Difference between revisions of "AP Statistics Curriculum 2007 ANOVA 1Way"

From SOCR
Jump to: navigation, search
(→‎ANOVA Table)
m (Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/")
 
(26 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - One-Way Analysis of Variance (ANOVA) ==
 
==[[AP_Statistics_Curriculum_2007 | General Advance-Placement (AP) Statistics Curriculum]] - One-Way Analysis of Variance (ANOVA) ==
  
−
In the [[EBook#Chapter_IX:_Inferences_from_Two_Samples |two-sample inference chapter]] we considered the comparisons of two independent group means using the [[AP_Statistics_Curriculum_2007_Infer_2Means_Indep#Independent_T-test_Validity |independent T-test]]. Now, we expand our inference methods to study and compare ''k'' independent samples. In this case, we will be decomposing the entire variation in the data into (independent/orthogonal) components - i.e., we'll be analyzing the variance of the data. Hence, this procedure called '''Analysis of Variance (ANOVA)'''.
+
In the [[EBook#Chapter_IX:_Inferences_From_Two_Samples |two-sample inference chapter]], we considered the comparisons of two independent group means using the [[AP_Statistics_Curriculum_2007_Infer_2Means_Indep#Independent_T-test_Validity |Independent T-test]]. Now, we expand our inference methods to study and compare ''k'' independent samples. In this case, we will be decomposing the entire variation in the data into (independent/orthogonal) components - i.e., we'll be analyzing the variance of the data. Hence, this procedure called '''Analysis of Variance (ANOVA)'''.
  
 
===Motivational Example===
 
===Motivational Example===
Line 36: Line 36:
 
</center>
 
</center>
  
−
Using the [http://socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts] (see [[SOCR_EduMaterials_Activities_BoxAndWhiskerChart | SOCR Box-and-Whisker Plot Activity]] and [[SOCR_EduMaterials_Activities_DotChart | Dot Plot Activity]]) we can generate plots that enable us to compare visually the yields of the 5 different types peas.
+
Using the [http://socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts] (see [[SOCR_EduMaterials_Activities_BoxAndWhiskerChart | SOCR Box-and-Whisker Plot Activity]] and [[SOCR_EduMaterials_Activities_DotChart | Dot Plot Activity]]), we can generate plots that enable us to compare visually the yields of the 5 different types peas.
  
 
<center>[[Image:SOCR_EBook_Dinov_ANOVA1_021708_Fig1.jpg|500px]]</center>
 
<center>[[Image:SOCR_EBook_Dinov_ANOVA1_021708_Fig1.jpg|500px]]</center>
Line 42: Line 42:
 
Using ANOVA, the data are regarded as random samples from ''k'' populations. Suppose the population means of the samples are <math>\mu_1, \mu_2, \mu_3, \mu_4, \mu_5</math> and their population standard deviations are: <math>\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5</math>.  We have 5 group means to compare. Why not just carry out <math>{5\choose 2}=10</math> T-tests comparing all (independent) pairs of groups?
 
Using ANOVA, the data are regarded as random samples from ''k'' populations. Suppose the population means of the samples are <math>\mu_1, \mu_2, \mu_3, \mu_4, \mu_5</math> and their population standard deviations are: <math>\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5</math>.  We have 5 group means to compare. Why not just carry out <math>{5\choose 2}=10</math> T-tests comparing all (independent) pairs of groups?
  
−
Repeated T-tests would mean testing hull hypotheses of the type <math>H_o: \mu_i = \mu_j, \forall i\not= j</math>. What is the problem with this approach? Suppose each test is carried out at <math>\alpha = 0.05</math>, so a [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | type I error]] is 5% for each test.
+
Repeated T-tests would mean testing null hypotheses of the type <math>H_o: \mu_i = \mu_j, \forall i\not= j</math>. What is the problem with this approach? Suppose each test is carried out at <math>\alpha = 0.05</math>, so a [[AP_Statistics_Curriculum_2007_Hypothesis_Basics | type I error]] is 5% for each test.
 
Then, the overall risk of a type I error is larger than 0.05 and gets much larger as the number of groups (''k'') gets larger. To solve this problem, we need to make multiple comparisons with an overall error of <math>\alpha = 0.05</math> (or whichever level is specified initially).
 
Then, the overall risk of a type I error is larger than 0.05 and gets much larger as the number of groups (''k'') gets larger. To solve this problem, we need to make multiple comparisons with an overall error of <math>\alpha = 0.05</math> (or whichever level is specified initially).
  
  
−
The main idea behind ANOVA is that we need to know how much inherent variability there is in the data before we can judge whether there is a difference in the sample means - i.e., presence of a grouping effect.  To make an inference about means we compare two types of variability:
+
The main idea behind ANOVA is that we need to know how much inherent variability exists is in the data before we can judge whether there is a difference in the sample means - i.e., presence of a grouping effect.  To make an inference about means we compare two types of variability:
  
−
: variability between sample means
+
: Variability between sample means
  
−
: variability within each group
+
: Variability within each group
  
 
It is very important that we keep these two types of variability in mind as we work through the following formulas. It is our goal to come up with a numerical recipe that describes/computes each of these variabilities.
 
It is very important that we keep these two types of variability in mind as we work through the following formulas. It is our goal to come up with a numerical recipe that describes/computes each of these variabilities.
−
 
  
 
===One-Way ANOVA Calculations===
 
===One-Way ANOVA Calculations===
Line 65: Line 64:
 
: The grand mean is: <math>\bar{y}=\bar{y}_{.,.} = {\sum_{i=1}^k {\sum_{j=1}^{n_i}{y_{i,j}}} \over n}</math>
 
: The grand mean is: <math>\bar{y}=\bar{y}_{.,.} = {\sum_{i=1}^k {\sum_{j=1}^{n_i}{y_{i,j}}} \over n}</math>
  
−
To compute the difference between the means we will compare each group mean to the grand mean.
+
To compute the difference between the means, we will compare each group mean to the grand mean.
  
−
* SST (Sum Square due to Treatment):
+
* SST (Sum Square due to Treatment, or ''Between'' Group variation): First, we describe the variation between the group means. For the independent T-test we described the difference between two group means as <math>\bar{y_1} - \bar{y_1}</math>. In ANOVA we describe the difference between ''k'' means as sums of squares due to treatments (or between-group variance):
−
First, we describe the variation between the group means. For the independent T-test we described the difference between two group means as <math>\bar{y_1} - \bar{y_1}</math>. In ANOVA we describe the difference between ''k'' means as sums of squares due to treatments (or between-group variance):
 
  
−
: SST(between) = <math>\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}</math>. SST can be though of as the difference between each group mean and the grand mean.
+
: SST(Between) = <math>\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}</math>. SST can be thought of as the difference between each group mean and the grand mean.
  
 
: Degrees of Freedom: ''df (Between) = k – 1''.
 
: Degrees of Freedom: ''df (Between) = k – 1''.
  
−
: Mean Sum Square due to Treatment (Between): c This measures variability between the sample means.
+
: Mean Sum Square due to Treatment (Between): <math>MST(Between) = {SST(Between)\over df(Between)}.</math> This measures variability between the sample means.
  
−
* SSE (Sum Square due to Error):
+
* SSE (Sum Square due to Error, or ''Within'' Group Variation): Second, we assess the within group variation.  Recall that to measure the variability within a single sample we used:  <math>\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2} \over n-1}</math>.  In ANOVA to describe the combined variation within the ''k'' groups we use sums of squares due to error (within-group variation):
−
Second, we assess the within group variation.  Recall that to measure the variability within a single sample we used:  <math>\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2} \over n-1}</math>.  In ANOVA to describe the combined variation within the ''k'' groups we use sums of squares due to error (within-group variation):
 
 
   
 
   
−
: SSE(within) = <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}</math>, which can be though of as the combination of variation within the ''k'' groups.  
+
: SSE(Within) = <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}</math>, which can be thought of as the combination of variation within the ''k'' groups.  
  
 
:SSE(within)degrees of freedom: ''df (Within) = n - k''
 
:SSE(within)degrees of freedom: ''df (Within) = n - k''
  
−
: Mean variability within groups: <math>MSE(within) = {SSE(Between)\over df(Within)}.</math> This is a measure of variability within the groups.
+
: Mean variability within groups: <math>MSE(Within) = {SSE(Within)\over df(Within)}.</math> This is a measure of variability within the groups.
  
 
* Decomposition of Variance: Note that we have the following decomposition of the total variability in the data:
 
* Decomposition of Variance: Note that we have the following decomposition of the total variability in the data:
Line 89: Line 86:
 
: (Verbal): Deviation of an observation from the grand mean (Total variability) = Variation-Within + Variation_Between!
 
: (Verbal): Deviation of an observation from the grand mean (Total variability) = Variation-Within + Variation_Between!
  
−
: (Mathematics): <math>y_{i,j} - \bar{y} = (y_{i,j} - \bar{y}_{i,.}) + (\bar{y}_{i,.} - \bar{y})</math>. Ans summing up over all observations we get the desired ANOVA decomposition.
+
: (Mathematics): <math>y_{i,j} - \bar{y} = (y_{i,j} - \bar{y}_{i,.}) + (\bar{y}_{i,.} - \bar{y})</math>. And summing up over all observations we get the desired ANOVA decomposition.
  
 
: ('''ANOVA Decomposition'''): <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}} = \sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y}_{i,.})^2}} + \sum_{i=1}^{k}{n_i(\bar{y}_{i,.} - \bar{y})^2}</math>.
 
: ('''ANOVA Decomposition'''): <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}} = \sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y}_{i,.})^2}} + \sum_{i=1}^{k}{n_i(\bar{y}_{i,.} - \bar{y})^2}</math>.
Line 95: Line 92:
 
* Interpretations:
 
* Interpretations:
  
−
: This means S''ST(total) = SSE(within) + SST(between)''
+
: This means ''SST(total) = SSE(within) + SST(between)''
  
 
: ''df(Total) = df(Within) + df(Between) = (n - k) + (k – 1) = n - 1''
 
: ''df(Total) = df(Within) + df(Between) = (n - k) + (k – 1) = n - 1''
Line 158: Line 155:
 
{| class="wikitable" style="text-align:center; width:50%" border="1"
 
{| class="wikitable" style="text-align:center; width:50%" border="1"
 
|-
 
|-
−
| Variance Source || Degrees of Freedom (df) || Sum  of Squares (SS) || Mean Sum  of Squares (MS) || F-Statistics || P-value
+
| Variance Source || Degrees of Freedom (df) || Sum  of Squares (SS) || Mean Sum  of Squares (MS) || F-Statistics || [http://socr.ucla.edu/htmls/SOCR_Distributions.html P-value]
 
|-
 
|-
−
| Treatment Effect (Between Group) || k-1 || <math>\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}</math> || <math>{SST(Between)\over df(Between)}</math> || <math>{MST(Between)\over MSE(Within)}</math> || <math>P(F_{df(Between), df(Within)} > F)</math>
+
| Treatment Effect (Between Group) || k-1 || <math>\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}</math> || <math>MST(Between)={SST(Between)\over df(Between)}</math> || <math>F_o = {MST(Between)\over MSE(Within)}</math> || <math>P(F_{(df(Between), df(Within))} > F_o)</math>
 
|-
 
|-
−
| Error (Within Group) || n-k || <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}</math> || <math>{SSE(Within)\over df(Within)}</math> ||  ||
+
| Error (Within Group) || n-k || <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}</math> || <math>MSE(Within)={SSE(Within)\over df(Within)}</math> ||  || [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm F-Distribution Calculator]
 
|-
 
|-
−
| Total || n-1 || <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}}</math> ||  ||  ||
+
| Total || n-1 || <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}}</math> ||  ||  || [[SOCR_EduMaterials_AnalysisActivities_ANOVA_1 | ANOVA Activity]]
 
|}
 
|}
 
</center>
 
</center>
 +
 +
===ANOVA Hypotheses===
 +
The general form of the ANOVA hypotheses is:
 +
: <math>H_o: \mu_1=\mu_2=\mu_3 = \cdots = \mu_k</math>
 +
: <math>H_a: \mu_i \not= \mu_j</math> for some <math>i\not=j</math>.
 +
 +
Note that <math>H_o</math> is compound hypothesis, when k > 2, so rejecting <math>H_o</math> doesn't tell us which <math>\mu_i</math>'s are different. It only tells us that some two are not equal.
 +
 +
* Test Statistics: The test statistic: <math>F_o = {MST(Between) \over MSE(Within)}</math>. If <math>F_o</math> is large, then there is a lot of between group variation, relative to the within group variation. Therefore, the discrepancies between the group means are large compared to the variability within the groups (error). Hence, large values of <math>F_o</math> provide strong evidence against <math>H_o</math>.
 +
 +
===Examples===
 +
 +
====Hands-on Example====
 +
This example is provided as a hands-on demonstration of the calculations performed in 1-way ANOVA. Clearly the value added by this example is conceptual understanding and appreciation of the methodological protocol for computing critical values and probabilities in ANOVA designs.
 +
 +
<center>
 +
{| class="wikitable" style="text-align:center; width:30%" border="1"
 +
|-
 +
| || colspan=3| Groups
 +
|-
 +
| Index || A || B || C
 +
|-
 +
| 1 || 0 || 1 || 4
 +
|-
 +
| 2 || 1 || 0 || 5
 +
|-
 +
| 3 ||  || 2 ||
 +
|-
 +
| <math>n_i</math> || 2 || 3 || 2
 +
|-
 +
| s || 1 || 3 || 9
 +
|-
 +
| <math>\bar{y}_i</math> || 0.5 || 1 || 4.5
 +
|}
 +
</center>
 +
 +
Now use the [[AP_Statistics_Curriculum_2007_ANOVA_1Way#ANOVA_Table |ANOVA table]] to calculate the F-statistics and the corresponding p-value. Then use [http://socr.ucla.edu/htmls/SOCR_Analyses.html SOCR Analyses (ANOVA)] to validate your manual calculations.
 +
 +
<center>
 +
{| class="wikitable" style="text-align:center; width:50%" border="1"
 +
|-
 +
| Variance Source || Degrees of Freedom (df) || Sum  of Squares (SS) || Mean Sum  of Squares (MS) || F-Statistics || [http://socr.ucla.edu/htmls/dist/Fisher_Distribution.html P-value]
 +
|-
 +
| Treatment Effect (Between Group) || 3-1 || <math>\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}=19.86</math> || <math>{SST(Between)\over df(Between)}={19.86\over 2}</math> || <math>F_o = {MST(Between)\over MSE(Within)}=13.24</math> || <math>P(F_{(df(Between), df(Within))} > F_o)=0.017</math>
 +
|-
 +
| Error (Within Group) || 7-3 || <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}=3</math> || <math>{SSE(Within)\over df(Within)}={3\over 4}</math> ||  || [http://socr.ucla.edu/Applets.dir/Normal_T_Chi2_F_Tables.htm F-Distribution Calculator]
 +
|-
 +
| Total || 7-1 || <math>\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}}=22.86</math> ||  ||  || [[SOCR_EduMaterials_AnalysisActivities_ANOVA_1 | Anova Activity]]
 +
|}
 +
</center>
 +
 +
<center>[[Image:SOCR_EBook_Dinov_ANOVA1_021708_Fig3.jpg|500px]]</center>
 +
 +
Therefore, in this case we can reject the null hypothesis (at <math>\alpha = 0.05</math>).
 +
 +
====Walking Age====
 +
Parents are frequently concerned when their child seems slow to begin walking. In 1972, [http://www.sciencemag.org/cgi/content/abstract/176/4032/314 Science reported on an experiment in which the effects of several different treatments on the age at which a child’s first walks were compared]. Children in the ''first group'' were given special walking exercises for 12 minutes daily beginning at the age 1 week and lasting 7 weeks. The ''second group'' of children received daily exercises, but not the walking exercises administered to the first group. The ''third'' and ''forth groups'' received no special treatment and differed only in that the third group’s progress was checked weekly and the forth was checked only at the end of the study. Observations on age (months) when the child began to walk are on the next slide.
 +
 +
<center>
 +
{| class="wikitable" style="text-align:center; width:30%" border="1"
 +
|-
 +
| Age_(months) || Treatment_Group
 +
|-
 +
| 9 || 1
 +
|-
 +
| 9.5 || 1
 +
|-
 +
| 9.75 || 1
 +
|-
 +
| 10 || 1
 +
|-
 +
| 13 || 1
 +
|-
 +
| 9.5 || 1
 +
|-
 +
|11 || 2
 +
|-
 +
|10 || 2
 +
|-
 +
|10 || 2
 +
|-
 +
|11.75 || 2
 +
|-
 +
|10.5 || 2
 +
|-
 +
|15 || 2
 +
|-
 +
|11.5 || 3
 +
|-
 +
|12 || 3
 +
|-
 +
|9 || 3
 +
|-
 +
|11.5 || 3
 +
|-
 +
|13.25 || 3
 +
|-
 +
|13 || 3
 +
|-
 +
|13.25 || 4
 +
|-
 +
|11.5 || 4
 +
|-
 +
|12 || 4
 +
|-
 +
|13.5 || 4
 +
|-
 +
|11.5 || 4
 +
|}
 +
</center>
 +
 +
<center>[[Image:SOCR_EBook_Dinov_ANOVA1_021708_Fig4.jpg|500px]]</center>
 +
 +
* '''Conclusion''': These data show that a child's true mean walking age is not statistically significantly different among any of the four treatment groups (p = 0.128546).
 +
 +
===ANOVA Conditions===
 +
ANOVA is valid if:
 +
* Design conditions: All groups of observations represent random samples from their respective populations. Also required is that the samples, and observations within each group, are independent of one another.
 +
 +
* Population conditions: The ''k'' population distributions must be approximately [[EBook#Chapter_V:_Normal_Probability_Distribution |Normal]]. Normality is less crucial if the sample sizes are large. Also it requires that the standard deviations of all populations are equal: <math>\sigma_1=\sigma_2=\sigma_3=\cdots = \sigma_k</math>. This condition can be slightly relaxed in practice to <math>0.5 \leq {\sigma_i \over \sigma_j} \leq 2</math>, for all ''i'' and ''j'', i.e., none of the population variances is twice larger than any of the other ones.
  
 
<hr>
 
<hr>
  
−
===References===
+
===[[EBook_Problems_ANOVA_1Way|Problems]]===
  
 
<hr>
 
<hr>
 
* SOCR Home page: http://www.socr.ucla.edu
 
* SOCR Home page: http://www.socr.ucla.edu
  
−
{{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=AP_Statistics_Curriculum_2007_ANOVA_1Way}}
+
"{{translate|pageName=http://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_ANOVA_1Way}}

Latest revision as of 11:35, 3 March 2020

General Advance-Placement (AP) Statistics Curriculum - One-Way Analysis of Variance (ANOVA)

In the two-sample inference chapter, we considered the comparisons of two independent group means using the Independent T-test. Now, we expand our inference methods to study and compare k independent samples. In this case, we will be decomposing the entire variation in the data into (independent/orthogonal) components - i.e., we'll be analyzing the variance of the data. Hence, this procedure called Analysis of Variance (ANOVA).

Motivational Example

Suppose 5 varieties of peas are currently being tested by a large agribusiness cooperative to determine which is best suited for production. A field was divided into 20 plots, with each variety of peas planted in four plots. The yields (in bushels of peas) produced from each plot are shown in two identical forms in the tables below.

Variety of Pea
A B C D E
26.2 29.2 29.1 21.3 20.1
24.3 28.1 30.8 22.4 19.3
21.8 27.3 33.9 24.3 19.9
28.1 31.2 32.8 21.8 22.1



A 26.2,24.3,21.8,28.1
B 29.2,28.1,27.3,31.2
C 29.1,30.8,33.9,32.8
D 21.3,22.4,24.3,21.8
E 20.1,19.3,19.9,22.1

Using the SOCR Charts (see SOCR Box-and-Whisker Plot Activity and Dot Plot Activity), we can generate plots that enable us to compare visually the yields of the 5 different types peas.

SOCR EBook Dinov ANOVA1 021708 Fig1.jpg

Using ANOVA, the data are regarded as random samples from k populations. Suppose the population means of the samples are \(\mu_1, \mu_2, \mu_3, \mu_4, \mu_5\) and their population standard deviations are\[\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5\]. We have 5 group means to compare. Why not just carry out \({5\choose 2}=10\) T-tests comparing all (independent) pairs of groups?

Repeated T-tests would mean testing null hypotheses of the type \(H_o: \mu_i = \mu_j, \forall i\not= j\). What is the problem with this approach? Suppose each test is carried out at \(\alpha = 0.05\), so a type I error is 5% for each test. Then, the overall risk of a type I error is larger than 0.05 and gets much larger as the number of groups (k) gets larger. To solve this problem, we need to make multiple comparisons with an overall error of \(\alpha = 0.05\) (or whichever level is specified initially).


The main idea behind ANOVA is that we need to know how much inherent variability exists is in the data before we can judge whether there is a difference in the sample means - i.e., presence of a grouping effect. To make an inference about means we compare two types of variability:

Variability between sample means
Variability within each group

It is very important that we keep these two types of variability in mind as we work through the following formulas. It is our goal to come up with a numerical recipe that describes/computes each of these variabilities.

One-Way ANOVA Calculations

Let's make the following notation: \[y_{i,j}\] = the measurement from group i, observation-index j.

k = number of groups

\[n_i\] = number of observations in group i

n = total number of observations, \(n= n_1 + n_2 + \cdots + n_k\)
The group mean for group i is\[\bar{y}_{i,.} = {\sum_{j=1}^{n_i}{y_{i,j}} \over n_i}\]
The grand mean is\[\bar{y}=\bar{y}_{.,.} = {\sum_{i=1}^k {\sum_{j=1}^{n_i}{y_{i,j}}} \over n}\]

To compute the difference between the means, we will compare each group mean to the grand mean.

  • SST (Sum Square due to Treatment, or Between Group variation): First, we describe the variation between the group means. For the independent T-test we described the difference between two group means as \(\bar{y_1} - \bar{y_1}\). In ANOVA we describe the difference between k means as sums of squares due to treatments (or between-group variance):
SST(Between) = \(\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}\). SST can be thought of as the difference between each group mean and the grand mean.
Degrees of Freedom: df (Between) = k – 1.
Mean Sum Square due to Treatment (Between)\[MST(Between) = {SST(Between)\over df(Between)}.\] This measures variability between the sample means.
  • SSE (Sum Square due to Error, or Within Group Variation): Second, we assess the within group variation. Recall that to measure the variability within a single sample we used\[\sqrt{\sum_{i=1}^n{(y_i - \bar{y})^2} \over n-1}\]. In ANOVA to describe the combined variation within the k groups we use sums of squares due to error (within-group variation):
SSE(Within) = \(\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}\), which can be thought of as the combination of variation within the k groups.
SSE(within)degrees of freedom: df (Within) = n - k
Mean variability within groups\[MSE(Within) = {SSE(Within)\over df(Within)}.\] This is a measure of variability within the groups.
  • Decomposition of Variance: Note that we have the following decomposition of the total variability in the data:
(Verbal): Deviation of an observation from the grand mean (Total variability) = Variation-Within + Variation_Between!
(Mathematics)\[y_{i,j} - \bar{y} = (y_{i,j} - \bar{y}_{i,.}) + (\bar{y}_{i,.} - \bar{y})\]. And summing up over all observations we get the desired ANOVA decomposition.
(ANOVA Decomposition)\[\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}} = \sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y}_{i,.})^2}} + \sum_{i=1}^{k}{n_i(\bar{y}_{i,.} - \bar{y})^2}\].
  • Interpretations:
This means SST(total) = SSE(within) + SST(between)
df(Total) = df(Within) + df(Between) = (n - k) + (k – 1) = n - 1

SOCR ANOVA Calculations

SOCR Analyses provide the tools to compute the 1-way ANOVA. For example, the ANOVA for the peas data above may be easily computed - see the image below. Note that SOCR ANOVA requires the data to be entered in this format:

Variety of Pea
Value Group-Index
26.2 A
24.3 A
21.8 A
28.1 A
29.2 B
28.1 B
27.3 B
31.2 B
29.1 C
30.8 C
33.9 C
32.8 C
21.3 D
22.4 D
24.3 D
21.8 D
20.1 E
19.3 E
19.9 E
22.1 E
SOCR EBook Dinov ANOVA1 021708 Fig2.jpg
F-Value = 23.966
P-Value = 2.2855121203368967E-6

ANOVA Table

Variance Source Degrees of Freedom (df) Sum of Squares (SS) Mean Sum of Squares (MS) F-Statistics P-value
Treatment Effect (Between Group) k-1 \(\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}\) \(MST(Between)={SST(Between)\over df(Between)}\) \(F_o = {MST(Between)\over MSE(Within)}\) \(P(F_{(df(Between), df(Within))} > F_o)\)
Error (Within Group) n-k \(\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}\) \(MSE(Within)={SSE(Within)\over df(Within)}\) F-Distribution Calculator
Total n-1 \(\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}}\) ANOVA Activity

ANOVA Hypotheses

The general form of the ANOVA hypotheses is: \[H_o: \mu_1=\mu_2=\mu_3 = \cdots = \mu_k\] \[H_a: \mu_i \not= \mu_j\] for some \(i\not=j\).

Note that \(H_o\) is compound hypothesis, when k > 2, so rejecting \(H_o\) doesn't tell us which \(\mu_i\)'s are different. It only tells us that some two are not equal.

  • Test Statistics: The test statistic\[F_o = {MST(Between) \over MSE(Within)}\]. If \(F_o\) is large, then there is a lot of between group variation, relative to the within group variation. Therefore, the discrepancies between the group means are large compared to the variability within the groups (error). Hence, large values of \(F_o\) provide strong evidence against \(H_o\).

Examples

Hands-on Example

This example is provided as a hands-on demonstration of the calculations performed in 1-way ANOVA. Clearly the value added by this example is conceptual understanding and appreciation of the methodological protocol for computing critical values and probabilities in ANOVA designs.

Groups
Index A B C
1 0 1 4
2 1 0 5
3 2
\(n_i\) 2 3 2
s 1 3 9
\(\bar{y}_i\) 0.5 1 4.5

Now use the ANOVA table to calculate the F-statistics and the corresponding p-value. Then use SOCR Analyses (ANOVA) to validate your manual calculations.

Variance Source Degrees of Freedom (df) Sum of Squares (SS) Mean Sum of Squares (MS) F-Statistics P-value
Treatment Effect (Between Group) 3-1 \(\sum_{i=1}^{k}{n_i(\bar{y}_{i,.}-\bar{y})^2}=19.86\) \({SST(Between)\over df(Between)}={19.86\over 2}\) \(F_o = {MST(Between)\over MSE(Within)}=13.24\) \(P(F_{(df(Between), df(Within))} > F_o)=0.017\)
Error (Within Group) 7-3 \(\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j}-\bar{y}_{i,.})^2}}=3\) \({SSE(Within)\over df(Within)}={3\over 4}\) F-Distribution Calculator
Total 7-1 \(\sum_{i=1}^{k}{\sum_{j=1}^{n_i}{(y_{i,j} - \bar{y})^2}}=22.86\) Anova Activity
SOCR EBook Dinov ANOVA1 021708 Fig3.jpg

Therefore, in this case we can reject the null hypothesis (at \(\alpha = 0.05\)).

Walking Age

Parents are frequently concerned when their child seems slow to begin walking. In 1972, Science reported on an experiment in which the effects of several different treatments on the age at which a child’s first walks were compared. Children in the first group were given special walking exercises for 12 minutes daily beginning at the age 1 week and lasting 7 weeks. The second group of children received daily exercises, but not the walking exercises administered to the first group. The third and forth groups received no special treatment and differed only in that the third group’s progress was checked weekly and the forth was checked only at the end of the study. Observations on age (months) when the child began to walk are on the next slide.

Age_(months) Treatment_Group
9 1
9.5 1
9.75 1
10 1
13 1
9.5 1
11 2
10 2
10 2
11.75 2
10.5 2
15 2
11.5 3
12 3
9 3
11.5 3
13.25 3
13 3
13.25 4
11.5 4
12 4
13.5 4
11.5 4
SOCR EBook Dinov ANOVA1 021708 Fig4.jpg
  • Conclusion: These data show that a child's true mean walking age is not statistically significantly different among any of the four treatment groups (p = 0.128546).

ANOVA Conditions

ANOVA is valid if:

  • Design conditions: All groups of observations represent random samples from their respective populations. Also required is that the samples, and observations within each group, are independent of one another.
  • Population conditions: The k population distributions must be approximately Normal. Normality is less crucial if the sample sizes are large. Also it requires that the standard deviations of all populations are equal\[\sigma_1=\sigma_2=\sigma_3=\cdots = \sigma_k\]. This condition can be slightly relaxed in practice to \(0.5 \leq {\sigma_i \over \sigma_j} \leq 2\), for all i and j, i.e., none of the population variances is twice larger than any of the other ones.

Problems

"-----

Translate this page:

(default)
Uk flag.gif

Deutsch
De flag.gif

Español
Es flag.gif

Français
Fr flag.gif

Italiano
It flag.gif

Português
Pt flag.gif

日本語
Jp flag.gif

България
Bg flag.gif

الامارات العربية المتحدة
Ae flag.gif

Suomi
Fi flag.gif

इस भाषा में
In flag.gif

Norge
No flag.png

한국어
Kr flag.gif

中文
Cn flag.gif

繁体中文
Cn flag.gif

Русский
Ru flag.gif

Nederlands
Nl flag.gif

Ελληνικά
Gr flag.gif

Hrvatska
Hr flag.gif

Česká republika
Cz flag.gif

Danmark
Dk flag.gif

Polska
Pl flag.png

România
Ro flag.png

Sverige
Se flag.gif

Retrieved from "https://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_ANOVA_1Way&oldid=16744"