🦹🏼‍♀️Student's t-Test

A breeder develops a new greengram variety expected to yield 13 quintals per hectare. She tests it on only 12 plots — too few for a Z-test. With a small sample and unknown population standard deviation, the t-test is her go-to tool for determining whether the observed yield matches the expected value.

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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.

The t-test (also called Student’s t-test) is one of the most widely used statistical tests in agricultural research. It is specifically designed for situations where the sample size is small and the population standard deviation is unknown.

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t-test for One Samples

• 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 +∞. Unlike the Z-distribution, the t-distribution has heavier tails, meaning extreme values are more likely. As the sample size increases, the t-distribution approaches the normal distribution.
• ‘T’ test is applicable when number of treatments is 2.
• When sample size is small (n < 30) and S.D. is unknown, use ‘T’ test. This is the key condition that distinguishes the t-test from the Z-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.

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Example:

• 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:

• Null Hypothesis: H₀ : μ = μ₀ = 13 i.e. the results conform the expectation
• The test statistic becomes

• 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.

Since the calculated t-value is less than the critical value, we do not have sufficient evidence to say the yield is different from 13 quintals/hectare. The new variety’s performance is consistent with the expectation.

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t-test for Two Samples

The two-sample t-test (also called the independent samples t-test) is used to compare the means of two independent groups when the sample sizes are small and the population standard deviation is unknown.

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.

Note that the degrees of freedom for the two-sample t-test is (n₁ + n₂ - 2), which accounts for the two constraints (the two sample means).

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Example:

• 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”.

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Paired t – test

The paired t-test is used when the two samples are not independent but are related — typically measurements taken on the same subject before and after a treatment or under two different conditions.

• 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

• Populations are distributed normally
• Samples are drawn independently and at random

Conditions

• Samples are related with each other
• Sizes of the samples are small and equal
• Standard deviations in the populations are equal and not known

Hypothesis

• H₀: μ_d = 0

This null hypothesis states that the mean difference between paired observations is zero, i.e., there is no effect of the treatment.

• 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.

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Example

• 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.

Since the calculated t (1.65) is well below the critical value (2.26), the observed difference is likely due to random variation rather than a genuine difference in seed setting between top and bottom flowers.

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Summary Table

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Summary Cheat Sheet