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1.
State your research hypothesis in the form of a
relation between two variables.
2.
Find a statistic to summarize your sample data and
convert the above into statistical hypothesis:
Statistical hypothesis: r > 0
m1 - m2 < 0
3.
Set a straw man, i.e., null hypothesis
Null hypothesis:
r=0
4.
Set the alpha level and conduct the statistical test
with the assumption that the null is true.
5.
Make a decision with potential errors.
m1 - m2 = 0.
Sampling Distribution of a Statistic
Imagined and theoretical
X 
μ=72
Population
μ=72
Sampling Distribution

n
X 
Sample size N = 36
μ=72
μ=72
μ=72

n
X 
Sample Size N = 16
μ=72
μ=72

n
X 
Sample Size N = 36
μ=72
μ=72

n
Central Limit Theorem
The mean of the sampling distribution of
means (any statistic) equals the population mean
(any parameter).
The standard deviation of the sampling
distribution of means (any statistic) equals the
population standard deviation divided by the square
root of sample size. This is called the standard
error of means.
The sampling distribution of means is normal
independent of the pattern of the population
distribution, given a large enough sample size (e.g.,
n = 30)
An example:
Hypothesis: Chinese children today are overweight.
Choose a statistic: Mean weight
Past records: m = 50 lb;  = 30 lb
H1: m > 50 lb
H0: m = 50 lb
a<.01
n = 225 children ages 7 to 9; X  55
X  2

30
X 

2
n
225
zX 
X m
X

55  50
 2.5
2
Reject Null
μ=50
2.32
z X  2.5
X  55
Point estimate: X  55
Interval estimates: CI90
 X  1.64   X
 55  1.64  2
 51.72,58.28
-1.64
X  55
1.64
An example:
Hypothesis: Children’s weight differs from past.
Choose a statistic: Mean weight
Past records: m = 50 lb;  = 30 lb
H1: m
 50 lb
H0: m = 50 lb
a<.01; two tails, a<.01/2 or a<.005 at each tail
n = 225 children ages 7 to 9; X  55
X  2
-2.58
μ=50
z X  2.5 2.58
X  55
Null Hypothesis
Actually True
Decision NOT reject
Reject
Actually False

: Type II error
a: Type I error
 Power
H0: μ = 50
Reject Null
.05
z = 1.96
μ= 50
H1: μ > 50
power
β
H0: μ = 50
Reject Null
.01
z = 1.96
μ= 50
H1: μ > 50
power
β
Large N
H0: μ = 50
Reject Null
.05
μ= 50
z = 1.96
H1: μ > 50
power
β
Small N
H0: μ = 50
Reject Null
.05
μ= 50
z = 1.96
H1: μ > 50
power
β