AP Statistics Curriculum 2007 ANOVA 1Way
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[hide]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 anzlyzing 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.
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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 hull 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 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:
- 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\[y_{i,.} = {\sum_{j=1}^{n_i}{y_{i,j}} \over n_i}\]
- The grand mean is\[\bar{y}=y_{.,.} = {\sum_i=1}^k {\sum_{j=1}^{n_i}{y_{i,j}} \over n}}\]
To compute the difference between the means we will compare each group mean to the grand mean
Approach
Models & strategies for solving the problem, data understanding & inference.
- TBD
Model Validation
Checking/affirming underlying assumptions.
- TBD
Computational Resources: Internet-based SOCR Tools
- TBD
Examples
Computer simulations and real observed data.
- TBD
Hands-on activities
Step-by-step practice problems.
- TBD
References
- TBD
- SOCR Home page: http://www.socr.ucla.edu
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