.937
P Value is between .02 and .025.
Exact P Value = .0234 ( from computer).
Conclusion: if α = .05: Reject Ho. There is sufficient evidence to
conclude that vitamin C is lower after cooking.
Using SPSS:
> Analyze > Compare Means > Paired – Sample T test.
Move “before and after” to “paired variable box” (whichever variable is listed first
will come first in the subtraction)
Click “OK”
Paired Samples Statistics
Mean
Pair 1
N
Std. Deviation
Std. Error Mean
Before
80.8000
5
6.14003
2.74591
After
75.8000
5
7.52994
3.36749
Paired Samples Test
Paired Differences
95% Confidence Interval of the
Mean
Pair
Before -
1
After
5.00000
Std.
Std. Error
Deviation
Mean
3.93700
1.76068
Difference
Lower
.11156
Sig. (2Upper
t
9.88844 2.840
df
tailed)
4
.047
(P-Values for t tests are given by SPSS as Sig (2 tailed) and must be
divided by 2 for one-side tests). One side P Value = .0235.
A confidence interval or statistical test is called robust if the confidence level or Pvalue does not change very much when the assumptions of the procedure are
violated. The t procedures are robust against non-normality of the population when
there are no outliers, especially when the distribution is roughly symmetric and
unimodal.
Robustness and use of the One-Sample t and Matched Pair t procedures:
 Unless a small sample is used, the assumption that the data comes from a SRS
is more important than the assumption that the population distribution is
normal.
Lecture 9, Section 7.1 & 7.2
Page 8
 n<15: Use t procedures only if the data are close to normal with no outliers.
 n is 15 - 39: The t procedure can be used except in the presence of outliers or
strong skewness.
 n≥40: The t procedure can be used even for clearly skewed distributions.
Comparing Two Means:
Two-Sample Problems: Sec 7.2
A situation in which two populations or two treatments based on separate samples
are compared.
A two-sample problem can arise:
 from a randomized comparative experiment which randomly divides the units
into two groups and imposes a different treatment on each group.
 From a comparison of two random samples selected separately from two
different populations.
Note: Do not confuse two-sample designs with matched pair designs! In the twosample t problems, each group is composed of separate subjects, ie, no subject is in
both groups, and each subject furnishes only one piece of data, ie, no subject is
tested twice.
Assumptions for Comparing Two Means:
 Two independent simple random samples, from two distinct populations are
compared. The same variable is measured on both samples. The sample
observations are independent, ie, neither sample has an influence on the other.
 Both populations are normally distributed.
 The means 1 and 2 and standard deviations 1 and  2 of both populations
are unknown.
Typically we want to compare two population means by giving a confidence interval
for their difference, 1  2 , or by testing the hypothesis of no difference,
H0 : 1  2  0.
The Two-Sample t Confidence Interval:
Suppose that an SRS of size n1 is drawn from a normal population with unknown
mean  and that an independent SRS of size n2 is drawn from another normal
1
Lecture 9, Section 7.1 & 7.2
Page 9
population with unknown mean  2 . The confidence interval for the difference
between population means, 1  2 is given by
s12 s22
( x1  x2 )  t *

n1 n2
The value of t* is determined for the confidence level, C, desired. The confidence
interval formula is still composed of the same two components, (1) a point estimate
for the difference between population means, and (2) a margin of error which
expresses the uncertainty involved.
Note that the two sample sizes do not have to be equal.
Here, t* is the value for the t(k) density curve with area C between –t* and t*. The
value of the degrees of freedom, k, is approximated by software, or if we do the
calculations by hand, we determine the df using the smaller of the two sample
sizes.
Two-Sample t Procedure For Tests Of Significance:
1. Write the hypotheses in terms of the difference between means.
H0 : u1  2  0
H a : 1  2  0
H a : 1  2  0
H a : 1  2  0
one side right or
one side left
two side
or
2. Calculate the test statistic. A SRS of size n is drawn from a normal
1
1 and a SRS of size n2 is drawn from another
normal population with unknown mean 2 . Sigma 1 and sigma 2 are also
population with unknown mean
unknown.
To test the hypothesis H 0 : u1  2  0 the two-sample t statistic is:
t
x1  x 2
s12 s22

n1 n2
and use P-values or critical values for the t(k) distribution, where the degrees of
freedom k is either approximated by software, or is based on the df for the
smaller of the the two sample sizes. Note: SPSS calculates a value for
degrees of freedom using a complicated formula which takes the sample sizes
Lecture 9, Section 7.1 & 7.2
Page 10
and sample standard deviations into account. This value does not have to be an
integer. The df value can be as small as the df for the smaller sample, or it can
be as large as the sum of the two dfs. The closer the sample sizes are, and the
closer the two sample standard deviations are, the higher the calculated df value
will be.
3. Calculate the P-value.
Note: Unless we use software, we can only get a range for the P-value. We use
the following formulas:
H a : 1  2  0 use P(T  t ) one side right test
H a : 1  2  0 use P(T  t ) one side left test
H a : 1  2  0 use 2 P(T | t |) two side test
Note: If doing the calculations by hand you should use the df for the smaller of
the two sample sizes. The resulting procedure is conservative.
4. State the conclusions in terms of the problem. Choose a significance level
such as α = 0.05, then compare the P-value to the α level.
If P-value  α, then reject H0 ; we have sufficient evidence to reject Ho
If P-value > α, then fail to reject H0 ; we lack sufficient evidence ……..
Robustness and use of the Two-Sample t Procedures:
The two-sample t procedures are more robust than the one-sample t methods,
particularly when the distributions are not symmetric. They are robust in the
following circumstances:
 If two samples are equal size and the two populations that the samples come
from have similar distributions then the t distribution is accurate for a variety
of distributions even when the sample sizes are as small as n1  n2  5 .
 When the two population distributions are different, larger samples are
needed.
 n  n  15 : Use two-sample t procedures if the data are close to normal. If
1
2
the data are clearly non normal or if outliers are present, do not use t.
 n1 + n2 is between 15-39: The t procedures can be used except in the
presence of outliers or strong skewness.
 n  n  40 : The t procedures can be used even for clearly skewed
1
2
distributions.
Lecture 9, Section 7.1 & 7.2
Page 11
Examples:
1. The U.S. Department of Agriculture (USDA) uses many types of surveys to
obtain important economic estimates. In one pilot study they estimated wheat
prices in July and in September using independent samples. Here is a brief
summary from the report:
Month
July
September
n
90
45
x
s / n (SEM)
3.50
3.61
0.023
0.029
a. Note that the standard error of the sample mean,(SEM) was reported,
instead of sample standard deviations. We find the standard deviation for each of
the samples as follows:
b. Use a significance test to examine whether or not the price of wheat was
the same in July and September. Be sure to give details and carefully state
your conclusion.
Ho: µ july = µ sept
Ha: µ july ≠ µ sept two side test because direction not indicated
Lecture 9, Section 7.1 & 7.2
Page 12
c. Give a 95% confidence interval for the increase in the population mean
between July and September.
2. The survey for Study Habits and Attitudes (SSHA) is a psychological test
designed to measure the motivation, study habits, and attitudes toward learning
of college students. These factors, along with ability are important in explaining
success in school. Scores on the SSHA range from 0 to 200. A selective private
college gives the SSHA to a SRS of both male and female first-year students.
Here are the data for 18 women:
154 109 137 115 152 140 145 178 101
103 126 126 137 165 165 129 200 148
Here are the data for 20 men:
108 140 114 91
180
109 132 75
88
113
115
151
126
70
92
115
169
187
146
104
a. Examine each sample graphically, with special attention to outliers and
skewness. Is use of a t procedure acceptable for these data?
STEMPLOT
MEN
STEM WOMEN
50 7
8 8
21 9
984 10
139
5543 11
5
6 12
669
2 13
77
60 14
058
1 15
24
9 16
55
17
8
70 18
19
20
0
Lecture 9, Section 7.1 & 7.2
Page 13
6
5
Frequency
4
3
2
1
Mean = 140.56
Std. Dev. = 26.262
N = 18
0
100
120
140
160
Women scores
Lecture 9, Section 7.1 & 7.2
Page 14
180
200
7
6
Frequency
5
4
3
2
1
Mean = 121.25
Std. Dev. = 32.852
N = 20
0
60
80
100
120
140
160
180
200
Men scores
b. Most studies have found that the mean SSHA score for men is lower than
the mean score in a comparable group of women. Test this supposition
here. That is, state the hypotheses, carry out the test and obtain a P-value,
and give your conclusions.
Lecture 9, Section 7.1 & 7.2
Page 15
Using SPSS:
Note: The data needs to be typed in using two columns. In the first column you
need to put all the scores. In the second column define the grouping variable as
gender and enter “ women” next to the women’s scores and “men” next to the
men’s scores.
Analyze > Compare means > Independent Sample T test
Move “score” to “Test Variable” box and “gender” to “grouping variable” box.
Click “define groups” and enter “women” for group 1 and “men” for group 2.
Click “continue” followed by “OK”.
Group Statistics
score
group
women
N
men
Mean
Std. Deviation
18
140.56
26.262
6.190
20
121.25
32.852
7.346
Levene's Test for
Equality of
Variances
F
score
Equal
variances
assumed
Equal
variances
not
assumed
Std. Error
Mean
1.030
Sig.
.317
t-test for Equality of Means
t
Sig. (2tailed)
df
Mean
Differenc
e
Std. Error
Differenc
e
95% Confidence
Interval of the
Difference
Lower
Upper
1.986
36
.055
19.306
9.721
-.410
39.021
2.010
35.537
.052
19.306
9.606
-.185
38.797
Independent Samples Test
We ALWAYS use the second line, “Equal variances not assumed” to get the t
test statistic, p-value, etc. Never assume equal variances in this test.
NOTE: PValue in the above example is .052/2 = .026
Give a 95% confidence interval for the mean difference between the SSHA scores
of male and female first-year students at this college.
Lecture 9, Section 7.1 & 7.2
Page 16
3. Suppose we wanted to compare how students performed on test 1 versus
test 2 in stat 301. Below is data for a random sample of 10 students taking stat 301
along with the printout of the results from running A MATCHED PAIR T TEST.
Subject
Test 1
1
2
3
4
5
6
7
8
9
10
Test 2
60
59
90
87
99
100
92
82
75
84
55
60
82
85
100
98
90
76
79
82
Paired Samples Statistics
Pair
1
Tes t 1
Tes t 2
Mean
82.80
80.70
N
10
10
Std. Deviation
14.382
14.507
Std. Error
Mean
4.548
4.588
Paired Samples Test
Paired Differences
Pair 1
Tes t 1 - Test 2
Mean
2.100
Std. Deviation
3.573
Std. Error
Mean
1.130
95% Confidence
Interval of the
Difference
Lower
Upper
-.456
4.656
t
1.859
df
9
Sig. (2-tailed)
.096
Answer the questions below based on the test.
a. Why was a matched pairs test used as opposed to a two sample t-test?
Lecture 9, Section 7.1 & 7.2
Page 17
b. Is there a difference between test 1 and test 2 scores. Write out the two-side
matched pair hypotheses to test this and give the P-value. Are our results
significant at the 5% significance level?
c. Suppose one of our friends thought test 1 was easier and the students
generally did better on it. We want to test whether the student is correct. Write
out the one-side matched pair hypotheses to test this and give the P-value. Are
our results significant at the 5% significance level?
Lecture 9, Section 7.1 & 7.2
Page 18
4. An instructor thought the material for the second test was more difficult and
hence the students may have done worse on the second test. She decided to
randomly sample 10 Exam 1 tests and a different 10 Exam 2 tests, with each sample
taken from all the tests taken, ie the two groups of 10 are different students. Below
is data for the two random samples and the analysis.
Sample
Test 1
Sample
Test 2
60
59
90
87
99
100
92
82
75
84
55
60
82
85
100
98
90
76
79
82
Group Statistics
s core
test
1
2
N
Mean
82.80
80.70
10
10
Std. Error
Mean
4.548
4.588
Std. Deviation
14.382
14.507
Independent Samples Test
Levene's Test for
Equality of Variances
F
s core
Equal variances
ass umed
Equal variances
not as sumed
.015
Sig.
.905
t-tes t for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
.325
18
.749
2.100
6.460
-11.472
15.672
.325
17.999
.749
2.100
6.460
-11.472
15.672
a. What procedure was used and why?
Two sample T test because the two groups are made up of different
students.
b. Write out the one-side hypotheses for this two-sample t test and give the Pvalue. Are your results significant at the 5% significance level?
Lecture 9, Section 7.1 & 7.2
Page 19