Math Section MW 1-2:30pm SR 117. Bekki George 206 PGH

Transcription

1 Math 3339 Section MW 1-2:30pm SR 117 Bekki George 206 PGH Office Hours: M 11-12:30pm & T,TH 10:00 11:00 am and by appointment

2 More than Two Independent Samples: Single Factor Analysis of Variance (more than 2 proportion means: µ 1,µ 2,µ 3,...,µ I ) Single-factor ANOVA focuses on comparison of more than two population or treatment means. ANOVA stands for Analysis of Variance. Assumptions: 1. Each group comes from an SRS 2. Each population is normally distributed 3. Assumed equal variances in all populations We are testing how much variance should be present from sampling error alone if H 0 is true.

3 Let M = the number of populations or treatments being compared (I>2) µ 1 = the mean of population 1 or the true average response when treatment 1 is applied µ M = the mean of population M or the true avg response when treatment M is applied The relevant hypotheses are: H 0 : µ 1 = µ 2 =... = µ M versus H a : at least two of the µ i 's are different Some new terms: M n i Grand Mean: X.. = 1 X ij N i=1 j=1 N = total number of observations in all M groups. Treatment Sum of Squares between groups: Error Sum of Squares: SSE = (X X ij i.)2 i=1 j=1 M n i SS( betw) = SSTr = M n (X. X..)2 1 i i=1

4 The F test statistic In order to calculate the F test statistic by hand, we also need the Mean Square for Treatments MSTr = SSTr M 1 Mean Square for Error MSE = SSE N M And F = MSTr MSE Now, F has two different degrees of freedom parameters: d1=m 1 and d2=n M

5 The ANOVA table:

6 ANOVA command in R Example: An experiment was conducted to measure and compare the effectiveness of various feed supplements on the growth rate of chickens. The data is in the chickwts data package. Open R Studio, make sure the package datasets is checked. Type: chickwts to see the data. Now let s make the model: > chick.lm=lm(weight~feed,data = chickwts) > summary(chick.lm) > anova(chick.lm)

7 Calcium is an essential mineral that regulates the heart, is important for blood clotting and for building healthy bones. The National Osteoporosis Foundation recommends a daily calcium intake of mg/day for adult men and women. While calcium is contained in some foods, most adults do not get enough calcium in their diets and take supplements. Unfortunately some of the supplements have side effects such as gastric distress, making them difficult for some patients to take on a regular basis. A study is designed to test whether there is a difference in mean daily calcium intake in adults with normal bone density, adults with osteopenia (a low bone density which may lead to osteoporosis) and adults with osteoporosis. Adults 60 years of age with normal bone density, osteopenia and osteoporosis are selected at random from hospital records and invited to participate in the study. Each participant's daily calcium intake is measured based on reported food intake and supplements. The data are shown below. Normal Bone Density Osteopenia Osteoporosis Is there a statistically significant difference in mean calcium intake in patients with normal bone density as compared to patients with osteopenia and osteoporosis?

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9 Popper 18 Suppose you are carrying out a test of significance at the α = 0.01 level and your null hypothesis is H 0 : µ 1 = µ 2 = µ 3. You find only a portion of the ANOVA table for this test: Source of Variation df Sum of Squares Mean Square Treatments Error Total What is the value of f? a b c d. cannot be determined 2. Conclusion? a. Reject H 0 b. Fail to reject H 0

10 Suppose you are carrying out a test of significance at the α = 0.05 level and your null hypothesis is H 0 : µ 1 = µ 2 = µ If each of the three treatments has 5 levels, what is the value of F*? a b c d e. none of these Now suppose (for the same problem) you find only a portion of the ANOVA table Source of Variation df Sum of Squares Mean Square Treatments Error Total What is the value of f? 5. Conclusion? a a. Reject H 0 b b. Fail to reject H 0 c d. cannot be determined

11 Another example: The data in the accompanying table resulted from an experiment run in a completely randomized design in which each of four treatments was replicated five times. Total Mean Group Group Group Group All Groups

12 Part of the resulting ANOVA table is Source SS DF MS Treatments Error Complete the ANOVA table. Perform a significance test to see if at least two of the µ i 's are different.

13 Tukey s Method (The T Method) The T-method is used to determine which pair (or pairs) of means differ significantly.

14 Let s use this method on previous example to determine which pairs differ significantly:

15 Pairwise t test An alternative to Tukey s method (and the most common method used) is to do a t-test for each pair of means. There is an issue with this though, our P(Type 1 error) becomes greater than α for multiple comparisons. So we need to do a method of adjustments to reduce the significance level of the pairwise test enough so that the probability of one or more type I errors in the whole set of comparisons is less than α. The Bonferroni Method uses α/k instead of α for the testing of k comparisons. R code: >pairwise.t.test(data,categories,"bonferroni")

16 Let s revisit this example: Example: An experiment was conducted to measure and compare the effectiveness of various feed supplements on the growth rate of chickens. The data is in the chickwts data package. Open R Studio, make sure the package datasets is checked. Type: chickwts to see the data. Now let s make the model: > chick.lm=lm(weight~feed,data = chickwts) > summary(chick.lm) > anova(chick.lm) > pairwise.t.test(chickwts$weight,chickwts$feed,"bonferroni")

17 Popper One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples and calculate the given summary statistics of the boys and the girls for comparable math classes. In Calculus, 15 boys had a mean percentage of 82.3 with standard deviation of 5.6 while 12 girls had a mean percentage of 81.2 with standard deviation of 6.7. Which of the following would be the most appropriate test for establishing whether boys do better in math classes than girls? A. ANOVA test B. two-sample t-test for means C. two-sample z-test for proportions D. none of these tests would be appropriate

18 6. Suppose we have two SRSs from two distinct populations and the samples are independent. We measure the same variable for both samples. Suppose both populations of the values of these variables are normally distributed but the means and standard deviations are unknown. For purposes of comparing the two means, we use a. Two-sample t procedures b. Matched pairs t procedures c. z procedures d. none of these