Class Notes
April 07
The general steps in any hypothesis test:
1. Check the appropriate conditions for a valid significance test.
2. Determine the null and alternative hypothesis;
3. Summarize the data into an appropriate test statistic (t-statistic for the
mean and z-statistic for the proportion);
4. Assuming the null hypothesis is true, find the p-value;
5. Decide whether or not the result is statistically significant based on
the p-value.
13.2 Testing Hypotheses about One Population Mean
The usual procedure for testing this hypothesis is called a one-sample t-test
(that’s why we use t* for CI).
Conditions under which a “t-test” for one population mean is valid:
1. The population of the measurements of interest is bell shaped and a
random sample of any size is measured. In practice, we can use this
method as long as there is no evidence that the shape is notably
skewed or that there are extreme outliers.
2. The population of measurements of interest is not bell-shaped, but a
large random sample is measured. 30 is usually used as “large”.
The statistic we use is t-statistic:
t=
sample mean  null value
standard error
Example: Suppose we want to see if the average height of Stat 200 students
is shorter than 67 inches. The following is the Minitab output:
One-Sample T: Height
Test of mu = 67 vs mu < 67
Variable
Height
Variable
Height
N
418
Mean
66.682
95.0% Upper Bound
67.074
StDev
4.863
T
-1.34
SE Mean
0.238
P
0.091
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Determine the p-value: most software programs can do the work for you.
You can also find p-value from “Tables of the student t-distribution.”
p-value and conclusion: In our case, the p-value is 0.091 > 0.05. Hence, we
do not reject the null hypothesis. We therefore claim that the average height
of Stat 200 students is 67 inches.
Matched Pair Analysis
Most often, this type of data occurs when the researcher collects two
measurements from each observational unit. For instance, if we record
weights both before and after a diet program for each person in a sample, we
have paired data. With paired data, we calculate the difference between the
measurements for each pair, and then we use procedures for analyzing a
single sample.
Note: We use d as the difference between the 2 treatments.
The following are 5 steps that we use to solve a Matched Pair problem.
1. Compute difference for each unit;
2. Set up null and alternative hypothesis;
3. Calculate average difference and standard deviation of difference.
4. Check conditions that difference is normal and paired analysis.
5. Calculate confidence interval: d  t*
sd
n
, where d is the average
difference between 2 measurements; sd is the sample standard
deviation of difference; n is sample size.
Or hypothesis test with t-statistic: t =
d  null value
.
sd
n
Example: Stichler, Richey, and Mandel compared two methods of measuring
treadwear in their paper “Measurement of Treadwear of Commercial Tires”
(Rubber Age, 73:2, 1953). 11 tires were each measured for treadwear by
two methods, one based on weight and the other on groove wear. We want
to see if there is any difference between these two methods.
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weight
groove
difference
30.5
28.7
1.8
30.9
25.9
5.0
31.9
23.3
8.6
30.4
23.1
7.3
27.3
23.7
3.6
20.4
20.9
-0.5
24.5
16.1
8.4
20.9
19.9
1.0
18.9
15.2
3.7
13.7
11.5
2.2
11.4
11.2
0.2
The Minitab output is the following:
Test of mu = 0 vs mu not = 0
Variable
difference
Variable
difference
N
11
(
Mean
3.755
95.0% CI
1.590,
5.919)
StDev
3.221
T
3.87
SE Mean
0.971
P
0.003
How do you calculate the difference between “weight” and “groove”? What
is the value for d ? What is the value for sd? What is the value for tstatistic, and p-value?
p-value and conclusion: In this problem, we have p-value = 0.003 < 0.05.
Hence, we reject the null hypothesis. Therefore we claim that there is a
difference between the 2 methods.
13.3 The Difference between Two Means (Independent
Sample)
It is often of interest to determine whether or not the mean of populations
represented by two independent samples of a quantitative variable differ.
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In most cases when comparing two means, the null hypothesis is that they
are equal:
H0: 1 - 2 = 0
Note: Again, in order to carry out this hypothesis testing, we need to check
the 2 conditions on Page 1 for both samples. Moreover, the samples must
be independent.
In the 2-sample t test case, we have:
t=
sample mean  null value
=
standard error
x1  x2
2
2
s
s1
 2
n1
n2
This is the t statistic for unpooled version, where we do not assume equal
variance.
Example: Suppose we want to see if on average, the female students in Stat
200 students study more than male students do, we do a 2-sample t test using
Minitab and the following is the Minitab output.
Two-Sample T-Test and CI: StudyHrs, Gender
Two-sample T for StudyHrs
Gender
female
male
N
258
159
Mean
15.93
14.19
StDev
9.53
8.57
SE Mean
0.59
0.68
Difference = mu (female) - mu (male )
Estimate for difference: 1.735
95% lower bound for difference: 0.247
T-Test of difference = 0 (vs >): T-Value = 1.92
P-Value = 0.028
DF = 361
P-value and conclusion: from the output, we see the p-value is 0.028 < 0.05.
We have evidence to reject the null hypothesis. Hence we conclude that on
average, female students do study more than male students do.
13.4 The Difference between Two Proportions
Conditions for a confidence interval for the difference in two proportions:
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1. Sample proportions are available based on independent samples from
the two populations.
2. All of the quantities n1 pˆ 1, n2 pˆ 2, n1 (1  pˆ 1 ), and n2 (1  pˆ 2 ) are at least 10.
We will use z statistic for proportions.
z=
p̂ =
sample mean  null value
=
standard error
pˆ 1  pˆ 2
, where
pˆ (1  pˆ ) pˆ (1  pˆ )

n1
n2
n1 pˆ 1  n2 pˆ 2
.
n1  n2
For example, if we want to see if there is any difference between the
proportions of right-handedness among male and female Stat 200 students,
we can use 95% confidence interval to solve this question. The Minitab
output is the following:
Test and CI for Two Proportions: Handed, Gender
Success = right-handed
Gender
female
male
X
236
129
N
258
156
Sample p
0.914729
0.826923
Estimate for p(female) - p(male): 0.0878056
95% CI for p(female) - p(male): (0.0193535, 0.156258)
Test for p(female) - p(male) = 0 (vs not = 0): Z = 2.51
P-Value = 0.012
p-value and conclusion: from the output, we see the p-value is 0.012 < 0.05.
We have evidence to reject the null hypothesis. Hence we conclude that
there is a difference between the proportions of right-handedness among
male and female Stat 200 students. One thing to notice is that although the
conclusion is “there is a difference”, we can in fact tell something more
specific: the proportion of females is higher than the proportion of males
(how?).
13.5 The Relationship between Significance Tests and
Confidence Interval
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So far, we have introduced the confidence interval and hypothesis testing for
one mean, the difference between 2 means, one proportion, and the
difference between 2 proportions. In this section, we will concentrate on the
important relationship between hypothesis testing and confidence interval.
Confidence Intervals and Two-sided Alternatives
When testing one population mean or the difference in two populations
means, with the hypothesis
H0: parameter = null value vs. Ha: parameter  null value
1. If the null value is covered by a (1 - )100% confidence interval, the
null hypothesis is not rejected and the test is not statistically
significant at level .
2. If the null value is not covered by a (1 - )100% confidence interval,
the null hypothesis is rejected and the test is statistically significant at
level .
Question 1: Suppose in order to compare the average time spent to drive
from State College to NYC and the average time spent to drive from NYC to
State College, we ask 100 students in random to drive from here to NYC and
then drive back. After careful calculation, we get the p-value is 0.02 at
significant level 0.05. Which one of the following interval is the possible
95% confidence interval?
a. (0.342, 1.987)
b. (-0.056, 0.743)
c. (-1.256, 0.245)
d. (-2.583, 3.872)
(*The Following Material is Not Required*)
Confidence Intervals and One-sided Alternatives
When testing the hypotheses H0: parameter = null value versus a one-sided
alternative (either “>” or “<”), compare the null value to a (1- 2)100%
confidence interval:
1. If the null value is covered by the interval, the test is not statistically
significant at level ;
2. For the alternative Ha: parameter > null value, the test is statistically
significant at level  if the entire interval falls above the null value;
3. For the alternative Ha: parameter < null value, the test is statistically
significant at level  if the entire interval falls below the null value.
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Question 2: Suppose in order to see if the average time spent to drive from
State College to NYC is less than 5 hours, we ask 100 students in random to
drive from here to NYC. Decide if we can reject the null hypothesis at the
significance level of  = 0.05, using the following 90% confidence interval.
1. (4.375, 5.551)
2. (3.985, 4.891)
3. (5.075, 5.876)
4. (4.577, 5.425)
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