Download Chapter 24: One-Way Analysis of Variance: Comparing Several

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Bootstrapping (statistics) wikipedia , lookup

Psychometrics wikipedia , lookup

History of statistics wikipedia , lookup

Taylor's law wikipedia , lookup

Degrees of freedom (statistics) wikipedia , lookup

Misuse of statistics wikipedia , lookup

Resampling (statistics) wikipedia , lookup

Time series wikipedia , lookup

Student's t-test wikipedia , lookup

Analysis of variance wikipedia , lookup

Transcript
CHAPTER 24 One‐Way Analysis of Variance: Comparing Several Means One‐Way Analysis of Variance Excel can perform a one‐way analysis of variance test to compare means of different populations. The response variable must be numeric. The data should be entered with each population in separate columns (or rows) on a worksheet. To perform a one‐way analysis of variance with stacked data, choose Data h Data Analysis h ANOVA: Single Factor from the Excel menu. Our example comes from a study conducted to compare three educational approaches (basal, DRTA, and strategies) to improve children’s reading comprehension. We want to test the null hypothesis that the three groups represent three populations that all have the same mean score on the pretest. The data are given in CA14_001.MTW. The data are arranged with one row for each student. The teaching method is given in one column and the student’s pretest score given in a second column. To analyze these data, select Stat h ANOVA h One‐way from the menu. In the dialog box enter the columns containing the test scores in the Input Range box, check the Chart Titles box, indicate where the output should go, click OK. 122 Chapter 24: One‐Way Analysis of Variance: Comparing Several Means It is important to check that the assumptions of one‐way analysis of variance are satisfied. Specifically, the populations are normal with possibly different means and the same variance. A histogram of the variables serves to detect outliers or extreme deviations from Normality. Compute the ratio of the largest to the smallest sample standard deviation. If this ratio is less than 2 and the histograms are satisfactory, the assumptions of ANOVA are satisfied. Call the mean test scores for the three educational approaches μ 1 , μ 2 , and μ 3 . We want to test the null hypothesis that there are no differences among the test scores for the three groups: H0: μ 1 = μ 2 = μ 3 The alternative is that there is some difference. Ha: not all of μ 1 , μ 2 , and μ 3 are equal Chapter 24: One‐Way Analysis of Variance: Comparing Several Means 123
The output provides the ANOVA table. The columns in this table are labeled Source of Variation, SS (sum of squares), df (degrees of freedom), MS (mean square), F, P‐value, and F crit. The rows in the table are labeled Between Groups, Within Groups, and Total. Consider our model DATA = FIT + RESIDUAL The Between Groups row corresponds to the FIT term, the Within Groups row corresponds to the RESIDUAL term, and the Total row corresponds to the DATA term. Notice that both the degrees of freedom and the sum of squares add to the value in the Total row. The pooled standard deviation can be computed from the ANOVA table using the sum of squares and degrees of freedom for the Within Groups row. That is, s 2p =
SS 572.45
=
= 9.09 DF
63
which implies that s p = 3.014. The F statistic is given in the ANOVA table. If H 0 is true, the F statistic has an F(DFG,DFE) distribution, where DFG stands for degrees of freedom for groups and DFE stands for degrees of freedom for error. DFG = I − 1, the number of groups minus 124 Chapter 24: One‐Way Analysis of Variance: Comparing Several Means 1. DFE = N − I , the number of observations minus the number of groups. The P‐value for this distribution is also given above. In this example, the P‐value is given as 0.329. A P‐value this large does not give us evidence against the hypothesis that the means are all equal. For information purposes, the output from one‐way analysis of variance provides the mean and variance for each group. Individual 95% confidence intervals for the means are of the form ⎛
s
s
⎜ x − t* p ,x + t* p
i
i
⎜
ni
ni
⎝
⎞
⎟ ⎟
⎠
where xi and ni are the sample mean and sample size for level i, s p Pooled StDev is the pooled estimate of the common standard deviation, and t* is the value from a t table corresponding to 95% confidence and the degrees of freedom for within groups. Alternatively, the exact same results can be obtained by selecting Add‐Ins hWHFStat h Analysis of Variance ‐ ANOVA from the Excel menu. Select the radio button for One way Analysis of Variance and fill in the dialog box as shown below. Chapter 24: One‐Way Analysis of Variance: Comparing Several Means 125
As another example, retrieve the data from Exercise 24.35 from the files and arrange it as shown below. Select Data > Data Analysis >Anova: Single Factor and fill in the dialog boxes as shown: We can see from this that the untreated fabric is the strongest. Now we compare the three treatments. Here note that the df = 2 or 12, F = 5.32 and the P‐value = 0.026. Fabrics treated with Permafresh 55 have a higher degree of strength than those treated with the others. 126 Chapter 24: One‐Way Analysis of Variance: Comparing Several Means Selected exercises. Try the following exercises using Excel. 24.9 Fertilizing bromeliads 24.13 Exercise and weight loss 24.33 Does nature heal best? 24.39 The results for 2001