Example of One Sample Student's t Test

Step 1) State the biological question.

Is the internal body temperature of 25 intertidal crabs different from the air temperature of 24.3 degrees C?

 

Step 2) Translate the biological question into statistical hypotheses.

 

 

 

 

Step 3) State the alpha level: p<0.05 is considered significant

 

 

Step 4) State the statistical test.

Because we are working with a hypothetical population mean and a sample mean we use the two-tailed One Sample t test.

 

 

Step 5) State the assumptions of the test.

The subjects in the sample must be randomly selected from a population. The sample data must come from a normally distributed population of observations for the variable under study.

 

 

Step 6) Calculate the observed t statistics from your data.

 

Internal

Temp.

d

d2

ss

var

sd

se

t

25.8000

.7720

.5960

43.21

1.8004

1.34

.2684

2.7128

24.6000

-.4280

.1832

.

.

.

.

.

26.1000

1.0720

1.1492

.

.

.

.

.

22.9000

-2.1280

4.5284

.

.

.

.

.

25.1000

.0720

.0052

.

.

.

.

.

27.3000

2.2720

5.1620

.

.

.

.

.

24.0000

-1.0280

1.0568

.

.

.

.

.

24.5000

-.5280

.2788

.

.

.

.

.

23.9000

-1.1280

1.2724

.

.

.

.

.

26.2000

1.1720

1.3736

.

.

.

.

.

24.3000

-.7280

.5300

.

.

.

.

.

24.6000

-.4280

.1832

.

.

.

.

.

23.3000

-1.7280

2.9860

.

.

.

.

.

25.5000

.4720

.2228

.

.

.

.

.

28.1000

3.0720

9.4372

.

.

.

.

.

24.8000

-.2280

.0520

.

.

.

.

.

23.5000

-1.5280

2.3348

.

.

.

.

.

26.3000

1.2720

1.6180

.

.

.

.

.

25.4000

.3720

.1384

.

.

.

.

.

25.5000

.4720

.2228

.

.

.

.

.

23.9000

-1.1280

1.2724

.

.

.

.

.

27.0000

1.9720

3.8888

.

.

.

.

.

24.8000

-.2280

.0520

.

.

.

.

.

22.9000

-2.1280

4.5284

.

.

.

.

.

25.4000

.3720

.1384

.

.

.

.

.

               
Sample Mean 25.03            
Hypothetical Mean 24.3            
n 25            
df=v=n-1 24            

 

tobs=(Sample Mean - Hypothetical Mean)/se

tobs=(25.03-24.3)/.2684

tobs=2.7128

Step 7) Find the df and critical value of t from the table of critical values.

 

 

 

 

Step 8) Compare the critical and observed t values and reject or do not reject the null hypothesis.

 

 

 

 

Find the actual P value

t.05(2)24=2.064<t.02(2)24=2.492<tobs=2.7128< t.01(2)24=2.797

0.01<P<0.02<0.05

 

 

Step 9) Interpret the results of the analysis as it relates to the biological question raised in 1).

We rejected the null hypothesis in Step 8. The internal body temperature of intertidal crabs is different from the ambient air temperature (t=2.7128, df=24, two-tailed, p<0.02).