Part IVA

1
Part IVA

• Analysis of Variance
• (ANOVA)

Dr. Stephen H. Russell
Weber State University
2
Introduction to the concept of ANOVA

• I wonder if there is a difference in the average
  amount of beef contained in 32-ounce jars of
  Prago (well call population A) and Ragu (well
  call population B) spaghetti sauces.
• HO µA µB
• HA µA ? µB
• Consider the data and do a t test of hypotheses
  at the .05 level of significance. We will
  assume the populations are normally distributed
  and have equal variances.

3
A note on two-sample t tests . . .

• The degrees of freedom for a one-sample problem
  is n 1, as you know.
• The degrees of freedom for a two-sample problem
  is
• n1 1 n2 1 or n1 n2 2
• In the spaghetti sauce problem, the two sample t
  test the degrees of freedom would be
  5 6 2 9

4

• Grams of Beef in .
• A 32-ounce jar of Prago A 32-ounce
  jar of Ragu
• 27 29
• 24 27
• 27 31
• 25 32
• 27 30
• 31

These sample results yield a P-value of
.003strong evidence against the null and in
favor of the alternative that these two brands
are not equal. Ragu gives us more beef!
5
Lets look at this problem again

• in terms of variation among samples (between
  columns) and within samples
• Grams of Beef in .
• A 32-ounce jar of Prago A 32-ounce
  jar of Ragu
• 27 29
• 24 27
• 27 31
• 25 32
• 27 30
• 31

6
Whats the influencing factor?

• The brand!
• Does the brand matter when it comes to the amount
  of beef? Yes! So we say the factor matters!
• Grams of Beef in .
• A 32-ounce jar of Prago A 32-ounce
  jar of Ragu
• 27 29
• 24 27
• 27 31
• 25 32
• 27 30
• 31

7
Dependent Independent Variables

• The Dependent Variable (the variable that is
  acted upon) in this problem is the amount of
  meat in the spaghetti sauce.
• The Independent Variable (also called the
  factor) is the brand.
• We say Brand may influence the amount of meat.
  So meat is dependent on brand. Brand is the
  independent variable.

8
Comparing variances

• The variability among columns appears to be
  greater than the variability within columns. Is
  this observation consistent with the null or the
  alternative?
• The alternative! These brands are not equal when
  it comes to the amount of beef.
• Grams of Beef in .
• A 32-ounce jar of Prago A 32-ounce
  jar of Ragu
• 27 29
• 24 27
• 27 31
• 25 32
• 27 30
• 31

9
The F test

• We want to compute a ratio of variances

What would high values for this ratio
suggest? What is the expected value of this ratio
if the null is true? Ratios of two variances
follow a special distribution called the F
Distribution.
Comparing variances like this is called Analysis
of Variance (ANOVA)
10
The F Test

?
F tests are always right tailed in ANOVA problems.
11
The F test

Do the spaghetti sauce problem as an ANOVA
problem in MINITAB
12
Spaghetti Sauce Problem

• The test statistic (the calculated F) is 16.36 .
• The tail of rejection is found in an F Table
• Degrees of freedom for the numerator c 1
  (Levels of the factor minus one).
• Degrees of freedom for the denominator n c.
  (Total sample size minus levels of the factor.)
• For this problem Dfn 2 1 1 Dfd 11
  2 9
• What is the tail of rejection for an alpha of
  .05?
• 5.12.

13
The Spaghetti Sauce Problem

?
5.12
16.36
The calculated F is way out in the right tail.
We reject the null and conclude these two
spaghetti sauces do not have equal amounts of
beef.
14
The Spaghetti Sauce Problem

• A comparison of t-test and F-test results
• T test F test
• Calculated t -4.05 Calculated F 16.36
• Tail of rejection Tail of rejection (with
• with n-2 df 2.262 dfn 1 dfd
  9) 5.12
• P-value .003 P-value .003
• Decision Handily reject null Decision
  Handily reject null
• NOTE These results are the same.
• AND t2 F

15
Comparing population means

• Why in the heck do the complicated F test if the
  t test yields the same results?
• Because the F test can handle more than two
  population means comparisons e.g.,
• Ho µ1 µ2 µ3 µ4
• If we compared these means with t tests wed have
  to do 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs
  4 3 vs 4 or 6 different t tests.
• Heres the problem with doing 6 t tests . . .
  

16

• At an alpha of .05 the probability of a correct
  decision if the null is true on any one test is 1
  - .05 or .95.
• The probability of six correct decisions if the
  null is true is
• .95 raised to the sixth power or .735.
• This means that after doing six t-tests, the
  probability of a Type I error is not .05. Rather
  it is 1 - .735 or .265.
• Hence, when comparing the equality of more than
  two population means, we use the F test..

17
Additional comments on ANOVA

• ANOVA is a misleading term. ANOVA is not a test
  to compare population variances!
• ANOVA is a very complicated area in statistics.
  We have discussed only One-Way ANOVA (which means
  one factor).
• In MINITAB always click on Stat ? ANOVA ? One
  Way (Unstacked) in this class.
• ANOVA tests assume
• The sampled populations are normally distributed
• The sampled populations have equal variances (a
  critical assumption for correct results)
• Its a good idea to use equal sample sizeswhich
  minimizes the impact of violating the
  equal-variances assumption.

18
Example problem

• A furniture manufacturer wants to compare the
  mean drying times for four brands of stain. Each
  stain was applied to 10 chairs and the drying
  times in minutes were recorded.
• The hypotheses are
• HO µ1 µ2 µ3 µ4
• HA Not all population means are equal
• Lets use an alpha of .01.
• (1) What is the tail of rejection?
• (2) Solve the problem with MINITAB

dfn 3 dfd 36 F for rejection 4.39
19
Homework Assignment for ANOVA

• Problem Set 4

20
Summary of ANOVA

• Analysis of variance( ANOVA) statistical
  technique designed to test whether the means of
  more than two populations are equal
• Variation has two components
• variation among columns, explained by the factor
  measures explained variation
• variation within columns, attributed to random
  error measures unexplained variation
• We have covered only one-way ANOVA (also called
  one-factor ANOVA)
• ANOVA analysis assumes normal populations with
  equal variances.

21
Homework solutions

• 1. HO ?L ?M ?H
• HA Not all of the population means are equal
• dfn c 1 2 dfd n c 12
• The Tail of Rejection in a F distribution is
  defined as 5.10 for 2.5 percent level of
  significance.
• The F statistic is 1.92, which means the
  variability attributable to levels of the factor
  is 1.92 times greater than the random
  variability.
• P-Value is .189, which is interpreted as If
  the Null is true, there is a .189 chance of
  observing an F statistic as contradictory (or
  more contradictory) to the null as the value
  found here.
• We fail to reject the null. We do not have
  sufficiently strong evidence to run with the
  conclusion that housing prices are not the same
  for three areas with different levels of air
  pollution.

22
Homework solutions

• 2. HO ?Food A ?Food B ?Food C
• HA Not all of the population means are equal
• dfn c 1 2 dfd n c 15
• The Tail of Rejection in a F distribution is
  defined as 6.36 for .01 level of significance.
• The F statistic is .36, which means the
  variability attributable to levels of the factor
  is .36 of the random variabilityi.e., very
  little factor variability.
• P-Value is a HUGE .701, which is interpreted
  as If the Null is true, there is a .701 chance
  of observing an F statistic as contradictory (or
  more contradictory) to the null as the value
  found here.
• We fail to reject the null. We do not have
  sufficiently strong evidence to run with the
  conclusion that dogs do not like these three
  foods equally. (In fact, random variability is
  greater than explained variability!)

23
Homework solutions

• 3. HO ?Epsilon ?Chevron ?BP
• HA Not all of the population means are equal
• dfn c 1 2 dfd n c 15
• The Tail of Rejection in a F distribution is
  defined as 3.68 for 5 percent level of
  significance.
• The F statistic is 20.35, which means the
  variability attributable to levels of the factor
  is more than 20 times greater than the random
  variability.
• P-Value is 0.000, which is interpreted as If
  the Null is true, there is a zero chance of
  observing an F statistic as contradictory (or
  more contradictory) to the null as the value
  found here.
• We reject the null. We have very strong
  evidence that these brands do not yield the same
  flying time.