In: Statistics and Probability
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.
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 |
(a)
H0: Null Hypothesis: ( there will be no difference among the groups in the ability to maintain attention).
HA: Alternative Hypothesis: (At least one mean is different from other 3 means) ( there will be a difference among the groups in the ability to maintain attention) (Claim)
From the given data, the following statistics are calculated:
Placebo (0 mag) | Drug (2 mg) | Drug (4 mg) | Drug (6 mg) | Total | |
N | 6 | 6 | 6 | 6 | 24 |
50 | 52 | 81 | 100 | 283 | |
Mean | 50/6 = 8.3333 | 52/6 = 8.6667 | 81/6 = 13.5 | 100/6 = 16.6667 | 283/24= 11.792 |
456 | 480 | 1111 | 1696 | 3743 | |
Std. Dev. | 2.8048 | 2.4221 | 1.8708 | 2.4221 | 4.2012 |
.From the above Table, ANOVA Table is calculated as follows:
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F Stat |
Between treatments | 290.4583 | 3 | 290.4583/3 = 96.8194 | 96.8194/ 5.775 = 16.7653 |
Within treatments | 1115.5 | 20 | 1115.5/20 = 5.775 | |
Total | 405.9583 | 23 |
F stat = 96.8194/ 5.775 = 16.7653
Degrees of Freedom for numerator = 3
Degrees of Freedom for denominator = 20
By Technology, p - value = 0.000011
Since p - value = 0.000011 is less than = 0.05, the difference is significant. Reject null hypothesis.
Conclusion:
The data support the claim that there will be a difference among
the groups in the ability to maintain attention.
(b)
H0: Null Hypothesis: ( children with ADD who receive the drug will not perform significantly better than children with ADD who do not receive the drug. )
HA: Alternative Hypothesis: ( children with ADD who receive the drug will perform significantly better than children with ADD who do not receive the drug. ) (Claim)
From the given data, the following statistics are calculated:
n1 = 6
1 = 50/6 = 8.3333
s1 = 2.8048
n2 = 18
2 = 233/18 = 12.9444
s2 = 3.9922
= 0.05
Pooled Standard Deviation is given by:
Test Statistic is given by:
= 0.05
df = 6 + 18 - 2 = 22
From Table, critical value of t = -1.717
Since calculated value of t= - 2.6047 is less than critical value of t = - 1.717, the difference is significant. Reject null hypothesis.
Conclusion:
The data support the claim that children with ADD who receive the drug will perform significantly better than children with ADD who do not receive the drug.
By the above analysis, we are able to confirm your prediction that there will be a difference among the groups in the ability to maintain attention and children with ADD who receive the drug will perform significantly better than children with ADD who do not receive the drug.