Exam 16: Analysis of Variance

In recent years a controversy has arisen in major league baseball in the US. Some players have been accused of 'doctoring' their bats to increase the distance the ball travels. However, a physics professor claims that the effect of doctoring is negligible. A major league manager decides to test the professor's claim. He doctors two bats by inserting cork into one and rubber into another. He then tells five players on his team to hit a ball with an undoctored bat and with the doctored bats. The distances are measured, and are listed below. Do these data provide sufficient evidence at the 5% level of significance to refute the professor's claim? Distance ball travels (feet) Player Undoctored bat Bat with cork Bat with rubber 1 275 265 280 2 315 335 320 3 425 435 440 4 380 375 370 5 450 460 450

The number of degrees of freedom for the numerator or MST is 3 and that for the denominator or MSE is 18. The total number of observations in the completely randomised design must equal 22.

In the one-way ANOVA where k is the number of treatments and n is the number of observations in all samples, the number of degrees of freedom for error is given by: A. $k-1$ . B. $n-k$ . C. $n-1$ . D. $n-k+1$ .

In a completely randomised design, 7 experimental units were assigned to the first treatment, 13 units to the second treatment, and 10 units to the third treatment. A partial ANOVA table for this experiment is shown below. Source of Variation SS df MS F Treatments * * * 1.50 Error * * 4 Total * * Test at the 5% significance level to determine whether differences exist among the three treatment means.

In recent years the irradiation of food to reduce bacteria and preserve the food longer has become more common. A company that performs this service has developed four different methods of irradiating food. To determine which is best, it conducts an experiment where different foods are irradiated and the bacteria count is measured. As part of the experiment the following foods are irradiated: beef, chicken, turkey, eggs, and milk. The results are shown below. Can the company infer at the 1% significance level that differences in the bacteria count exist among the four irradiation methods? Bacteria count Food Method 1 Method 2 Method 3 Method 4 Beef 47 53 36 68 Chicken 53 61 48 75 Turkey 68 85 55 45 Eggs 25 24 20 27 Milk 44 48 38 46

The test statistic of the single-factor ANOVA equals: A. sum of squares for treatments/sum of squares for error. B. sum of squares for error/sum of squares for treatments. C. mean square for treatments/mean square for error. D. mean square for error/mean square for treatments.

In one-way analysis of variance, if all the sample means are equal, then the: A. total sum of squares is zero. B. sum of squares for error is zero. C. sum of squares for treatments is zero. D. sum of squares for error equals sum of squares for treatments.

In order to examine the differences in age of teachers among five school districts, an educational statistician took random samples of six teachers' ages in each district. The data are listed below. $\text { Ages of teachers am ong five school districts }$ 1 2 3 4 5 42 23 45 31 37 22 52 48 45 40 33 32 50 33 45 32 44 45 38 61 26 39 37 29 50 24 33 49 46 50 Assume that ages of teachers are normally distributed. Use Tukey's multiple comparison method to determine which means differ.

Which of the following best describes the between-treatments in single-factor analysis of variance? A. Sum of squares for error. B. Total sum of squares. C. Sum of squares for treatments. D. All of these choices are correct.

A partial ANOVA table in a randomised block design is shown below. Source of Variation SS df MS F Treatments * 3 * * Blocks 1256 2 * * Error * * 67.67 Total 2922 11 Fill in the missing values (identified by asterisks) in the above ANOVA table.

Which of the following is not a required condition for one-way ANOVA? A. The populations are normally distributed. B. The population variances are equal. C. The samples are sel ected independently of each other. D. The population means are equal.

Which of the following best describes the distribution of the test statistic for ANOVA? A. Normal distribution and the distribution. B. Student t distribution. C. Chi-squared distribution. D. F distribution.

The analysis of variance (ANOVA) technique analyses the variance of the data to determine whether differences exist between the population means.

Consider the following ANOVA table: Source of Variation SS df MS F Treatments 128 4 32 2.963 Error 270 25 10.8 Total 398 29 The number of observations in all samples is: A. 25. B. 29. C. 30. D. 32.

Three tennis players, one a beginner, one experienced and one a professional, have been randomly selected from the membership of a large city tennis club. Using the same ball, each person hits four serves with each of five racquet models, with the five racquet models selected randomly. Each serve is clocked with a radar gun and the result recorded. Among ANOVA models, this setup is most like the simple regression model.

Then the mean square for error, MSE, equals SSE/26.

The following statistics were calculated based on samples drawn from four normal populations. Treatment Statistic 1 2 3 4 4 7 5 5 52 69 71 61 753 798 1248 912 Test at the 5% level of significance to determine whether differences exist among the population means.

In ANOVA, the between-treatments variation is denoted by SST, which stands for sum of squares of treatments.

A randomised block design experiment produced the following data. Treatment Block 1 2 3 1 8 17 5 2 5 10 12 3 9 14 13 4 12 11 13 5 12 14 16 a. Test to determine whether the treatment means differ. (Use $\alpha$ = 0.05) b. Test to determine whether the block means differ. (Use $\alpha$ = 0.05)

The objective of designing a randomized block experiment is to decrease the within-treatments variation to detect differences between the treatment means.