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, � 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 t-distribution is generally used if any of the following conditions apply;

If the sample is symmetrical, uni-model 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.

The number of degrees of freedom as applied in the t-distribution =

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

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.

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 one-tailed 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

H₀: μ = μ_o
or
H₁: μ < μ_o
H₁: 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
H₁:μ ≠ μ

To use the table below for a two tailed test with a confidence level α = say 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 then use the table with confidence levels of (0,5 + 0,90/2) = 0,95

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 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 H₁: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 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 t-score is calculated as

The resulting t-score is -1,73 and therefore the masses of the samples are within the confidence limits.   The null hypothesis is not rejected.

**** 0,975 = F(2,201) then 1 - 0,25 = 0,975 = F(-2,201).

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.

By simple arithmetic the sample mean x_m = 269,90
By simple arithmetic the variance s_x² = 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.

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

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 ]/(8-1) = 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 = (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.

Note: F(-z) = 1- F(z)