Independent T-Test for Two Samples (Unequal Variance)

Hypothesis

Statistical Test

If σ₁ ≠ σ₂ and both are not known in value, the value is approached with the approximate value, i.e. the average deviation of the example, s. because the two populations have different variations, the standard deviation of the example is approached by:

 $$s_{{\bar{y}}_1-{\bar{y}}_2}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$

with 

 $$ df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\left(\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}\right)}$$

The df value is rounded to the nearest number.

The_calculated value of t is determined by the following formula:

 $$t=\frac{{\bar{y}}_1-{\bar{y}}_2}{SE_{{\bar{y}}_1-{\bar{y}}_2}}$$ $$t=\frac{{\bar{y}}_1-{\bar{y}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$

Example 1:

There are two groups of land types, namely arable land and barren land, from each of these lands 7 pieces were sampled for the size of fine gravel. Test at the level of 5% is the fine gravel content in fertile land different from barren land? Fine and coarse gravel data are presented in the following Table:

Table.Fine gravel in surface soil (Torrie, 1980)

Manual Calculation (Classical/Traditional Method)

1. Step 1: Claim: the alleged average difference between the two lands can be denoted by: μ₁ ≠ μ₂, if the conjecture is false, then μ₁ = μ₂
2. Step 2: From the two equations above, the μ₁ = μ₂ contains an element of equality, thus becoming the null hypothesis and alternative hypothesis(H₁) μ₁ ≠ μ₂.
3. H₀: μ₁ = μ₂
4. H₁: μ₁ ≠ μ₂
5. Significant level: α = 0.05
6. Determine the statistical test: the sample is taken randomly; n < 30; normally distributed (Kolmogorov-Smirnov/Shapiro-Wilk Test) and σ₁ ≠ σ₂ (Levene Test). From these conditions, the appropriate statistical test is an independent t-test with different variations.
7. Calculate t_stat and determine t_critical:
8. Calculation of the _calculated value of t

 $$s_{{\bar{y}}_1-{\bar{y}}_2}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$

 $$\begin{matrix}s_{{\bar{y}}_1-{\bar{y}}_2}=\sqrt{\frac{(6.33441)^2}{7}+\frac{(2.63619)^2}{7}}\\=2.59324\\\end{matrix}$$

With:

 $$ df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\left(\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}\right)}$$

 $$=\frac{\left(\frac{(6.33441)^2}{7}+\frac{(2.63619)^2}{7}\right)^2}{\frac{\left(\frac{(6.33441)^2}{7}\right)^2}{7-1}+\frac{\left(\frac{(2.63619)^2}{7}\right)^2}{7-1}} \\=8.018 \approx 8$$

 $$ t=\frac{{\bar{y}}_1-{\bar{y}}_2}{s_{{\bar{y}}_1-{\bar{y}}_2}}=\frac{10.9143-3.9429}{2.59324}=2.688297$$

1. Determine thet_critica with df = 8 and α = 0.05. From the t-student table obtained the value of t_kritis for a Two-Tailed test = 2.305
2. Because |t_stat| > t_critical = |2,688| > 2.305, then H₀ is rejected!
3. From the results of the t-test, it was concluded that at a significant level of 5%, the average fine gravel content on arable and barren land is different.

Calculations by using the SmartstatXL Excel Add-In

Analysis Results

Calculation using SPSS software v.16:

Normality Test

The normality test (Kolmogorov-Smirnov^a/ Shapiro-Wilk) for both groups was not significant. This shows that the two samples are normally distributed.

Variance homogeneity test:

The results of the Levene's Test showed that the two populations had unequal variance (p = 0.001 < 0.05), so the t-test used was a test assuming unequal variance  (Equal variance not assumed). This Hal suggests that the two samples came from different populations.

Conclusion:

Hypothesis:

1. H₀: μ₁ = μ₂
2. H₁: μ₁ ≠ μ₂

Classic Method

t_stat = t_stat = T-Value = 2.688

t_critica = t_table = t_{(0.05.9) = 2,305} (obtained from the table value t-student)

Since |t_stat| > |t_critical| = 2,688 > 2,305 then H0 is rejected!

Modern Methods:

Tests with modern methods use p-value in determining whether or not a statistical test is significant.

If: P-Value < Significant Level then the significant test or H₀ is rejected

If: P-Value > Significant Level then the non-significant test or H₀ is accepted

Conclusion: 

See the output of the analysis results using SmartstatXL and SPSS software above! 

Since its signification value (p-value or Sig. = 0.028) < 0.05, then H₀ is rejected and H_A is accepted. This shows that at a significant level of 5%, the average fine gravel content on arable and barren land is different.

In the output the Table also includes the value of the confidence interval.

P(0.99372 < μ₁ - μ₂ < 12.94914) = 95%

Which means that 95% of us believe that the difference between the two average lands is in the range between 0.99372 and 12.94914.