21 students d each with a different caffeine e each of them a plain capuule cont

ID: 3367787 • Letter: 2

Question

21 students d each with a different caffeine e each of them a plain capuule containing of three drinks, are seven students assigned to each drink. You then that would be consumed in their designated drink y course t whether caffeine enhances test cest-taking abilities. You choose you randomly assign each student to one of three drinks and have them take an arthmetic best 1s n your study. y test scores Cola 2.9 85 86 82 75 water 13.4 92 87 x2141,479 G-1,715 N 21 k 3 Caffeine Content (mg/oz) 72 72 89 96 78 81 82 92 T-5379 T 619 Ss-381.43 SS 338.86 SS 185.71 n2-7 M 76.7143 M2 79.8571 M 88.4286 taking abilities. What is to test the impact of drinks with different caffeine contents on students' test You plan to use an ANOVA the null hypothesis? mean test scores for all three treatments a mean test are equal. O The popuiation O The popuiation O The population mean scores for all three treatments are not all equal. test scores for all three treatments are different mean test score for the water population is different from the population mean test score for the cola Calculate the degrees of freedom and the variances for the following ANOVA table: df MS 906.00 Total 1,420.67 The formula for the F-ratio is: using words (chosen from the dropdown menu), the formula for the F-ratio can be written as

Explanation / Answer

Solution

Let xij = The arithmetic test score of the jth student under ith drink; j = 1 to 7, i = 1 for water, 3 for cola and 3for coffee. Then the ANOVA model is: xij = µ + ?i + ?ij, where µ = common effect, ?i = effect of ith drink, and ?ij is the error component which is assumed to be Normally Distributed with mean 0 and variance ?2.

Null hypothesis: H0: ?1 = ?2 = ?3 = 0 Vs Alternative: HA: H0 is false [at least one ?i is different from others]

i.e., population mean test scores for all three treatments are equal. Option 1 ANSWER

ANOVA TABLE

[Given figures are in bold. Others are all derived]

Source

Between

514.67*1

2 *3

207.335 *5

Within

906.00

18 *4

50.333 *6

Total

1420.67

20 *2

710.335 *7

Explanations

*1: SS: Between + Within = Total or Between = Total - Within

*2: DF: Total: n (total number of observations) – 1 = 21 – 1 = 20

*3: DF: Between: (total number of treatments) – 1 = 3 – 1 = 2

*4: DF: Within: DF: Total - DF: Between = 20 – 2 = 18

*5, 6, 7 : MS = SS/DF ANSWER 2

In words, F = Treatment variation/Error variation. ANSWER 3

Using given data, F = 207.335/50.333 ANSWER 4

= 4.1183 ANSWER 5

Critical value (given ? = 0.05) = upper 5% point of F2, 18 = 3.55 ANSWER 6

Conclusion: Since 4.1183 > 3.55, the hypothesis can be rejected. Option 1 ANSWER 7

DONE