Two Sample T Hypothesis Tests

What is a Two Sample T Hypothesis Test?

A two sample t hypothesis tests also known as independent t-test is used to analyze the difference between two unknown population means. The Two-sample T-test is used when the two small samples (n< 30) are taken from two different populations and compared. The underlying chart makes use of the T distribution.

Assumptions of Two Sample T Hypothesis Tests

• The sample should be randomly selected from the two population
• Samples are independent to each other
• Two sample sizes must me less than 30
• Samples collected from the population are normally distributed

When Would You Use a Two Sample T Hypothesis Tests?

The two sample t test most likely used to compare two process means, when the data is having one nominal variable and one measurement variable. It is a hypothesis test of means. Use two sample Z test if the sample size is more than 30.

The two sample hypothesis t tests is used to compare two population means, while analysis of variance (ANOVA) is the best option if more than two group means to be compared.

Two sample T hypotheis tests are performed when the two group samples are statistically independent to each other, while the paired t-test is used to compare the means of two dependent or paired groups. 

Note: There are (2) types of Two Sample T Hypothesis tests!

• Two Sample T Hypothesis Test (Equal Variance)
  • Variance of two populations are equal
• Two Sample T Hypothesis Test (Unequal Variance)
  • Variance of two populations are NOT equal

Methods to determine population varince equal or unequal?

The best method to determine population variance is equal or unequal by using an appropriate F-test.

Hypothesis Testing

A tailed hypothesis is an assumption about a population parameter. The assumption may or may not be true. One-tailed hypothesis is a test of hypothesis where the area of rejection is only in one direction. Whereas two-tailed, the area of rejection is in two directions. The selection of one or two-tailed tests depends upon the problem.

• Null hypothesis- H_0: The population means are same alternatively the difference between two population means are equal to hypothesized difference (d). So, µ₁ = µ₂  orµ₁– µ₂ = d
• Alternative hypothesis: µ₁ ≠ µ₂  orµ₁– µ₂ ≠ d (Two-tailed test)
• µ₁ < µ₂  orµ₁– µ₂ < d (left-tailed)
• µ₁ > µ₂  orµ₁– µ₂ > d (Right-tailed)

Two Sample T Hypothesis Test (Equal Variance) formula

• Where n1 and n2 are sample sizes
• x̅1 and x̅2 are means of sample sizes
• Sp is the pooled standard deviation

Steps to Calculate Two Sample T Hypothesis Test (Equal Variance)

• State the claim of the test and determine the null hypothesis and alternative hypothesis
• Determine the level of significance
• Calculate degrees of freedom
• Find the critical value
• Calculate the test statistics
• Make a decision, the null hypothesis will be rejected if the test statistic is less than or equal to the critical value
• Finally, Interpret the decision in the context of the original claim.

Example of a Two Sample T Hypothesis Test (Equal Variance) in a DMAIC Project

Two Sample T test mostly performed in Analyze phase of DMAIC to evaluate the difference between two process means are really significant or due to random chance, this is basically used to validate the root cause(s) or Critical Xs (see the below example for more detail)

Two-tailed (Equal variance)

Example: Apple orchard farm owner wants to compare the two farms to see if there are any weight difference in the apples. From farm A, randomly collected 15 apples with an average weight of 86 gms, and the standard deviation is 7. From farm B, collected 10 apples with an average weight of 80 gms and standard deviation of 8. With a 95% confidence level, is there any difference in the farms?

• Null Hypothesis (H₀) : Mean apple weight of farm A is equal to farm B
• Alternative Hypothesis (H₁) : Mean apple weight of farm A is not equal to farm B
• n1=15
• n2=10
• S1²=49
• S2² =64
• X̅1 =86
• X̅2 = 80

Significance level: α=0.05

Degrees of freedom df: 15+10-2= 23

Calculate critical value

Refer two tailed t table for 23 degrees of freedom

If the calculated t value is less than -2.069 or greater than 2.069, then reject the null hypothesis.

Test Statistic

Interpret the results:

Compare t _calc to t _critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value less than the critical value, hence failed to reject null hypothesis ( H₀). So, there is no significant difference between mean weights of apples in farm A and farm B.

Two Sample T 2 Tailed Equal Variance template file

Right-Tailed (Equal Variance)

Example: The thickness of ceramic tile of vendor A is slightly greater than the vendor 2. The researcher randomly collected samples from both vendors. Assuming the population variances are equal, at 95% confidence level, is there enough evidence to support the claim?

• Null Hypothesis (H₀) : Mean tile thickness of vendor A is equal to vendor B
• Alternative Hypothesis (H₁) : Mean tile thickness of vendor A is greater than vendor B
• Significance level: α=0.05

• n1=12
• n2=12

Degrees of freedom df: n1+n2-2=12+12-2= 22

Calculate mean (X̅) of each sample

• X̅1 = (x₁ + x₂ + ….+ x_n) / n =18.7
• X̅2 = (x₁ + x₂ + ….+ x_n) / n =18.5

Calculate variance (S²) of each sample

• S1² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1 =8.61
• S2² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1=16.09

Calculate critical value

Refer one tailed t table for 22 degrees of freedom

If the calculated t value is greater than 1.717, then reject the null hypothesis.

Test Statistic

Interpret the results:

Compare t _calc to t _critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is less than the critical value and it is not in rejection region, hence failed to reject null hypothesis ( H₀). So, there is no significant difference between mean tile thickness of vendor A and vendor B.

Two Sample T Right Tailed Equal Variance template file

Left-Tailed (Equal Variance)

Example: Samples of sales in similar shops in two different US states are taken for a new product with the following results. Is there any evidence that sales in state A is less than the state B. Assuming the population variances are equal, at 95% confidence level, is there enough evidence to support the claim?

• Null Hypothesis (H₀) : Mean sales in state A is equal to the state B
• Alternative Hypothesis (H₁) : Mean sales in state A is lesser than state B
• Significance level: α=0.05

• n1=12
• n2=12

Degrees of freedom df: n1+n2-2=12+12-2= 22

Calculate mean (X̅) of each sample

• X̅1 = (x₁ + x₂ + ….+ x_n) / n =44.7
• X̅2 = (x₁ + x₂ + ….+ x_n) / n =42.8

Calculate variance (S²) of each sample

• S1² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1 =53.3
• S2² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1=28.2

Calculate critical value

Refer one tailed t table for 22 degrees of freedom

If the calculated t value is less than 1.717, then reject the null hypothesis.

Test Statistic

Interpret the results:

Compare t _calc to t _critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is less than the critical value and it is in the rejection region, hence we reject the null hypothesis ( H₀). So, mean sales in state A is less than the state B.

Two Sample T Left Tailed Equal Variance template file

Two Sample T Hypothesis Test (Equal Variance) Videos

2 SAMPLE T-TEST FOR MEANS (UNEQUAL VARIANCE)

Test for equality of two means.  This test is used to determine if the mean of one sample is not equal to the mean of another sample.

Two Sample T Hypothesis Test (Unequal Variance) formula

• Where n1 and n2 are sample sizes
• S₁² & S₂² are variances of sample 1 and sample 2
• x̅₁ and x̅₂ are means of sample sizes

ASSUMPTIONS:

• Observations (data):
  • Are Normally distributed.
    • Note: Although this method is relatively robust against non-normality the shape of distribution should at least be symmetric.
  • Within and between samples (groups) are independent of each other.
  • Are randomly “drawn”.
  • Are continuous.
    • Depending on some conditions (e.g. wide data range and field of applications), data that is not continuous (e.g. count) may still be used.
    • Not to use certain ratio type data (e.g. proportions and percentages).
• Variances are unknown and estimated by sample variances (S²).
• Variances are not equal, S₁² ≠ S₂²

Steps to Calculate Two Sample T Hypothesis Test (Unequal Variance):

Step 1 – Calculate mean (X̅) of each sample

•  
  X̅ = (x₁ + x₂ + ….+ x_n) / n
• Where, x₁, x₂, …x_n are observations in each sample (“n₁” = sample size in sample 1 and “n₂” = sample size in sample 2)

Step 2 – Calculate variance (S²) of each sample

• S² = {(X̅– x₁)² + (
  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1

Step 3 – Calculate t value

Step 4 – Calculate degrees of freedom (DF)

• Round to nearest integer

Step 5 – Compare t-calc with critical value in the t-distribution table.

• Look up the t-critical value in t-distribution table given calculated degrees of freedom in column α (significance level):
  • For 1 tailed test the t-critical is based on the given α
    • The commonly used T-critical value for a 1 tailed test is in the 0.05 α column
  • For 2 tailed test the t-critical is based on the given α divided by 2
    • The commonly used T-critical value for a 2 tailed test is in the 0.025 α column

INTERPRETATION:

• Finally, interpret the decision in the context of the original claim.

Example of a Two Sample T Hypothesis Test (Unequal Variance) in a DMAIC Project

Two-tailed (Unequal Variance)

Example: Dogs are fed with two different diets. Test if the two diets differ significantly as regards their effect on their weight. Assuming the population variance is unequal, at 95% confidence level, is there enough evidence to support the claim?

• Null Hypothesis (H₀) : The mean dog weights are the same with Diet 1 and Diet 2
• Alternative Hypothesis (H₁) : The mean dog weights are not the same with Diet 1 and Diet 2
• Significance level: α=0.05

• n1=10
• n2=10

Calculate mean (X̅) of each sample

• X̅1 = (x₁ + x₂ + ….+ x_n) / n =24.4
• X̅2 = (x₁ + x₂ + ….+ x_n) / n =20.00

Calculate variance (S²) of each sample

• S1² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1 =25.6
• S2² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1=56.22
• S1²/n1=2.56
• S2²/n2=5.62

Calculate critical value

Calculate the degrees of freedom

Refer two tailed t table for 16 degrees of freedom

If the calculated t value is less than -2.120 or greater than 2.120, then reject the null hypothesis.

Test Statistic

Interpret the results:

Compare t _calc to t _critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is less than the critical value and it is not in the rejection region, hence we failed to reject the null hypothesis ( H₀). So, the mean dog weights are the same with Diet 1 and Diet 2.

Two Sample T 2 Tailed Unqual Variance template file

Right-tailed (Unequal Variance)

Example: A researcher believes that students in school A have better computer knowledge than school B. Randomly computer marks were collected from the two schools for the same curriculum. Assuming the population variance is unequal, at 95% confidence level, is there enough evidence to support the claim?

• Null Hypothesis (H₀) : Mean students marks in school A is equal to the school B
• Alternative Hypothesis (H₁) : Mean students marks in school A is greater than the school B
• Significance level: α=0.05

• n1=19
• n2=17

Calculate mean (X̅) of each sample

• X̅1 = (x₁ + x₂ + ….+ x_n) / n =26.19
• X̅2 = (x₁ + x₂ + ….+ x_n) / n =20.65

Calculate variance (S²) of each sample

• S1² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1 =29.23
• S2² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1=84.95
• S1²/n1=1.54
• S2²/n2=5.00

Calculate critical value

Calculate the degrees of freedom

Refer one tailed t table for 25 degrees of freedom

If the calculated t value is greater than 1.708, then reject the null hypothesis.

Test Statistic

Interpret the results:

Compare t _calc to t _critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is greater than the critical value and it is in the rejection region, hence we reject the null hypothesis ( H₀). So, the mean students marks in school A is greater than the school B.

Two Sample T Right Tailed Unequal Variance template file

Left-tailed (Unequal Variance)

Example: A manufacturer believes that breaking strength of plastic parts of supplier A is less than the supplier B. Operator randomly checked the breaking strength of plastic parts of two suppliers. Assuming the population variance is unequal, at 95% confidence level, is there enough evidence to support the claim?

• Null Hypothesis (H₀) : Mean breaking strength of plastic parts of supplier A is equal to the supplier B
• Alternative Hypothesis (H₁) : Mean breaking strength of plastic parts of supplier A is less than supplier B
• Significance level: α=0.05

• n1=12
• n2=12

Calculate mean (X̅) of each sample

• X̅1 = (x₁ + x₂ + ….+ x_n) / n =161.67
• X̅2 = (x₁ + x₂ + ….+ x_n) / n =158.08

Calculate variance (S²) of each sample

• S1² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1 =10.97
• S2² = {(X̅– x₁)² + (

  

  X̅ – x₂)² + … +(X̅ – x_n)²} / n-1=25.72
• S1²/n1=0.91
• S2²/n2=2.14

Calculate critical value

Calculate the degrees of freedom

Refer one tailed t table for 19 degrees of freedom

If the calculated t value is less than 1.728, then reject the null hypothesis.

Test Statistic

Interpret the results:

Compare t _calc to t _critical . In hypothesis testing, a critical value is a point on the test distribution compares to the test statistic to determine whether to reject the null hypothesis. Calculated t statistic value is greater than the critical value and it is not in the rejection region, hence we failed to reject the null hypothesis ( H₀). So, Mean breaking strength of plastic parts of supplier A is equal to the supplier B.

Two Sample T Left Tailed Unequal Variance template file

Two Sample T Hypothesis Test (Unequal Variance) Videos

Additional Two Sample T Hypothesis Tests Resources

• Good example of  two sample T tests here.

• http://www.cliffsnotes.com/math/statistics/univariate-inferential-tests/two-sample-t-test-for-comparing-two-means (Two sample T test for comparing two means / DF for separate s: the smaller of n 1– 1 and n 2– 1 DF for pooled s: df = n 1+ n 2– 2)

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