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Applied Statistics Using SAS and SPSS Topic: One Way ANOVA By Prof Kelly Fan, Cal State Univ, East Bay

Statistical Tools vs. Variable Types Response (output) Predictor (input) NumericalCategorical/Mixed Numerical Simple and Multiple Regression Analysis of Variance (ANOVA) Analysis of Covariance (ANCOVA) CategoricalCategorical data analysis

Example: Battery Lifetime 8 brands of battery are studied. We would like to find out whether or not the brand of a battery will affect its lifetime. If so, of which brand the batteries can last longer than the other brands. Data collection: For each brand, 3 batteries are tested for their lifetime. What is Y variable? X variable?

Data: Y = LIFETIME (HOURS) BRAND 3 replications per level

Statistical Model “LEVEL” OF BRAND (Brand is, of course, represented as “categorical”) Y 11 Y 12 Y 1c Y ij Y 21 Y nI 1 2 n 1 2 C Y ij =  i +  ij i = 1,....., C j = 1,....., n Y nc

Hypotheses Setup H O : Level of X has no impact on Y H I : Level of X does have impact on Y H O :  1 =  2 =  8 H I : not all  j are EQUAL

ONE WAY ANOVA Analysis of Variance for life Source DF SS MS F P brand Error Total Estimate of the common variance  ^2 S = R-Sq = 59.67% R-Sq(adj) = 42.02%

Review Fitted value = Predicted value Residual = Observed value – fitted value

Diagnosis: Normality The points on the normality plot must more or less follow a line to claim “normal distributed”. There are statistic tests to verify it scientifically. The ANOVA method we learn here is not sensitive to the normality assumption. That is, a mild departure from the normal distribution will not change our conclusions much. Normality plot: normal scores vs. residuals

From the Battery lifetime data:

Diagnosis: Equal Variances The points on the residual plot must be more or less within a horizontal band to claim “constant variances”. There are statistic tests to verify it scientifically. The ANOVA method we learn here is not sensitive to the constant variances assumption. That is, slightly different variances within groups will not change our conclusions much. Residual plot: fitted values vs. residuals

From the Battery lifetime data:

Multiple Comparison Procedures Once we reject H 0 :   =   =...  c in favor of H 1 : NOT all  ’s are equal, we don’t yet know the way in which they’re not all equal, but simply that they’re not all the same. If there are 4 columns, are all 4  ’s different? Are 3 the same and one different? If so, which one? etc.

These “more detailed” inquiries into the process are called MULTIPLE COMPARISON PROCEDURES. Errors (Type I): We set up “  ” as the significance level for a hypothesis test. Suppose we test 3 independent hypotheses, each at  =.05; each test has type I error (rej H 0 when it’s true) of.05. However, P(at least one type I error in the 3 tests) = 1-P( accept all ) = 1 - (.95) 3 .14 3, given true

In other words, Probability is.14 that at least one type one error is made. For 5 tests, prob =.23. Question - Should we choose  =.05, and suffer (for 5 tests) a.23 OVERALL Error rate (or “a” or  experimentwise )? OR Should we choose/control the overall error rate, “a”, to be.05, and find the individual test  by 1 - (1-  ) 5 =.05, (which gives us  =.011)?

The formula 1 - (1-  ) 5 =.05 would be valid only if the tests are independent; often they’re not. [ e.g.,  1 =  2  2 =  3,  1 =  3 IF accepted & rejected, isn’t it more likely that rejected? ]

When the tests are not independent, it’s usually very difficult to arrive at the correct  for an individual test so that a specified value results for the overall error rate.

Categories of multiple comparison tests - “Planned”/ “a priori” comparisons (stated in advance, usually a linear combination of the column means equal to zero.) - “Post hoc”/ “a posteriori” comparisons (decided after a look at the data - which comparisons “look interesting”) - “Post hoc” multiple comparisons (every column mean compared with each other column mean)

There are many multiple comparison procedures. We’ll cover only a few. Post hoc multiple comparisons 1)Pairwise comparisons: Do a series of pairwise tests; Duncan and SNK tests 2)(Optional) Comparisons to control: Dunnett tests

Example: Broker Study A financial firm would like to determine if brokers they use to execute trades differ with respect to their ability to provide a stock purchase for the firm at a low buying price per share. To measure cost, an index, Y, is used. Y=1000(A-P)/A where P=per share price paid for the stock; A=average of high price and low price per share, for the day. “The higher Y is the better the trade is.”

} R=6 CoL: broker Five brokers were in the study and six trades were randomly assigned to each broker.

SPSS Output Analyze>>General Linear Model>>Univariate…

Homogeneous Subsets

Conclusion : 3, 1 2, 4, 5 Conclusion : 3, ???

Broker 1 and 3 are not significantly different but they are significantly different to the other 3 brokers. Broker 2 and 4 are not significantly different, and broker 4 and 5 are not significantly different, but broker 2 is different to (smaller than) broker 5 significantly. Conclusion : 3,

Comparisons to Control Dunnett’s test Designed specifically for (and incorporating the interdependencies of) comparing several “treatments” to a “control.” Example: Col } R=6 CONTROL

- Cols 4 and 5 differ from the control [ 1 ]. - Cols 2 and 3 are not significantly different from control. In our example: CONTROL

Exercise: Sales Data Sales

Exercise. 1.Find the Anova table. 2.Perform SNK tests at a = 5% to group treatments. 3.Perform Duncan tests at a = 5% to group treatments. 4.Which treatment would you use?

Post Hoc and Priori comparisons F test for linear combination of column means (contrast) Scheffe test: To test all linear combinations at once. Very conservative; not to be used for a few of comparisons.