Question

1. 
   
2. You are a public health nurse working in an elementary school in Hamilton, Ontario. You have seen a substantial increase in the number of children in your school presenting with Attention Deficit Disorder (ADD). You are interested in whether a new drug developed by a local drug company improves the ability of children with ADD to maintain attention. You obtain ethics approval to study the drug, informed consent from the parents of 24 fifth grade children with ADD, administer the drug to the 24 children, and administer a test to the children to determine their ability to maintain attention while performing a task. Scores are continuous, can range from 0 to 25 with higher scores reflecting a better ability to maintain attention, and you know from previous research that scores tend to be normally distributed. Six of these students receive a placebo containing none of the drug. Six students receive 2 mg of the drug, six students receive 4 mg of the drug, and six students receive 6 mg of the drug. The table below summarizes the results from your study. Use these data to answer the following questions. Assume α=0.05.

3. Placebo (0 mg)	Drug (2 mg)	Drug (4 mg)	Drug (6 mg)
   4	7	16	17
   7	8	14	18
   11	13	12	13
   11	6	11	17
   7	9	15	20
   10	9	13	15

4. a.You have one further prediction, and that is that children with ADD who receive the drug will perform significantly better than children with ADD who do not receive the drug. Conduct the appropriate test to confirm this prediction, if the results from question a support doing this additional testing. Are you able to confirm your prediction? Explain, including the calculations and/or SPSS output to support your answer. [2 Marks]

Answer 1

Null and Alternative Hypothesis:

H₀: µ_Placebo = µ_{Drug 2mg} = µ_{Drug 4mg}= µ_{Drug 6mg}

H₁: Not all Means are equal

Alpha = 0.05

N = 24

n = 6

Degress of Freedom:

df_Between = a – 1 = 4-1 =3

df_Within = N-a = 24-4 = 20

df_Total = N-1 = 24-1 = 23

Critical Values:

Time (df_Between, df_Within): (3,20) = 3.10

Decision Rule:

If F is greater than 3.10, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 290.46

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 115.50

SS_Total = SS_{Betwen ­­}+ SS_Within = 405.96

MS = SS/df

               

F = MS_effect / MS_error

Hence,

F = 96.82/5.78 = 16.77

Source of Variation	SS	df	MS	F
Between Groups	290.46	3	96.82	16.77
Within Groups	115.50	20	5.78
Total	405.96	23

Result:

Our F = 16.77, we reject the null hypothesis

Conclusion:

Not all means are equal.

Post- Hoc Test:

I will be using Tukey HSD (Honestly Significant Difference):

HSD = q *Sqrt (MS _Within /n)          

q is calculated from q table using (No of Groups vs degrees of freedom of error)

where q is 3.96,

Hence, HSD = 3.89

µ_Placebo = 8.33

µ_{Drug 2mg} = 8.67

µ_{Drug 4mg} = 13.50

µ_{Drug 6mg} = 16.67

Pairs:

Placebo and Drug 2mg

Mod(µ_Placebo - µ_{Drug 2mg})= mod(8.33 – 8.67) = 0.33 which is less than HSD, hence they are not different

Placebo and Drug 4mg

Mod(µ_Placebo - µ_{Drug 4mg})= mod(8.33 – 13.5) = 5.17 which is greater than HSD, hence they are different

Placebo and Drug 6mg

Mod(µ_Placebo - µ_{Drug 6mg})= mod(8.33 – 16.67) = 8.33 which is greater than HSD, hence they are different

Drug 2mg and Drug 4mg

Mod(µ_{Drug 2mg} - µ_{Drug 4mg})= mod(8.67 – 13.5) = 4.83 which is greater than HSD, hence they are different different

Drug 2mg and Drug 6mg

Mod(µ_{Drug 2mg} - µ_{Drug 6mg})= mod(8.67 – 16.67) = 8 which is greater than HSD, hence they are different different

Drug 4mg and Drug 6mg

Mod(µ_{Drug 4mg} - µ_{Drug 6mg})= mod(13.5 – 16.67) = 3.17 which is less than HSD, hence they are not different

So, Students with Drug 4mg and Drug 6mg perform better than Placebo