Winter, 2010 Tuesday, March 2Stat 217 – Day 24Analysis of Variance (ANOVA)Last Time: Chi-square Procedures (Topic 25)The “chi-square” test can be applied anytime you have a two way table. When it is a 2 × 2 table (and thealternative is two sided), it is equivalent to the two-sample z-test for two proportions. But chi-square tests also allow you to - Compare more than two population proportions (H0: -1 = -2 = … = -I) when you have collected datafrom several independent random samples- Compare two or more population distributions on a categorical response variable (H0: all the population distributions are the same) when you have collected data from several independent random samples (Activity 25-2)- Compare two or more treatment groups on a categorical response variable (H0: all treatment effects are the same or the probabilities of success are the same for each treatment) when you have a randomized comparative experiment (Activity 25-3)- Decide whether two categorical variables are associated (H0: no association between the two variables) when you have a random sample and simultaneously record two variables on each observational unit (Activity 25-4)The main thing for you to worry about is deciding when to use the chi-square test and how to draw legitimate conclusions (significance, causation, and what population you are willing to generalize to).Follow-up Analysis: If you want to say a bit more than “there is a difference somewhere,” you can examine the “cell contributions” (the bottom row in each cell of the Minitab output, find 1-2 that are larger than the others and then compare the observed counts to the expected counts to discuss the “direction” of the difference(s). (See also part d of Activity 25-2.)Example: A study published in 1990 randomly assigned 70 undergraduates to watch a videotaped job interview and to rate the candidate’s qualifications on a 1-10 scale. The videos used the same actors andscript each time but the job applicant appeared with different disability: no disability, leg amputation, crutches, hearing impairment, and wheelchair confinement. (a) Identify the observational units and sample:Identify and classify the explanatory variable:Identify and classify the response variable:(b) Let -nd represent the “population” mean rating for the no disability group. Similarly for -leg, -crutches, -hearing, -wheelchair.. Analogous to what we learned yesterday, state a null hypothesis and an alternative hypothesis about these parameters.Ho:Ha:Winter, 2010 Tuesday, March 2Now let’s look at the data. Below are descriptive statistics for this sample.Numerical Summaries Graphical Summary (Boxplots)n mean SDno disability 14 4.90 1.79leg amputation 14 4.43 1.59Crutches 14 5.92 1.48hearing impairment 14 4.05 1.53wheelchair confinement 14 5.34 1.75(c) Briefly describe what the numerical and graphical summaries reveal about how these 5 treatments compare. Also, what symbols could we use to refer to these means and standard deviations?Statistical Inference: Again we want to ask are the observed differences in the sample means large enough to conclude that the population treatment means differ, or are the observed differences what we might see by chance under the null hypothesis?(d) The figure to the right shows the mean score given to each disability in the study. Construct hypothetical boxplots around these means, such that you believe a statistical analysis would reveal a statistically significant difference among the five disabilities. (I’m just asking for a rough idea of thedistribution, in comparison to the next question.)(e) Repeat part (d) where you would be fairly certain the differences between the five disabilities would not be statistically significant.To measure the collective “discrepancy” between the five group means, we will look at how much they deviate from the overall mean, standardizing by how much variability there is in the data values themselves. Your bottom figure should have much wider distributions, much larger s values, than the top figure, which should not such more overlap between the distributions.Winter, 2010 Tuesday, March 2This leads to a rather complicated test statistic formula (There are more details in Topic 33 online):Test Statistic =INsnIxxnyvariabilitgroupwithinyvariabilitgroupbetweenFiiiiIiii1212)1(1)(But you should notice:- The numerator is larger when you have bigger differences (variances) between the group means- The denominator is larger when you have more “within group” variability. This represents our “random variation.”- This test is large when the variability between the groups is larger than our random variation.So again our p-value will just be how often randomness alone (assuming the null hypothesis is true) would lead to a test statistic value this large or larger. This p-value comes from an “F distribution” which is skewed to the right just like the Chi-square distribution.Minitab output: One-way ANOVA: SCORE versus DISABILITYSource DF SS MS F PDISABILITY 4 30.52 7.63 2.86 0.030Error 65 173.32 2.67Total 69 203.84(f) Report the test statistic and p-value from this output.(g) What conclusions do you draw in terms of significance, causation, and generalizability.Note: Whenever you reject H0, can do follow-up analyses (“multiple comparison procedures”) to decide which means are different. For now, describe the differences exhibited in the boxplots and dotplots.Example 2: Open the “ANOVA” applet from the Stat 217 Applets page. Press the Draw Samples button.This select three samples at random from three populations, each with mean - = 2.0 and standard deviation - = 1.1.(a) Should you expect a large or a small p-value in this case?(b) Continue to press the Draw Samples button until you have a fairly large p-value (greater than .8). Now, press the right or left arrow on one of the slider bars to change the value of that population mean. This will shift over the sample data that you already had accordingly. What do you notice about how the p-value changes? (EC: In the ANOVA table, which values are changing?)Winter, 2010 Tuesday, March 2(c) Put the slider back to 2 and now slowly decrease the population standard deviation -. What do you notice about how the p-value changes? (EC: In the ANOVA table, which values are changing?)Technical conditions: To compare several sample means, we really need