Question

Recent results suggest that children with ADHD also tend to watch more TV than children who are not diagnosed with the disorder.  To examine this relationship, a researcher obtains a random sample of n = 36 children, 8 to 12 years old, who have been diagnosed with ADHD.  Each child is asked to keep a journal recording how much time each day is spent watching TV.  The average daily time for the sample is M = 4.9 hours.  It is known that the average time for the general population of 8 to 12 year-old children is µ = 4.1 hours with σ = 1.8.  

a.  Are the data sufficient to conclude that children with ADHD watch significantly           

     more   TV than children without the disorder?  Use a two-tailed test with α = .05.

1. State the Independent and Dependent Variables
2. State the Null Hypothesis in words and symbols.

b State the alternative Hypothesis in words and symbols

3. Compute the appropriate statistic.
4. What is the decision (Retain or Reject    
5. State the full conclusion in words

Effect

Direction

Size of Effect. (Cohen’s d) mean difference/standard deviation

b.  If the researcher had used a sample of n = 9 children and obtained the same

               sample mean, would the results be sufficient to reject H₀?

Answer 1

Given that,
population mean(u)=4.1
standard deviation, σ =1.8
sample mean, x =4.9
number (n)=36
null, Ho: μ=4.1
alternate, H1: μ>4.1
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 4.9-4.1/(1.8/sqrt(36)
zo = 2.667
| zo | = 2.667
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =2.667 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : right tail - ha : ( p > 2.667 ) = 0.004
hence value of p0.05 > 0.004, here we reject Ho
ANSWERS
---------------
1.
independent variables
2.
null, Ho: μ=4.1
alternate, H1: μ>4.1
1.
test statistic: 2.667
critical value: 1.645
2.
decision: reject Ho
p-value: 0.004
3.
we have enough evidence to support the claim that children with ADHD watch significantly   
more TV than children without the disorder.
a.
size of effect = mean difference/standard deviation
size of effect =(4.9-4.1)/1.8 =0.444
medium effect
b.
Given that,
population mean(u)=4.1
standard deviation, σ =1.8
sample mean, x =4.9
number (n)=9
null, Ho: μ=4.1
alternate, H1: μ>4.1
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 4.9-4.1/(1.8/sqrt(9)
zo = 1.333
| zo | = 1.333
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.333 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : right tail - ha : ( p > 1.333 ) = 0.091
hence value of p0.05 < 0.091, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=4.1
alternate, H1: μ>4.1
test statistic: 1.333
critical value: 1.645
decision: do not reject Ho
p-value: 0.091
we do not have enough evidence to support the claim that children with ADHD watch significantly   
more TV than children without the disorder