ADVERTISEMENTS:

The below mentioned article provides a study note on test of significance.

**Test of Significance****:**

In biological research when we compare any character of two samples, we calculate the significance of difference in the mean and variance to draw a meaningful conclusion.

ADVERTISEMENTS:

For example, if we are interested to compare the protein content of two seed varieties of maize or yield of two crop varieties, then it is necessary to assess the significance of the difference in mean or variance. Test of significance helps us in determining whether the difference between the two samples are actually due to chance factor or the difference is really significant among the samples.

**Null Hypothesis and Alternative Hypothesis****:**

Testing of hypothesis is the procedure which approaches the comparison between means. In biological science statistical test of hypothesis plays an important role. The test of a sample mean (x̅) is done against a population mean (µ) or between the two means of two samples (x̅_{1} and x̅_{2}).

The hypothesis or a statement is that either µ = x̅ or x̅_{1} = x̅_{2}, i.e., examination should be done to determine whether there is no difference between p and x or between x̅_{1} and x̅_{2}.The hypothesis is that µ and x̅ are equal or x̅_{1}, and x̅_{2} are equal which is called null hypothesis.

ADVERTISEMENTS:

**It is expressed in the following way:**

Null hypothesis (H_{0}) can be defined as the hypothesis which is a test for possible rejection under the assumption that it is true, i.e., it is accepted.

Alternative hypothesis (H_{1} or H_{A}) can be defined as the hypothesis which is complementary to the null hypothesis. Rejection of null hypothesis leads to the acceptance of alternative hypothesis.

Generally it can be written as,

which states that x̅ is not equal to population mean (µ) or there is difference between two sample means, i.e., either x̅_{1} > x̅_{2} or x̅_{1} < x̅_{2}.

**Student’s ‘t’-Test****: **

To test the significance of difference of means of two samples, W.S. Gosset applied a statistical tool called ‘t’-test. According to nick name of Gosset, the test has been named as Student’s ‘t’-test.

ADVERTISEMENTS:

**In this test we make a choice between two alternatives: **

1. To accept the null hypothesis (no difference between the two means);

2. To reject the null hypothesis (the difference between the means of two samples significant).

The test is applied to small sample (n<30), and samples must be drawn randomly from normal population. The ‘t’ is defined as quantity representing the difference between the sample mean or true means or population mean expressed in terms of the standard error.

ADVERTISEMENTS:

T = Difference between sample means/Standard error of the difference between means

x̅_{1} and x̅_{2} = Mean of two samples

SE_{D} = Standard error of the difference between means

ADVERTISEMENTS:

**Degree of Freedom****: **

To calculate the standard error of the difference, the degree of freedom can be obtained by one less than the number of observations of sample. In case of unpaired ‘t’ test, i.e., when the comparison between two samples is done, degree of freedom is calculated by the formula

(n_{1} + n_{2}) – 2, where n_{1} = no. of observations in sample I

n_{2} = no. of observations in sample II.

ADVERTISEMENTS:

Whereas, in case of paired ‘t’ test, the degree of freedom is (n – 1), as there is the same number of observations.

**Determination of Significance****:**

Probability of occurrence of any calculated value of ‘t’ is determined by comparing it with the value given in the ‘t’-table corresponding to the calculated degree of freedom, derived from the number of observations in the samples under study. If the calculated value of ‘t’ exceeds the value given at p = 0.05 (5% level) in the table, it is said to be significant. If the calculated value of ‘t’ is less than the value given in ‘t’-table, it is not significant.

The ‘t’-test procedure is done to locate the observed value of ‘t’ in the student’s ‘t’- distribution curve. The ‘t’ distribution curve is a symmetrical curve with mean zero, it extends to infinity on either sides. When degree of freedom is less in number, ‘t’- distribution resembles to normal distribution curve. If the calculated value of ‘t’ is near the centre then the data is regarded as compatible with H_{0}, which concludes that the observed deviation is due to chance factor or only due to sampling. If the ‘t’ value is situated at tail region then H_{0} is not accepted.

**Unpaired V-Test****: **

This test is applied to unpaired data of independent observations made on individuals of two different or separate groups or samples drawn from two different populations.

ADVERTISEMENTS:

**Steps to be followed for calculation of ‘t’:**

1. The means of two samples are calculated and the differences between the means of two samples are calculated.

(x̅_{1} – x̅_{2})

2. The standard error of the difference between two means is calculated.

3. ‘t’ value is calculated by the ratio between the observed difference of means and its standard error.

ADVERTISEMENTS:

For Large Sample,

4. The calculated ‘t’ value is compared with ‘t’ table to find out the significance at that particular degree of freedom.

**Practical sheet for**** ‘I’ analysis: **

Degrees of freedom = n_{1} + n_{2} – 2

Calculated t =

Tabulated t =

Significance level, p =

Inference

**Example 1: **

In an experiment to find out the effect of a hormone spray on the seed yield of dwarf French bean, the following results were obtained. Analyse the data using the ’t’ test for inference on the effect of hormonal spray on the seed yield.

Degrees of freedom = n_{1} + n_{2} – 2 = 50 + 50 – 2 = 98

Calculated t = 14.829

Tabulated t = 3.37 (p = 0.001)

Since the calculated t (14.829) far exceeds the tabulated t (3.37) with 98 d.f. at p = 0.001, the null hypothesis stating that there is no difference in the two sample means, is rejected. Alternatively there is a highly significant difference between the two sample means, i.e., the hormonal spray has a very significant effect on the seed yield.

**Example 2: **

Two varieties of potato plants (A and B) yielded tubers as shown in the following table. Does the mean weight of tubers of the variety A significantly differ from that of variety B.

As the calculated t (4.108) for the difference of the two sample means is greater than the tabulated t (3.92) for 18 d.f. at p ≤ 0.001, the difference between the two means is therefore highly significant. The null hypothesis stating that the two sample means are not different is rejected at p ≤ 0.001.

**Example 3: **

Application of fertilizer (NPK) was tested on the yield of rice grown in 10 plots. Another set of 10 plots of similar size and conditions was taken as the control. Determine whether the fertilizer has any significant effect on the yield of rice.

Tabulated t = 3.92 (p = 0.001) for 18 d.f. is found to be less than the calculated t (4.58). Thus the difference between the means of two samples is highly significant. The null hypothesis stating that the two means have no difference is rejected at p = 0.001.

**Paired ‘T’-****Test:**

Paired ‘t’-test is applied when each individual gives a pair of observations. Here the paired data of independent observation from one sample only to be compared.

**This kind of observations is made available in biological sciences, such as: **

(a) To study the effect of fertilizer, pesticide, drug on plants.

(b) To compare the effect of two different fertilizers or drugs.

(c) To compare the result of two techniques or the accuracy of two different instruments.

**Steps to be followed for calculation of**** ‘t’: **

1. The difference in each set of paired observations are made.

2. The mean of difference D̅ is calculated,

3. Standard deviation of the difference (S_{D}) and then standard error of mean from the same is calculated,

4. The ‘t’ value is determined by substituting the above values in the formula,

5. The calculated value of ‘t’ is compared with ‘t’-table to find out the significance at that particular degree of freedom.

**Example 4: **

Determine the effect of a new growth regulator (x) on the callus growth of Nicotiana in laboratory condition. The observations are made on particular 10 calli.

Tabulated t = 4.78 at p .001 with 9 degrees of freedom.

The calculated ‘t’ far exceeds the tabulated ‘t’ at p = 0.001 with 9 degrees of freedom. So the result is highly significant, i.e., the hormone has the positive effect on callus growth.