In earlier blogs we discussed about two sample z test,t test with equal and unequal variance.
Today we will talk about two sample paired t test and two sample p test.
Paired t Test:
- Two samples in which observations in one sample can be paired with observations in other sample.
- The samples are dependent, means values in one sample affect the other sample.
- Example: Blood sample before and after some medication.
Below is the formula to calulate the paired t test:
Where d1, d2, d3..dn are difference of two sample and find the mean and standard deviation of these differences.
Let’s solve one problem to get clear understanding of paired t test:
Problem: BP is measured for few patients before and after some medication. Are there any difference at 95% confidence level?
| Patient Name | Before Medication | After Medication |
| John | 115 | 116 |
| Harry | 110 | 108 |
| Jason | 145 | 140 |
| Bob | 130 | 128 |
| Andrew | 140 | 140 |
Solution:
Null Hypothesis: Mean of sample before medication = Mean of sample after medication
Alternate Hypothesis: Mean of sample before medication not equal to the mean of sample after medication.
| Patient Name | Before Medication | After Medication | differences |
| John | 115 | 116 | -1 |
| Harry | 110 | 108 | 2 |
| Jason | 145 | 140 | 5 |
| Bob | 130 | 128 | 2 |
| Andrew | 140 | 140 | 0 |
d-bar= mean of differences=1.6
standard deviation(s)=2.3
n=5
Put all the values in paired t Test formula
t calculated=1.6
t critical=2.766
t calculated < t critical meaning failed to reject the null hypothesis.
There is no significant difference in BP after specific medication.
Two sample p Test: Here we calculate the proportion and find the z statistics. If z calculated is gerater than z critical value, can reject the null hypothesis otherwise failed to reject the null hypothesis.
Let’s try to understand the concept solving one of the practical problem.
Problem: In a factory, we picked 100 samples that are produced by machine A in which 10 are defective. We also picked other 200 samples that are produced by machine B in which 30 are defective. Is there any significant difference between two machines at 95% confidence level?
Solution :
| Machine A | Machine B | Pooled | |
| Total Items | 100 | 200 | 300 |
| Defectives | 10 | 30 | 40 |
| Percentage defective | 0.1 | 0.15 | 0.13 |
p1-bar=0.1
p2-bar=0.15
p-bar=0.13
n1=100
n2=200
When we keep these values in the pooled formula mentioned above :
z calculated=1.2
z critical for 95% confidence level =1.96
z calculated < z critical
It means that we failed to reject the null hypothesis.
we can say that there is no significant difference between machine A and machine B.
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