Download AP_Statistics_Week_28_files/Bock - HT for 2

Assumptions and Conditions
• Independence Assumption (Each condition
needs to be checked for both groups.):
– Randomization Condition: Were the data collected
with suitable randomization (representative random
samples or a randomized experiment)?
– 10% Condition: We don’t usually check this condition
for differences of means. We will check it for means
only if we have a very small population or an
extremely large sample.
Assumptions and Conditions
(cont.)
• Normal Population Assumption:
– Nearly Normal Condition: This must be checked for
both groups. A violation by either one violates the
condition.
• Independent Groups Assumption: The two
groups we are comparing must be independent
of each other. If not, we will discuss this later in
the lesson.
Testing the Difference Between
Two Means
• The hypothesis test we use is the twosample t-test for means.
• The conditions for the two-sample t-test for
the difference between the means of two
independent groups are the same as for
the two-sample t-interval.
Testing the Difference Between
Two Means (cont.)
We test the hypothesis H0:1 – 2 = 0, where
the hypothesized difference, 0, is almost always
0, using the statistic
x1  x2    0

t
s12 s22

n1 n2
When the conditions are met and the null hypothesis is
true, this statistic can be closely modeled by a Student’s
t-model with a number of degrees of freedom given by a
special formula. We use that model to obtain a P-value.
Example:
The Better Cookie Company claims its chocolate chip cookies have more chips than
another chocolate chip cookie. 120 Better Cookies and 100 of the other type of cookie
were randomly selected and the number of chips in each cookie was recorded. The
results are as follows.
Better Another
Mean number of chips
7.6
6.9
Standard deviation
1.4
1.7
At the 2% level of significance, test the claim that the population of Better Cookies has a
higher mean number of chips.
1. Hypothesis
1 = population mean number of chips from Better Cookie Co.
2 = population mean number of chips from another cookie company
Ho: 1 = 2
Ha: 1 > 2
Example:
The Better Cookie Company claims its chocolate chip cookies have more chips than
another chocolate chip cookie. 120 Better Cookies and 100 of the other type of cookie
were randomly selected and the number of chips in each cookie was recorded. The
results are as follows.
Better Another
Mean number of chips
7.6
6.9
Standard deviation
1.4
1.7
At the 2% level of significance, test the claim that the population of Better Cookies has a
higher mean number of chips.
2. Check Assumptions/Conditions
•
•
•
•
SRS is stated
 is unknown, use t-distribution
Population are independent
We will assume an approximately normal distribution
Example:
The Better Cookie Company claims its chocolate chip cookies have more chips than
another chocolate chip cookie. 120 Better Cookies and 100 of the other type of cookie
were randomly selected and the number of chips in each cookie was recorded. The
results are as follows.
Better Another
Mean number of chips
7.6
6.9
Standard deviation
1.4
1.7
At the 2% level of significance, test the claim that the population of Better Cookies has
a higher mean number of chips.
3. Calculate Test
x1  x2    0  7.6  6.9   0

t

 3.29
s12 s22

n1 n2
1.42 1.72

120 100
P  x1  x2  0.7 1  2  0   0.0006
0.0006
Example:
The Better Cookie Company claims its chocolate chip cookies have more chips than
another chocolate chip cookie. 120 Better Cookies and 100 of the other type of cookie
were randomly selected and the number of chips in each cookie was recorded. The
results are as follows.
Better Another
Mean number of chips
7.6
6.9
Standard deviation
1.4
1.7
At the 2% level of significance, test the claim that the population of Better Cookies has a
higher mean number of chips.
4. Conclusion
Since P-value is less than alpha, we reject that
there is no difference in population mean number
of chips between Better Cookie and another cookie
company.
Paired Data
• Data are paired when the observations are
collected in pairs or the observations in one
group are naturally related to observations in the
other group.
• Paired data arise in a number of ways. Perhaps
the most common is to compare subjects with
themselves before and after a treatment.
– When pairs arise from an experiment, the pairing is a
type of blocking.
– When they arise from an observational study, it is a
form of matching.
Paired Data (cont.)
• If you know the data are paired, you can (and
must!) take advantage of it.
– To decide if the data are paired, consider how they
were collected and what they mean (check the W’s).
– There is no test to determine whether the data are
paired.
• Once we know the data are paired, we can
examine the pairwise differences.
– Because it is the differences we care about, we treat
them as if they were the data and ignore the original
two sets of data.
Paired Data (cont.)
• Now that we have only one set of data to
consider, we can return to the simple onesample t-test.
• Mechanically, a paired t-test is just a onesample t-test for the mean of the pairwise
differences.
– The sample size is the number of pairs.
Assumptions and Conditions
• Paired Data Assumption: The data must be
paired.
• Independence Assumption: The differences
must be independent of each other. Check the:
– Randomization Condition
• Normal Population Assumption: We need to
assume that the population of differences follows
a Normal model.
– Nearly Normal Condition: Check this with a histogram
or Normal probability plot of the differences.
The Paired t-Test
• When the conditions are met, we are
ready to test whether the paired
differences differ significantly from zero.
• We test the hypothesis H0: d = 0, where
the d’s are the pairwise differences and 0
is almost always 0.
The Paired t-Test (cont.)
• We use the statistic
xd   0
t
sd
n
where n is the number of pairs.
• When the conditions are met and the null
hypothesis is true, this statistic follows a
Student’s t-model on n – 1 degrees of freedom,
so we can use that model to obtain a P-value.
Example:
A test of abstract reasoning is given to a random sample of students
before and after they completed a formal logic course. The results are
given below. Do the data suggest that the mean score after the course
differs from the mean score before the course? Perform a test at the
5% significance level.
Before 74 83 75 88 84 63 93 84 91 77
After 73 77 70 77 74 67 95 83 84 75
Before – After
1
6
5 11 10 -4 -2
1 7
2
1. Hypothesis
d = population mean test score difference before and after
completing logic course
Ho: d = 0
Ha: d  0
Example:
A test of abstract reasoning is given to a random sample of students
before and after they completed a formal logic course. The results are
given below. Do the data suggest that the mean score after the course
differs from the mean score before the course? Perform a test at the
5% significance level.
Before 74 83 75 88 84 63 93 84 91 77
After 73 77 70 77 74 67 95 83 84 75
Before – After
1
6
5 11 10 -4 -2
2. Check Assumptions/Conditions
• SRS is stated
•  is unknown, use t-distribution
• Populations are dependent
Based on the linearity of the normal
probability plot, we have an
approximately normal distribution.
1 7
2
Example:
A test of abstract reasoning is given to a random sample of students
before and after they completed a formal logic course. The results are
given below. Do the data suggest that the mean score after the course
differs from the mean score before the course? Perform a test at the
5% significance level.
Before 74 83 75 88 84 63 93 84 91 77
After 73 77 70 77 74 67 95 83 84 75
Before – After
1
6
5 11 10 -4 -2
1 7
2
3. Calculate Test
xd   3.7  0
t

 2.366
sd
4.945
10
n
P  xd  3.7 d  0   0.042
0.021
Example:
A test of abstract reasoning is given to a random sample of students
before and after they completed a formal logic course. The results are
given below. Do the data suggest that the mean score after the course
differs from the mean score before the course? Perform a test at the
5% significance level.
Before 74 83 75 88 84 63 93 84 91 77
After 73 77 70 77 74 67 95 83 84 75
Before – After
1
6
5 11 10 -4 -2
1 7
2
4. Conclusion
Since P-value is less than alpha, we reject that
there is no difference in population mean test score
difference before and after completing logic
course.