Question

In: Statistics and Probability

You are a public health nurse working in an elementary school in Hamilton, Ontario. You have...

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

    1. You hypothesize that there will be a difference among the groups in the ability to maintain attention. Run the appropriate statistical test to determine if your hypothesis is correct. Was your hypothesis confirmed by the study results? Provide the calculation and/or SPSS output to support your answer.  
    1. 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.

Solutions

Expert Solution

(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.


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