Question

In: Statistics and Probability

A dolphin trainer wants to test the effect of food motivation on the ability for captive...

A dolphin trainer wants to test the effect of food motivation on the ability for captive dolphins to learn tricks.  She studies a sample of dolphins from Sea World and randomly assigns individuals to one of three food motivation conditions: low, moderate or high.  The level of food motivation corresponds to the type of food given to the dolphins during training: a low-preference food, a moderately-preferred food, or a highly-preferred food.  Dolphins are trained to jump through hoops in their pool and are rewarded with one of the three types of food.  The trainer then measures the number of hoops successfully jumped the day after the training session.

The data for the three food motivation conditions are below.

Low

Moderate

High

14

2

9

8

6

7

8

3

10

10

2

5

9

5

5

5

6

6

9

7

7

11

6

7

13

3

7

6

7

8

Complete the following calculations BY HAND:

a. State the null and alternative hypotheses and identify the critical value of F.

b.  Calculate SS Total, SS Between, and SS Within.  Also calculate df Total, df Between and df Within.

c.  Calculate MS Between and MS Within.  Calculate the observed F-ratio, and create an ANOVA source table for the analysis.  

d.  What is your decision regarding the null hypothesis?  State the results of the main ANOVA in nonstatistical terms.

e.  Calculate the Tukey critical value for the analysis.  Which food motivation conditions significantly differ from one another?

f. Write a very short paragraph (2 - 3 sentences) summarizing your ANOVA and Tukey results in APA format

Solutions

Expert Solution

a)

Hypothesis:

Ho: 1 = 2 = 3 = 4

Ha: At least one of the mean is different

Number of Treatment, t = 3 n = 30

Critical value:

Ftabulate = F,(t-1,n-t) = F0.05,(2,27) = 3.3541                            ..............Using F table

b)

CALCULATION:

Number of Treatment, t = 3 n = 30

T1 (sum of Low) = 93, T2(Sum of Moderate) = 47, T3(Sum of High) = 71

G = Grand Total = 211

CF = Correction Factor = G2/N = 2112 / 30 = 1484.03

= (14)2 + (2)2 + ......................+ (8)2 - 1484.03

TSS = 236.97

SSBetween = (1 / 10) * [(93)2 + (47)2 + (71)2] - 1484.03

SSTR = 105.87

SSE = TSS - SSBetween

= 236.97 - 105.87

SSE = 131.1

C)

MSSBetween = SSBetween/t-1 = 105.87 / 3-1 = 52.93

MSSE = SSE / n-t = 131.1 /30-3 = 4.86

F = MSSBetween / MSSE = 52.93 / 4.86 = 10.90

P-value = 0.0003                            ..............Using F table

ANOVA TABLE:

d)

CONCLUSION:

P-value < α, i.e. 0.0003 < 0.05, That is Reject Ho at 5% level of significance.

Therefore, At least one of the mean of these three food motivation conditions are different.


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