Introduction
When sample sizes are small, and the standard deviation of the population is unknown
it is normal to use the distribution of the t statistic (also known as the t score),
whose values are given by:
where x_{m} is the sample mean, µ is the population mean, s is the standard deviation
of the sample, and n is the sample size. The distribution of the t statistic is
called the t distribution or Student's t distribution.
The tdistribution is generally used if any of the following conditions apply;
If the sample is symmetrical, unimodel and with a sample size less than say 15
If the sample is moderately skewed and the sample size is 15  40
If the sample size is over 40
It is not recommended that this distribution is used if there are any sample
values which are obviously way out compared to the other values.
The probability density function of t is given by
T_{ν} is a constant depending on ν . Thisis
an integer parameter identified as the number of degrees of freedom of the distribution.
Symbols
α = Significance (1% 5% etc)
γ = Confidence (95% 99% etc)
n = sample number
f(x) = probability function. (values between 0 and 1)
F(x) = probability distribution function.
ν = number of degrees of freedoms
X_{m} = Sample mean
var = sample variance

Φ (x) = Probability distribution function.(Standardised probability )
μ = population /random variable mean
σ ^{2} = population /random variable variance
σ = population /random variable standard deviation
x_{m} = arithmetic mean of sample
s_{x} ^{2} = variance of sample
s_{x} = Standard deviation of sample

Degrees of Freedom
The number of degrees of freedom as applied in the tdistribution =
ν = n  r
r = the number of population parameters which have to be estimated. In the case of evaluating t
x_{m} and s_{x} can be calculated from the data and μ has to be estimated therefore r = 1
ν = n  1
Properties
The t distribution has the following properties:
The mean of the distribution is equal to 0 .
The variance is equal to ν / (ν  2 ), where ν is the degrees of
freedom (see above) and ν > 2.
The variance is always greater than 1, although it is close to 1 when there are many degrees of freedom. With infinite degrees of freedom, the t distribution is the same as the standard normal distribution.
One or Tail tailed tests
The table below includes the value or F(z) and is a one tailed distribution, that it includes the area under
the curve from  to z.
The values in this table are applicable to onetailed tests. A one tailed test is used to
test if one mean is higher than another or if it is equal to another. The following hypothesis are being tested
H_{0}: μ = μ_{o}
or
H_{1}: μ < μ_{o}
H_{1}: μ > μ_{o}
If the calculated value is greater than z as shown as the unshaded region then any of the above
hypothesis can be accepted.
The hypothesis that μ ≠ μ_{o} can not be tested with a
one tailed test and the two tailed test should be used.
H_{O}:μ = μ
H_{1}:μ ≠ μ
To use the table below for a two tailed test with a confidence level α = say 95% then use table values for
a confidence level of 0,5 + α /2. and apply the resulting z value as a ± value as the curve is symmetrical.
Example:if the required confidence level is 90% then use the table with confidence levels of (0,5 + 0,90/2) = 0,95
Examples
Example 1:
A sample of twelve tubs of margarine gave a mean mass of 495 gm and a standard deviation
10 gm . The mass stated on the tub is 500 gm .
What is the 95% confidence that the mean mass of margarine tubs is the stated value.
Best estimate for the mean μ mass = 500
Null hypothesis H_{o}: μ =500
Alternative hypothesis H_{1}:μ ≠ 500
s_{x} = 5 ^{2}
n = 12 and ν = n  1 = 11
The table below F(z)is used to determine the +/ confidence limits.
The area under the t curve from  to + c = 0,5 + (0,95/2)= 0,975 From the table +c= 2,201
The area under the t curve from  to  c = 0,5  0,95/2 = 0,025. From the table c = 2,201 ****
Therefore if the mean value is 500 gm there is a 95% confidence that the t score will be within the
range 2,201 to 2,201 with a 5% significance that the mean value is not 500 gm
The tscore is calculated as
The resulting tscore is 1,73 and therefore the masses of the samples are within the 95% confidence limits. The null hypothesis is not rejected.
**** 0,975 = F(2,201) then 1  0,25 = 0,975 = F(2,201).
Example 2:
Consider 104 tensile tests on twine resulting in the following table of breaking loads (N).
The population is assumed to be normal.
Determine the 99% confidence interval for the corresponding population mean.
Breaking load ( Newtons) 
201 
234 
242 
250 
256 
261 
267 
271 
277 
282 
292 
300 
310 
203 
234 
243 
250 
256 
262 
267 
271 
277 
283 
293 
302 
312 
221 
237 
246 
252 
257 
264 
268 
272 
278 
284 
293 
302 
315 
224 
238 
246 
252 
258 
264 
268 
272 
278 
286 
294 
304 
316 
224 
239 
247 
252 
259 
265 
268 
273 
279 
287 
296 
304 
321 
229 
239 
247 
253 
259 
266 
269 
273 
279 
289 
297 
306 
326 
231 
241 
249 
254 
260 
266 
270 
276 
281 
291 
298 
307 
341 
231 
241 
249 
256 
261 
266 
271 
276 
282 
291 
299 
309 
342 
By simple arithmetic the sample mean x_{m} = 269,90
By simple arithmetic the variance s_{x}^{2} = 763.04.
Therefore Standard Deviation s_{x} = 27,62
n = 104 and ν = n  1 = 103:
Using the table below with F(z)= 0,5 + 0,99/2 = 0,995 therefore using ν = 100
c = 2,626
As the curve is symmetrical t = +/ 2,626
The 99% confidence limits for the sample mean is 262,72 and 277,01.
Example 3:
Eight hens were fed with different two different type grain
for a period of one month.
The eggs laid be each hen in the two monthly periods were as follows
Test the hypothesis that grain type A yields better results than grain type B
The null hypothesis is that the hens lays the same number of eggs on both types of grain .. H_{o}:μ_{A} = μ_{B}
The alternative hypothesis is that hens fed on grain A lay more eggs than the when fed with grain B.. H_{o}:μ_{A} > μ_{B}
Chicken No  1  2  3  4  5  6  7  v8 
Grain type A  18  17  19  14  15  17  11  14 
Grain type B  16  16  15  17  16  15  12  15 
The mean difference for each hen is (d _{m})
d_{m} = (2 + 1 + 4 + 3 + 1 + 2 + 1 + 1) /8 = 0,375
s _{d} ^{2} =
[(2  0,375) ^{2} + (1  0,375) ^{2} + (4  0,375) ^{2} + (3  0,375) ^{2} +
(1  0,375) ^{2} + (2  0,375) ^{2} + (1  0,375) ^{2} + (1  0,375) ^{2} ]/(81) = 5,124
d_{x} = 2,264
The degrees of freedom ν = n  1 = 7
The significance value α for the null hypothesis is 5%.
The associated confidence of the null hypothesis = 1 α = 95% (0,95).
For ν = 7 and f(z) = 0,95 the difference is 0 is 1,895 from the table below.
The hypothesis that the mean is greater than 0 is not rejected if t is greater than 1,895. As t is
0,438 there is no difference in the use of the two feed types.
Students t Distribution ..Table of z for given values of F(z)
Note: F(z) = 1 F(z)
Deg Of Freedom 
F(z) 
0,5 
0,6 
0,7 
0,8 
0,9 
0,95 
0,975 
0,99 
0,995 
1 
0 
0,3249 
0,7265 
1,3764 
3,0777 
6,3137 
12,7062 
31,821 
63,6559 
2 
0 
0,2887 
0,6172 
1,0607 
1,8856 
2,92 
4,3027 
6,9645 
9,925 
3 
0 
0,2767 
0,5844 
0,9785 
1,6377 
2,3534 
3,1824 
4,5407 
5,8408 
4 
0 
0,2707 
0,5686 
0,941 
1,5332 
2,1318 
2,7765 
3,7469 
4,6041 
5 
0 
0,2672 
0,5594 
0,9195 
1,4759 
2,015 
2,5706 
3,3649 
4,0321 
6 
0 
0,2648 
0,5534 
0,9057 
1,4398 
1,9432 
2,4469 
3,1427 
3,7074 
7 
0 
0,2632 
0,5491 
0,896 
1,4149 
1,8946 
2,3646 
2,9979 
3,4995 
8 
0 
0,2619 
0,5459 
0,8889 
1,3968 
1,8595 
2,306 
2,8965 
3,3554 
9 
0 
0,261 
0,5435 
0,8834 
1,383 
1,8331 
2,2622 
2,8214 
3,2498 
10 
0 
0,2602 
0,5415 
0,8791 
1,3722 
1,8125 
2,2281 
2,7638 
3,1693 
11 
0 
0,2596 
0,5399 
0,8755 
1,3634 
1,7959 
2,201 
2,7181 
3,1058 
12 
0 
0,259 
0,5386 
0,8726 
1,3562 
1,7823 
2,1788 
2,681 
3,0545 
13 
0 
0,2586 
0,5375 
0,8702 
1,3502 
1,7709 
2,1604 
2,6503 
3,0123 
14 
0 
0,2582 
0,5366 
0,8681 
1,345 
1,7613 
2,1448 
2,6245 
2,9768 
15 
0 
0,2579 
0,5357 
0,8662 
1,3406 
1,7531 
2,1315 
2,6025 
2,9467 
16 
0 
0,2576 
0,535 
0,8647 
1,3368 
1,7459 
2,1199 
2,5835 
2,9208 
17 
0 
0,2573 
0,5344 
0,8633 
1,3334 
1,7396 
2,1098 
2,5669 
2,8982 
18 
0 
0,2571 
0,5338 
0,862 
1,3304 
1,7341 
2,1009 
2,5524 
2,8784 
19 
0 
0,2569 
0,5333 
0,861 
1,3277 
1,7291 
2,093 
2,5395 
2,8609 
20 
0 
0,2567 
0,5329 
0,86 
1,3253 
1,7247 
2,086 
2,528 
2,8453 
21 
0 
0,2566 
0,5325 
0,8591 
1,3232 
1,7207 
2,0796 
2,5176 
2,8314 
22 
0 
0,2564 
0,5321 
0,8583 
1,3212 
1,7171 
2,0739 
2,5083 
2,8188 
23 
0 
0,2563 
0,5317 
0,8575 
1,3195 
1,7139 
2,0687 
2,4999 
2,8073 
24 
0 
0,2562 
0,5314 
0,8569 
1,3178 
1,7109 
2,0639 
2,4922 
2,797 
25 
0 
0,2561 
0,5312 
0,8562 
1,3163 
1,7081 
2,0595 
2,4851 
2,7874 
26 
0 
0,256 
0,5309 
0,8557 
1,315 
1,7056 
2,0555 
2,4786 
2,7787 
27 
0 
0,2559 
0,5306 
0,8551 
1,3137 
1,7033 
2,0518 
2,4727 
2,7707 
28 
0 
0,2558 
0,5304 
0,8546 
1,3125 
1,7011 
2,0484 
2,4671 
2,7633 
29 
0 
0,2557 
0,5302 
0,8542 
1,3114 
1,6991 
2,0452 
2,462 
2,7564 
30 
0 
0,2556 
0,53 
0,8538 
1,3104 
1,6973 
2,0423 
2,4573 
2,75 
40 
0 
0,255 
0,5286 
0,8507 
1,3031 
1,6839 
2,0211 
2,4233 
2,7045 
50 
0 
0,2547 
0,5278 
0,8489 
1,2987 
1,6759 
2,0086 
2,4033 
2,6778 
60 
0 
0,2545 
0,5272 
0,8477 
1,2958 
1,6706 
2,0003 
2,3901 
2,6603 
100 
0 
0,254 
0,5261 
0,8452 
1,2901 
1,6602 
1,984 
2,3642 
2,6259 
200 
0 
0,2537 
0,5252 
0,8434 
1,2858 
1,6525 
1,9719 
2,3451 
2,6006 
5000 
0 
0,2534 
0,5244 
0,8417 
1,2817 
1,6452 
1,9604 
2,3271 
2,5768 
