Student t-test

Student t-testUnit 8 strand

Student’s t-test

“Student” was W. S. Gossett. He published his test anonymously as ‘Student’ because he was working for the brewer’s Guinness and had to keep the fact they were suing statistics a secret

Student’s t-test

The test is used to compare samples from two different batches.

It is usually used with small (<30) samples that are normally distributed.

There are two types of t-test:

Matched pairs and independent pairs

If there is some link between the data then use the matched pairs test (eg a ‘before’ and ‘after’)

If there is no link between the data then use the independent pairs test

DFO 8 12 11 6 9 11 7 8 10 9DFO+ninhydrin

10 15 12 6 13 14 9 9 15 12

QuestionIn an investigation to determine the effectiveness of sequencing of fingerprints 10 prints are taken enhanced with DFO and then with ninhydrin. The points of detail at each stage are recorded. Is there a difference at the 95% confidence level?

t-test for matched pairs

1. Set up the null and alternative hypothesis:

H0 there is no difference in the number of minutae when using ninhydrin

HA there are more minutae observed after the enhancement of ninhydrin.

H0 there is no difference in the number of minutae when using ninhydrin

HA there are more minutae observed after the enhancement of ninhydrin.

This is a one-tail test.

We are testing at the 95% or 5% (0.05) level

1. Set up the null and alternative hypothesis

2. Calculate the difference between the pairs in the sample

DFO 8 12 11 6 9 11 7 8 10 9DFO+ninhydrin 10 15 12 6 13 14 9 9 15 12Difference (D)

DFO 8 12 11 6 9 11 7 8 10 9 TotalDFO+ninhydrin 10 15 12 6 13 14 9 9 15 12Difference (D) 2 3 1 0 4 3 2 1 5 3 24

3. Calculate the mean of the differences

4. Calculate the standard deviation of the difference

= 1.51

5. Calculate the standard error

= = 0.478

5. Calculate the standard error6. Calculate the value of t

= = 5.0

7. Calculate the number of degrees of freedom (DoF) and find the critical value

DoF = No of pairs of data – 1

10-1 = 9

= n – 1

5. Calculate the standard error6. Calculate the value of t7. Calculate the number of degrees of freedom (DoF) and 8. find the critical value

7. Calculate the number of degrees of freedom (DoF) and 8. find the critical value

9. Determine if there is a difference or not

t > tcritical (5.0 > 1.833)

So, the null hypothesis is rejected and the alternative hypothesis is accepted.

The ninhydrin does make a positive difference

t-test independent samples

If there is no before and after relationship between the samples then the independent samples test is used.

ExampleSome brown dog hairs were found on the clothing of a victim at a crime scene involving a dog.The five of the hairs were measured: 46, 57, 54, 51, 38 μm.A suspect is the owner of a dog with similar brown hairs. A sample of the hairs has been taken and their widths measured: 31, 35, 50, 35, 36 μm.Is it possible that the hairs found on the victim were left by the suspect’s dog? Test at the %5 level.

[From D. Lucy Introduction to Statistics for Forensic Scientists Chichester: Wiley, 2005 p. 44.]

1. Calculate the mean and standard deviation for the data sets

Dog A Dog B

Standard deviation

Dog A Dog B

Total 246 187

Mean 49.2 37.4

Standard deviation

7.463 7.301

2. Calculate the magnitude of the difference between the two means

49.2 – 37.4 = 11.8

3. Calculate the standard error in the difference

2. Calculate the magnitude of the difference between the two means.│

= 4.669 ≈ 4.67 (3 sf)

4. Calculate the value of t

t = difference between the means ÷ standard error in the difference

4. Calculate the value of t:

t = difference between the means ÷ standard error in the difference

4.669 = 2.527 ≈ 2.53 (3 sig fig)

5. Calculate the degrees of freedom

5. Calculate the degrees of freedom = n1 + n2 - 2

5 + 5 - 2 = 8

6. Find the critical value for the particular significance you are working to from the table

6. Find the critical value for the particular significance you are working to and find the critical value from the table

At the 0.05 level tcrit = 2.306

If t < critical value then there is no significant difference between the two sets of dataIf t > critical value then there is a significant difference between the two sets of data

So, at 0.05 level there is a significant difference between the two data sets.So it could not come from the same dog.

. Group A

(leading questions)Group B(control)

2 33 54 04 32 44 03

. Group A

2 33 54 04 32 44 03

Mean = 3.143SD= 0.8997

Mean = 2.50SD = 2.07

. Group A

2 33 54 04 32 44 03

Mean = 3.143SD= 0.8997

Mean = 2.50SD = 2.07

Diff of means = 0.64SE of diff = 0.861DoF = 11t = 0.7433

DoF = 11t = 0.7433

t < tcritical at 5% level, 2-tailThere is no significant difference between the data test

Student t-test

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