🦹🏼‍♀️ T Test

One Sample, Two Sample and Paired Tests

Small Sample Tests

The entire large sample theory was based on the application of “normal test”. However, if the sample size “n” is small, the distribution of the various statistics, e.g., Z are far from normality and as such “normal test” cannot be applied if “n” is small (n < 30).
In such cases exact sample tests, we use t-test pioneered by W.S. Gosset (1908) who wrote under the pen name of student, and later on developed and extended by Prof. R.A. Fisher.

Let x₁, x₂, ………. x_n be a random sample of size “n” has drawn from a normal population with mean μ and variance σ² then Student’s t – is defined by the statistic.

This test statistic follows a t – distribution with (n-1) degrees of freedom (d.f.). To get the critical value of t we have to refer the table for t-distribution against (n-1) d.f. and the specific level of significance. Comparing the calculated value of t with critical value, we can accept or reject the null hypothesis.
The Range of t – distribution is - ∞ to + ∞.
‘T’ test is applicable when number of treatments is 2.
When sample size is small and S.D. is unknown use ‘T’ test.
For testing the significance of correlation coefficient use ‘T’ test.
When S.D. of population is not known but sample size is more than 30. Then use ‘Z’ Test.

Based on field experiments, a new variety of greengram is expected to give yield of 13 quintals per hectare. The variety was tested on 12 randomly selected farmer fields. The yields (quintal/hectare) were recorded as 14.3, 12.6, 13.7, 10.9,13.7, 12.0, 11.4, 12.0, 13.1, 12.6, 13.4 and 13.1. Do the results conform the expectation?

Solution:

Let yield = x_i (say)
t-table value at (n-1) = 11 d.f. at 5 percent level of significance is 2.20. Calculated value of t < table value of t, H₀ is accepted and we may conclude that the results conform to the expectation.

Assumptions:

Populations are distributed normally
Samples are drawn independently and at random Conditions:
Standard deviations in the populations are same and not known
Size of the sample is small Procedure:
If two independent samples x_i (i = 1, 2, …., n₁) and yj (j = 1, 2, ….., n₂) of sizes n₁ and n₂ have been drawn from two normal populations with means μ₁ and μ₂ respectively.
Null hypothesis H₀: μ₁ = μ₂
The null hypothesis states that the population means of the two groups are identical, so their difference is zero.
where s₁² and s₂² are the variances of the first and second samples respectively.
Which follows Student’s t – distribution with (n₁ + n₂ - 2) d.f. If calculated value of |t| < table value of t with (n₁ + n₂ - 2) d.f. at specified level of significance, then the null hypothesis is accepted otherwise rejected.

Two verities of potato plants (A and B) yielded tubers are shown in the following table. Does the mean weight of tubers of the variety “A” significantly differ from that of variety “B”?

Solution:

Hypothesis H₀ : μ₁ = μ₂ i.e. the mean weight of tubers of the variety “A” do not significantly differ from the variety “B”.
Calculated value of t = 3.77. Table value of t for 19 d.f. at 5 % level of significance is 2.09 Since the calculated value of t > table value of t, the null hypothesis is rejected and hence we conclude that the mean number of tubes of the variety “A” significantly differ from the variety “B”.

The paired t-test is generally used when measurements are taken from the same subject before and after some manipulation such as injection of a drug.
For example, you can use a paired t test to determine the significance of a difference in blood pressure before and after administration of an experimental pressure substance.

Assumptions

Conditions

Hypothesis

H₀: μ_d = 0
Under H₀, the test statistic becomes,
Where
- S² is variance of the deviations
- n = sample size; where d_i = x_i - y_i (i = 1,2, ……, n)
If calculated value of |t| < table value of t for (n-1) d.f. at α % level of significance, then the null hypothesis is accepted and hence we conclude that the two samples may belong to the same population. Otherwise, the null hypothesis rejected.

The average number of seeds set per pod in Lucerne were determined for top flowers and bottom flowers in ten plants.
The values observed were as follows:

Test whether there is any significant difference between the top and bottom flowers with respect to average numbers of seeds set per pod.

Solution:

Null Hypothesis H₀: μ_d = 0
Under H₀ becomes, the test statistic is
Calculated value of t = 1.65
Table value of t for 9 d.f. at 5% level of significance is 2.26
Calculated value of t < table value of t, the null hypothesis is accepted and we conclude that there is no significant difference between the top and bottom flowers with respect to average numbers of seeds set per pod.