Question

Problem 1: A new program of imagery training is used to improve the performance of basketball players shooting free-throw shots. The first group did an hour imagery practice, and then shot 30 free throw basket shots with the number of shots made recorded. A second group
received no special practice, and also shot 30 free throw basket shots. The data are
below. Did the imagery training make a difference? Set alpha = .05.

Group 1: 15, 17, 20, 25 26, 27

Group 2: 5, 6, 10, 15, 18, 20

1. You must use all five steps in hypothesis testing:

1. Restate the question as a research hypothesis and a null hypothesis about the populations.
2. Determine the characteristics of the comparison distribution.
3. Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected.
4. Determine your sample’s score on the comparison distribution.
5. Decide whether to reject the null hypothesis.

2. Solve for and evaluate the effect size of this study using Cohen's D.

3. Create a 90% confidence interval for this problem.

Answer 1

1.
Given that,
mean(x)=21.6667
standard deviation , s.d1=5.0465
number(n1)=6
y(mean)=12.33
standard deviation, s.d2 =6.2823
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.571
since our test is two-tailed
reject Ho, if to < -2.571 OR if to > 2.571
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =21.6667-12.33/sqrt((25.46716/6)+(39.46729/6))
to =2.8381
| to | =2.8381
critical value
the value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 2.571
we got |to| = 2.83812 & | t α | = 2.571
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.8381 ) = 0.036
hence value of p0.05 > 0.036,here we reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.8381
critical value: -2.571 , 2.571
decision: reject Ho
p-value: 0.036
we have enough evidence to support the claim that imagery training make a difference of means of group 1 and 2.
2.
cohen's d size = mean difference / pooled standard deviation
pooled standard deviation = sqrt ((s.d1^2 +s.d2^2 )/2 )
pooled standard deviation = sqrt ((5.0465^2 +6.2823^2 )/2 ) =5.698
cohen'd size = (21.6667-12.333)/5.698 =1.638
large effect
3.
TRADITIONAL METHOD
given that,
mean(x)=21.6667
standard deviation , s.d1=5.0465
number(n1)=6
y(mean)=12.33
standard deviation, s.d2 =6.2823
number(n2)=6
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((25.467/6)+(39.467/6))
= 3.29
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 2.015
margin of error = 2.015 * 3.29
= 6.629
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (21.6667-12.33) ± 6.629 ]
= [2.708 , 15.966]
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DIRECT METHOD
given that,
mean(x)=21.6667
standard deviation , s.d1=5.0465
sample size, n1=6
y(mean)=12.33
standard deviation, s.d2 =6.2823
sample size,n2 =6
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 21.6667-12.33) ± t a/2 * sqrt((25.467/6)+(39.467/6)]
= [ (9.337) ± t a/2 * 3.29]
= [2.708 , 15.966]
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interpretations:
1. we are 90% sure that the interval [2.708 , 15.966] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion