Section 10-3 – Comparing Two Variances
The F-Distribution
Let 𝑠12 and 𝑠22 represent the sample variances of two populations.
If both populations are normal, and the population variances 𝜎12 and 𝜎22
are equal, then the sampling distribution of 𝐹 =
𝑠12
𝑠22
is called an
F-distribution.
The populations must be independent and normally distributed.
Guidelines for Using a Two-Sample F-Test to Compare 𝜎12 and 𝜎22 .
1) Write the hypotheses and identify the claim.
2) Identify α.
3) STAT-TEST-E.
Make sure to enter the values correctly.
The calculator asks for the sample standard deviations; if you are
given the variances you will need to convert to standard
deviations (take the square roots).
4) Make a decision to reject or fail to reject the null hypothesis.
Section 10-3 – Comparing Two Variances
The F-Distribution
Example 3 – Page 583
A restaurant manager is designing a system that is intended to decrease
the variance of the time customers wait before their meals are served.
Under the old system, a random sample of 10 customers had a variance
of 400. Under the new system, a random sample of 21 customers had a
variance of 256. At α = 0.10, is there enough evidence to convince the
manager to switch to the new system? Assume both populations are
normally distributed.
Section 10-3 – Comparing Two Variances
The F-Distribution
Example 3 – Page 583
SOLUTION
Remember that the larger variance is 𝑠12 , so 400 is 𝑠12 , and 256 is 𝑠22 .
In other words, 𝑠12 and σ12 are the sample variance and the population
variance of the old system.
To write the hypotheses, we need to understand that the claim is that
“the variance of waiting times under the new system is less than the
variance of waiting times under the old system”, 𝑠22 < 𝑠12
Since we want to write this with 𝑠12 first, we need to reverse the sign.
H0: 𝑠12 ≤ 𝑠22 Ha: 𝑠12 > 𝑠22 (claim)
This makes this a right-tailed test.
Section 10-3 – Comparing Two Variances
The F-Distribution
Example 3 – Page 583
STAT-TEST-E. (2-SampFTest)
Please be careful in entering the data!!
The calculator asks for the standard deviation, not the variance.
The standard deviation is the square root of the variance.
We would enter the square root of 400 (20) for sample 1
We would enter the square root of 256 (16) for sample 2.
Indicate whether it is a right, left, or two-tailed test.
Calculate.
If p ≤ α, reject H0.
If p > α, fail to reject H0.
In this case, p = .194, which is > .10, so we fail to reject 𝐻0 .
At the 10% confidence level, there is not enough evidence to convince the
manager to switch systems.
Section 10-3 – Comparing Two Variances
The F-Distribution
Example 4 – Page 584
You want to purchase stock in a company and are deciding between two
different stocks. Because a stock’s risk can be associated with the standard
deviation of its daily closing prices, you randomly select samples of the
daily closing prices for each stock to obtain the results shown below. At α =
0.05, can you conclude that one of the two stocks is a riskier investment?
Assume the stock closing prices are normally distributed.
Stock A
Stock B
𝑛2 = 30
𝑛1 = 31
𝑠2 = 3.5
𝑠1 = 5.7
Since 5.7 is greater than 3.5, 5.7 will be 𝑠12 , and 3.5 will be 𝑠22 .
𝑠12 = 5.72 , 𝑠22 = 3.52 (Stock B is represented by 𝑠12 and 𝜎12 .)
Section 10-3 – Comparing Two Variances
The F-Distribution
Example 4 – Page 584
To write the hypotheses, remember that the claim is that “one of these
stocks is a riskier investment”.
We are not interested in which one it is, just that they are not equal.
𝐻0 : 𝑠12 = 𝑠22
𝐻𝑎 : 𝑠12 ≠ 𝑠22 (claim)
STAT-TEST-E (2-SampFTest)
Since we were given the standard deviations in this problem, we
enter them as given.
Use 5.7 for Sx1 and 31 for n1.
Use 3.5 for Sx2 and 30 for n2.
Select two tailed test and Calculate.
The test gives us the p-value of .0102, which we can compare to α.
.0102 < .05, so we reject the null.
At the 5% level of significance, there is enough evidence to conclude that
one of these stocks is a riskier investment.
Classwork: Page 585 #3, 11-16 All
Homework: Pages 585-586 #17-24 All
Skip part b - don't need it to run the test on the
calculator.
Section 10-4 – Analysis of Variance
One-Way ANOVA
One-way analysis of variance (ANOVA) is a hypothesis-testing
technique that is used to compare means from three or more
populations.
To begin an ANOVA test, you should first state a null and an
alternate hypothesis.
For a one-way ANOVA test, the null and alternate hypotheses
are always similar to the following statements:
H0: µ1 = µ2 = µ3 = ….. µk. (All population means are equal)
Ha: At least one of the means is different from the others.
To reject the H0 means that at least one of the means is different
from the others.
It will take more testing to determine which one of the means is
different from the others.
Section 10-4 – Analysis of Variance
One-Way ANOVA
In an one-way ANOVA test, the following conditions must be true.
1) Each sample must be randomly selected from a normal, or
approximately normal, population.
2) The samples must be independent of each other.
3) Each population must have the same variance.
Guidelines For Finding the Test Statistic for a One-Way ANOVA Test.
This test can be run on the TI-84 by entering each set of data into
STAT-EDIT using L1, L2, L3, etc.
Once the data has been entered, go to STAT-TEST-H (ANOVA).
Enter the names of the lists into which you put the data,
separated by commas, and allow the calculator to do the work.
Section 10-4 – Analysis of Variance
One-Way ANOVA
Example 1 – Page 591
A medical researcher wants to determine whether there is a difference
in the mean length of time it takes three types of pain relievers to
provide relief from headache pain. Several headache sufferers are
randomly selected and given one of the three medications. Each
headache sufferer records the time (in minutes) it takes the medication
to begin working. The results are shown in the table. At α = 0.01, can
you conclude that the mean times are different? Assume that each
population of relief times is normally distributed and that the
population variances are equal.
H0: µ1 = µ2 = µ3
Ha: At least one mean is different from the others.
Section 10-4 – Analysis of Variance
One-Way ANOVA
Example 1 – Page 591
Med 1
Med 2
Med 3
12
16
14
15
14
17
17
21
20
12
15
15
19
Use the calculator to enter these
lists into STAT-EDIT.
Med 1 into L1
Med 2 into L2
Med 3 into L3
Once the lists are entered, go to
STAT-TEST-H (ANOVA)
Press “2nd 1, 2nd 2, 2nd 3” and close
the parenthesis.
You must use the commas!!
Press Enter.
Section 10-4 – Analysis of Variance
One-Way ANOVA
Example 1 – Page 591
The calculator gives us many values and numbers.
For our purposes, we are going to look at the p-value and compare it
to α to make our decision.
Since p = .269, which is greater than 0.01, we will fail to reject the
null.
Because we failed to reject the null, we can not support the claim.
There is not enough evidence at the 1% significance level to conclude that
there is a difference in the mean length of time it takes the three pain
relievers to provide relief from headache pain.
Section 10-4 – Analysis of Variance
One-Way ANOVA
Example 2 – Page 593
Three airline companies offer flights between Corydon and
Lincolnville. Several randomly selected flight times (in minutes)
between the towns for each airline are shown. Assume that the
populations of flight times are normally distributed, the samples
are independent, and the population variances are equal. At α =
0.01, can you conclude that there is a difference in the means of
flight times?
H0: µ1 = µ2 = µ3
Ha: At least one mean is different from the others.
Section 10-4 – Analysis of Variance
One-Way ANOVA
Example 2 – Page 593
Airline 1
Airline 2
Airline 3
122
119
120
135
133
158
126
143
155
131
149
126
125
114
147
116
124
164
120
126
134
108
131
151
142
140
131
113
136
141
Use the calculator to enter these
lists into STAT-EDIT.
Airline 1 into L1
Airline 2 into L2
Airline 3 into L3
Once the lists are entered, go to
STAT-TEST-H (ANOVA)
Press “2nd 1, 2nd 2, 2nd 3” and close
the parenthesis.
You must use the commas!!
Press Enter.
Section 10-4 – Analysis of Variance
One-Way ANOVA
Example 2 – Page 593
You can see from the calculator that the p-value of this test is
0.0064.
Since this is less than the α value of 0.01, we need to reject the
null hypothesis.
At least one of the flight times has a mean different from the
others.
Classwork: Page 595-597 #5 - 10 All
Homework: Pages 597-599 #11-16 All
Skip part b - you don't need to do that when you
run the test on the calculator.