Question

A dentist wants to know if the flavor of toothpaste affects how often children brush their teeth, and therefore the number of cavities they develop. She makes 3 different flavors of toothpaste: regular, bubble gum, and cherry. She randomly selects 9 children into one of three conditions and collects data about the number of cavities they have at a 60-month check-up. Here are the data (number of cavities detected at check-up): Regular Flavor Bubble Gum Flavor Cherry Flavor Child Cavities Child Cavities Child Cavities 8 4 2 7 3 951 8 2 6 6 3 9 1 s stokal - ss within 34. What is the value of the between-groups mean square? (3 points) Level 3 35. What is the value of the within-groups mean square? (3 points) 36. What is the Feb value? (2 points) Ms beteen Ms within - 1,5 37. What is the Ferit value, assuming a -0.05? (1 point) - 1 Foriti = 9-1=8 = n.or 38. What would you conclude? (1 point) - F C Fri retain the null - Why? the popution meens are 39. Using Tukey's HSD, determine if Group 1 (Regular Flavor) has significantly more cavities affrent than Group 3 (Cherry Flavor). (2 points) - need these n -n

Answer 1

The Summary Statistics are as below

	Regular	Bubble Gum	Cherry
Total	23	6	6
n	3	3	3
Mean	7.67	2.00	2.00
Sum Of Squares	4.6667	2	2
Variance	2.3334	1	1.000

N = 9, k = 3

dfBetween = k - 1 = 2, df Within = N - k = 9 - 3 = 6

Overall Mean = (23 + 6 + 6) / 3 = 35 / 9 = 3.89

SS Between = SUM [n * Mean - Overall Mean)²] = 4 * (7.67 - 3.89)² + 4 * (2 - 3.89)² + 4 * (2 - 3.89)² = 64.23

SS Within = SUM (Sum of squares) = 4.67 + 2 + 2 = 8.67

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(34) MS Between = SS between / df Between = 64.23 / 3.89 = 32.115

(35) MS Within = SS within / df Within = 8.67 / 6 = 1.444

(36) F + MS Between / MS within = 32.115 / 1.444 = 22.23

(37) Fcritical at $\alpha$ = 0.05, df Between = 2 and DF within = 6; Fcrit = 5.14

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(38) Conclusion: Since F test is > F critical. reject H0. There is sufficient evidence at the $\alpha$ = 0.05 level to conclude that at least one mean is different from the others.

(39) The Critical tukeys value for $\alpha$ = 0.05, df error = 6 and k = 3 is 4.34

HSD - Mi - MA MSE Vn

HSD (Regular Vs Cherry) = (7.67 - 2) / [Sqrt (1.444 / 3)] = 8.17

Since HSD (8.17) is > Tukeys critical value, we can conclude that there is a significant difference in the means.

This means the number of cavities for regular has more cavities than that of cherry flavor.

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