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Two Sample Hypothesis Tests
Comparative Studies
- In an experimental study , one or more factors are controlled so as to obtain information
about their influence on the variables of interest, and the influence of any uncontrolled factors is balanced through randomization.
- A completely randomized design in an experimental study in which the experimental units are assigned to the treatments on a completely random basis.
- A randomized block design is an experimental study in which the experimental units are grouped into homogeneous blocks and the treatments are randomly assigned to the units within each block.
- In an observational study , no experimental controls are exercised over factors influencing
the variables of interest.
- An independent samples design is an observational study in which statistically independent samples are drawn from the populations and comparative information about the populations is derived from a comparison of the samples.
- A matched samples design is an observational study in which the elements of the samples drawn from the two populations are carefully matched in pairs so that the two elements in each pair are as similar as possible with respect to the factors that might influence the variables of interest.
Hypothesis Tests for μ 1 =μ 2
- Null and alternative hypotheses
0 1 1 2
a :
H
H
or 0 1 2 1 2
a :^0
H
H
Independent Samples and Completely Randomized Designs
- Normal populations with equal variance
o When independent samples of sizes and are selected from normal populations with means
n 1 n 2
μ 1 and μ 2 and a common variance σ 2 , the sampling distribution of X 1 − X 2 is
normal with a mean μ X 1 − X 2 = μ 1 − μ 2 and a variance 21 2 2
1 2
σ X − X σ n n = +
. The test
statistic 1 2
1 2 X X
z^ X^ X σ (^) − = − has a standard normal distribution. The decision rule for a given
significance level α is
n 1 (^) n 2
X 1 X 1
Do not reject H 0 if− z (^) α 2 ≤ z ≤ z α 2
Reject H 0 if z < − z α 2 or z > z α 2.
o If the common population variance σ 2 is unknown, it is estimated with a pooled sample
variance, 2 2 2
p 2 s = n^ s^ n^ − s
2
or (^1 2 ) 2 2 (^2 1 ) 1 2
n n i i p i^ i
X X
s n n
= =
Then
1 2
2 1 2
sX (^) − X s (^) p n n = ^ +
The test statistic 1 2
1 X X
X X 2
s (^) −
t − has a t distribution with n 1 (^) + n 2 − 2 degrees of freedom.
The decision rule for a given significance level α is Do not reject H 0 if − t (^) α 2 , n 1 (^) + n 2 (^) − 2 ≤ t ≤ t α 2 , n 1 (^) + n 2 − 2
Reject H 0 if t < − t (^) α 2 , n 1 (^) + n 2 − 2 or t > t α (^) 2 , n 1 (^) + n 2 − 2.
o Example
A personnel training instructor of a retail company randomly assigned the trainees into two sections. In one section, she used a passive approach using PowerPoint presentations while giving lectures. In the other section, she used an active approach having trainees
approximately a normal distribution with a mean μ X 1 − X 2 = μ 1 − μ 2 and a variance
1 2 1 2
2 2 2 12 2 X X X X (^) n 1 (^) n 2
σ σ σ^ σ^ σ^2
− =^ +^ =^ +^. The test statistic 1 2
1 2 X X
z^ X^ X
= − has an approximate
standard normal distribution.
o If the population variances are unknown, σ 2 X 1 − X 2 is estimated with
1 2 1 1
2 2 2 2 2 1 2 1 2 X X X X s s s s^ s − (^) n n
Then the test statistic 1 2
1 X X
z X^ X^2 s (^) −
= − has approximately a standard normal distribution.
o The decision rule for a given significance level α is,
Do not reject H 0 if− z (^) α 2 ≤ z ≤ z α 2
Reject H 0 if z < − z α 2 or z > z α 2.
o Example
A market researcher randomly divided a sample of 500 households into two equal groups of and. One group was interviewed over telephone with the manual approach and the other group was interviewed with a computer assisted approach. Among other things, the researcher is interested in if the population means of interview times are different. The sample results are shown in the following table.
n 1 (^) = 250 n 2 (^) = 250
Approach Sample Mean (^) DeviationsStandard Sample Size Manual X 1 (^) = 15.3 s 1^ =^ 2.25 n 1^ =^250 Computer assisted X (^) 2 = 12.4 s 2 (^) = 3.36 n 2 (^) = 250
Test if there is any difference between the population mean times at a significance level
o Solution
The null and alternative hypotheses
0 1 1 2
a :
H
H
Because α = 0.05, 0.
α = , then
z (^) α 2 = z 0.025 = 1.96. The decision rule is
Do not reject H 0 if − 1.96 ≤ z ≤1. Reject H 0 if z < −1.96 or z >1.96.
1 2 1 1
2 2 2 12 22 2 2 1 2
X X X X 250 250
s s s s^ s − (^) n n
1 2
1 2 15.3^ 12.4^ 11.
X X 0.
z X^ X s (^) −
= −^ = − =
The conclusion is to reject H 0 because z >1.96.
Matched Samples and Randomized Block Design
- In this case, the test procedure simplifies to those for a single population mean.
- Let. Let represent the differences in the paired observations.
Then
D = X (^) 1 − X 2 D 1 (^) , D 2 ,..., Dn
μ D = μ 1 − μ 2.
- The sample statistics are
1
1 n i i
D D
n (^) =
2 2 1
n D i i
s D n (^) =
− ∑^
D
sD = (^) nsD.
- When the population difference D is normally distributed, the test statistic D
D
s t = has a
distribution with degrees of freedom. The decision rule for a given significance level
t
n − 1
α is
Do not reject H 0 if− t (^) α 2 , n − 1 ≤ t ≤ t α 2 , n − 1
Reject H 0 if t < − t (^) α 2 , n − 1 or t > t α (^) 2 , n − 1.
0 1 2 1 2
a :
H p p H p p
or 0 1 2 1 2
a :^0
H p p H p p
- If p 1 (^) = p 2 , the variance of p 1 (^) − p 2 , denoted by σ (^2) p 1 (^) − p 2 , is estimated by s^2 p 1 (^) − p 2 , while
1 2
2 1 2
(1 )^1
s (^) p − p p (^) p p (^) p n n = − ^ +
- p (^) p is the pooled sample proportion,
1 2 1 1 2 1 2 1 2 p^2 p X^ X^ n p^ n p n n n n
= +^ = +
where X (^) 1 is the number of successes in sample 1 and X (^) 2 is the number of successes in sample 2 while n 1 is the sample size of sample 1 and n 2 is the sample size of sample 2.
- If n 1 and n 2 are both large, the test statistic 1 2
1 2 p p
z p^ p s (^) −
= − has approximately a standard
normal distribution. The decision rule for a given significance level α is,
Do not reject H 0 if− z (^) α 2 ≤ z ≤ z α 2
Reject H 0 if z < − z α 2 or z > z α 2.
A market survey was carried out for a food company by a market research consultant. Independent samples of n 1 (^) = 250 and n 2 (^) = 300 consumers were selected in two cities. Each person in the samples was given two servings of breakfast cereal and asked which one tasted better. Unknown to the consumers, one serving was a new high protein breakfast cereal and the other one is an existing product. In the first sample, X 1 (^) = 160 indicated the new cereal tasted better while in the second sample X (^) 2 = 175 preferred the new cereal. At a
significance level of α = 0.02, test if the population proportion in the two cities are the same.
Null and alternative hypotheses 0 1 1 2
a :
H p p H p p
(^2) or 0 1 2 1 2
a :^0
H p p H p p
Because α = 0.02, 0.
α = , then
z (^) α 2 = z 0.01 = 2.33. The decision rule is
Do not reject H 0 if − 2.33 ≤ z ≤2. Reject H 0 if t < −2.33 or t >2.33.
1
p = = , 2 175 7 300 12 p = = , 160 175 67 p 250 300 110 p = + =
1 2
2 1 2
p p p p 110 110 250 300 s p p − n n
^
1 2
1 2
p p 0.
z p^ p s (^) −
The conclusion is to not reject H (^) 0 because−2.33 ≤ z ≤ 2.33.
