Paired Z test

AIM: To compare a means of paired observations of a sample.

To test whether a mean is significantly different / more / less than a mean of it's own paired observation (e.g. before and after studies).


Paired Z test compares means of two observations recorded from a single unit or individual. Examples of such "paired" observations can be
1. Observations recorded at different times (Pre-test and Post-test scores) from same individuals
2. Observations recorded under two different conditions (Resting pulse rate and during exercise pulse rate)
3. Observations recorded from two halves of an individul or unit (e.g. Fairness in right half and left half of face).

Steps:

1. Null hypothesis and Alternate Hypothesis

Null Hypothesis

A. Two tailed test: Average difference in paired observations is equal to average expected difference. OR Average difference in paired observations is not significantly different than average expected difference.

B. One tailed test: (Right tailed): Average difference in paired observations is not more than average expected difference. OR Average difference in paired observations is equal to or less than average expected difference.

C. One tailed test: (Left tailed): Average difference in paired observations is not less than average expected difference. OR Average difference in paired observations is equal to or more than average expected difference.

 Alternate Hypothesis

A. Two tailed test: Average difference in paired observations is significantly different than average expected difference.

B. One tailed test: (Right tailed): Average difference in paired observations is significantly more than average expected difference.

C. One tailed test: (Left tailed): Average difference in paired observations is significantly less than average expected difference.

2. Calculate Test statistics t

Paired Z test

where Z is the test statistics,
x̄ = Average difference in paired observations,
μ = Average expected difference,
s = Standard deviation,
N = Sample Size

3. Know p value from Z table

4. Interpret

If p < = alpha, then reject Null hypothesis, and accept alternate hypothesis.

If p > alpha, then the study has failed to reject Null hypothesis. So accept Null hypothesis.


Confidence Intervals

1. For two tailed test:

100 - alpha % confidence interval = x̄ ± critical_Z_value * SE mean
where critical_Z_value is the Z table value corrsponding to 2 tailed test and given alpha.
SE mean = sqrt (s2 / N)

1. For one tailed test:

A. Right tailed test (AH = Sample mean is significantly more than hypothesized mean)

100 - alpha % confidence interval = x̄ - critical_t_value * SE mean to infinity.
where critical_Z_value is the Z table value corrsponding to 1 tailed test and given alpha.

B. Left tailed test (AH = Sample mean is significantly less than hypothesized mean)

100 - alpha % confidence interval = minus Infinity to x̄ + critical_t_value * SE mean.
where critical_Z_value is the Z table value corrsponding to 1 tailed test and given alpha.


Example 1:

Following is the data of pre-test and post-test scores of 30 participants, recorded before and after a training session. Apply an approriate statistical test to test whether there is satistically significant difference in test scores, at alpha level of 5%.


Participant No Pre-test score Post-test score
11015
21213
3911
4810
51511
61314
71111
81212
9109
10815
11910
12109
1388
141010
15912
161011
171012
181014
191012
201014
21910
22109
2388
241110
25912
261211
271112
281114
29912
301214

These are 30 paired values, collected from 30 individuals. As N = 30, appropriate statistical test will be "Paired Z test."
Here,
x̄ = 1.3 (Average of difference between paired observations)
s = 2.1838
μ = 0 (Signifying, Null Hypothesis of no change in the paired observations)
N = 30
alpha = 5%
tails = 2 (significantly different)

Putting above values at the respective places, we get the output as follows:

p = 0.0011. P value is statistically significant at given alpha level of 0.05 (5%). Average difference in paired observations is significantly different than average expected difference.

Avearge difference between pairs ± SD = 1.3 ± 2.1838
Expected difference = 0
Standard Error = 0.398705507026594
Z = 3.26055190382231
p = 0.00111207780188916 (Two tailed)
Alpha = 0.05 (5 %). Critical Z value = 1.95996398451889
95 % Confidence interval = 0.518551565798531 to 2.08144843420147.

Example 2:

For above data, apply an approriate statistical test to test whether post-test score is significantly more than pre-test scores, at alpha level of 5%.

These are 30 paired values, collected from 30 individuals. As N = 30, appropriate statistical test will be "Paired Z test."
Here,
x̄ = 1.3 (Average of difference between paired observations)
s = 2.1838
μ = 0 (Signifying, Null Hypothesis of no change in the paired observations)
N = 30
alpha = 5%
tails = 1 (significantly more)

Putting above values at the respective places, we get the output as follows:

P = 0.0006. P value is statistically significant at given alpha level of 0.05 (5%). Average difference in paired observations is significantly more than average expected difference.

Avearge difference between pairs ± SD = 1.3 ± 2.1838
Expected difference = 0
Standard Error = 0.398705507026594
Z = 3.26055190382231
p = 0.000556038900944578 (One tailed)
Alpha = 0.05 (5 %). Critical Z value = 1.64485362699727
95 % Confidence interval = 0.644187800663522 to ∞.


@ Sachin Mumbare