Question

A researcher hypothesizes that distractions occur more for those with greater prior exposure to sugars when sugary snacks are placed in their work environment. Essentially, the researchers were attempting to see if the presence or absence of sugary treats affected the amount of time to distraction for three groups: those with low, moderate and high exposure to sugar. They measured time to distraction by timing the participants in completing a computer task – thus, longer times meant more distraction, while shorter times meant less distraction. To test this, the researchers conducted a two-way ANOVA with replication, for which the data and output are shown in the file “Sugar”. The researchers conducted the research by setting α = 0.01. Identify the independent variables and their levels. Identify the dependent variable. Interpret the ANOVA output for significance at α = 0.05 for each of the following: the effect of exposure on completion time; the effect of the absence or presence of sugary snacks on completion time; and the interaction effect of snack availability and prior exposure to sugars on completion time. Provide a layperson interpretation of the findings in (b). What is the next step in this study?

Previous Exposure to Sugar Low Moderate High Sugar at Worksite
Absent 8 10 13 7 12 9 9 15 11 10 8 8 12 6 13 8 9 12
Present 5 15 15 8 9 12 5 7 15 6 6 16 5 4 12 7 11 14

a)

Since the data is not given, in proper format, re-arranging the data as:

 Sugary Snacks Completion time Absent Present Low 8 5 10 15 13 15 7 8 12 9 9 12 Moderate 9 5 15 7 11 15 10 6 8 6 8 16 High 12 5 6 4 13 12 8 7 9 11 12 14

Independent Variables:

Sugary Snacks - 2 Levels (Absent, Present)

Completion Time – 3 Levels (Low, Moderate, High)

Dependent Variable: Time to distraction

Null and Alternative Hypothesis:

We will have three hypotheses:

Completion time

H0: µLow = µModerate= µHigh

H1: Not all Means are equal

Sugary Snacks

H0: µAbsent = µPresent

H1: Not all Means are equal

H0: An interaction is absent

H1: Interaction is present

Alpha = 0.05

Degress of Freedom:

DfCompletion Time(A) = a-1 = 3-1 = 2

DfSugary Snacks (B) = b-1 = 2-1 = 1

df Completion Time * Sugary Snacks (A*B) = (a-1) * (b-1) = 2*1 = 2

df error = N – ab = 36 – 3*2 = 30

df total= N – 1 = 36 – 1 = 35

Decision Rule (3):

We have three hypotheses, so we have three decision rules:

Critical Values:

Completion Time (dfCompletion Time(A), df error): (2,30) = 3.32

Sugary Snacks (dfSugary Snacks (B),df error): (1,30) = 4.17

Interaction (df Completion Time * Sugary Snacks (A*B), df error) : (2,30) = 3.32

[Completion Time] If F is greater than 3.32, reject the null hypothesis

[Sugary Snacks] If F is greater than 4.17, reject the null hypothesis

[Interaction] If F is greater than 3.32, reject the null hypothesis

Test Statistics:

SSCompletion Time = ∑(∑ai)2/b*n - T2/N = 4.39

SSSugary Snacks = ∑(∑bi)2/a*n - T2/N = 1.78

SSCompletion Time*Sugary Snacks = ∑(∑ai * bi)2/n - ∑(∑ai)2/b*n - ∑(∑bi)2/a*n + T2/N = 7.39

SSTotal = ∑(Y)2 - T2/N = 400.22

SSError = SSTotal - SSCompletion Time - SSSugary Snacks - SSCompletion Time*Sugary Snacks = 386.67

MS = SS/df

F = MSeffect / MSerror

Hence,

FCompletion Time = 2.19/12.89 = 0.170

FSugary Snacks = 1.78/12.89 = 0.138

FInteraction = 3.69/12.89 = 0.287

 SS Df MS F Completion Time 4.39 2 2.19 0.170 Sugary Snacks 1.78 1 1.78 0.138 Interaction 7.39 2 3.69 0.287 Error 386.67 30 12.89 Total 400.22 35

Results:

[Completion Time] If F is greater than 3.32, reject the null hypothesis

Our F = 0.170, we retain the null hypothesis.

[Sugary Snacks] If F is greater than 4.17, reject the null hypothesis

Our F = 0.138, we retain the null hypothesis.

[Interaction] If F is greater than 3.32, reject the null hypothesis

Our F = 0.287, we retain the null hypothesis.

b)

Had we rejected the null hypothesis for any effect, we should have done a Post-Hoc Analysis.

PS: If the data is different, use the above approach to get the desired results.

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