Solved Problems of T-Test Formula

Problem 1: Determine whether the average weight of a sample of 20 mangoes is significantly different from the population’s average weight of 70 grams. The sample mean weight is 70.55 grams, and the sample standard deviation is 2.82 grams. Use one sample T-test.

Solution:

To perform a T-test, first of all, we define two hypotheses:

• Null hypothesis: The sample mean weight of mangoes is equal to the known population mean. (i.e., 70 grams).
• Alternative hypothesis: The sample mean weight of mangoes is not equal to the known mean value.

Then, determine the degrees of freedom (d_f): d_f = n – 1 = 20 – 1 = 19 and define the level of significance(α) as 0.05 for this case. Next, determine the t-value from the formula,

t = (70.55 – 70) / (2.82 / √20)

⇒ t ≈ 1.041

From the t-distribution table, we find 1.041 < 2.093. (i.e. p-value for α = 0.05). So, the null hypothesis is true. Thus, we conclude that the sample does not vary significantly from the population.

Problem 2: Determine if there is a significant difference in the average scores between the two teams. The following data is given: 
Team A: Score: 65, 68, 70, 63, 67
Team B: Score: 62, 66, 69, 64, 68

Solution:

According to the question, we come to know that we need to perform an Independent Samples T-test. Set up the null hypothesis and alternative hypothesis:

• Null hypothesis: The means of the two groups are equal (μ_A = μ_B).
• Alternative hypothesis: The means of the two groups are not equal (μ_A ≠ μ_B).

Next, we calculate the sample means (x̄_A and x̄_B) and sample standard deviations (σ_A and σ_B):

Team A:

• Sample size (n_A) = 5
• Sample mean (x̄_A) = (65 + 68 + 70 + 63 + 67) / 5 = 66.6
• Sample standard deviation (σ_A) ≈ 2.607

Team B:

• Sample size (n_B) = 5
• Sample mean (x̄_B) = (62 + 66 + 69 + 64 + 68) / 5 = 65.8
• Sample standard deviation (σ_B) ≈ 2.588

Now, we calculate the t-value using the formula:

t = (x̄_A – x̄_B) / √((σ_A² / n_A) + (σ_B² / n_B))

⇒ t = (66.6 – 65.8) / √{(2.607²/5) + (2.588²/5)}

⇒ t ≈ 0.296

Then, determine the degrees of freedom (df): 

df = n_A + n_B – 2 = 5 + 5 – 2 = 8 

and set the level of significance as 0.05. 

From the table, we get the critical t-value as 2.306. As the calculated t-value is less than the critical t-value, we conclude that the null hypothesis is not rejected, which suggests that there is no significant difference between the average scores of the two teams.

Problem 3: You need to assess the effectiveness of a new teaching scheme by comparing the test scores of the same group of students before and after the implementation of the scheme. The following data is given:
Before scores: 76, 88, 65, 56, 76
After scores: 85, 95, 75, 60, 81
Determine if there is a significant difference in the average test scores before and after the implementation of the scheme.

Solution:

Here, we need to perform a Paired Samples T-test, as we need to compare data of the same sample. Set up the null hypothesis and alternative hypothesis:

• Null hypothesis: The population mean difference between the before and after scores are zero (μ_d = 0).
• Alternative hypothesis: The population mean difference between the before and after scores is not zero (μ_d ≠ 0).

Next, calculate the differences between the paired observations:

Difference (d) = After score – Before score

• d₁ = 85 – 76 = 9
• d₂ = 95 – 88 = 7
• d₃ = 75 – 65 = 10
• d₄ = 60 – 56 = 4
• d₅ = 81 – 76 = 5

Now, calculate the sample mean (x̄d) and sample standard deviation (σ_d) of the differences:

• Sample size (n) = 5
• Sample mean (x̄_d) = (d₁ + d₂ + d₃ + d₄ + d₅) / 5 = (9 + 7 + 10 + 4 + 5) / 5 = 7
• Sample standard deviation (σ_d) ≈ 2.828

Then, calculate the t-value using the formula:

t = (x̄_d – μ_d) / (σ_d / √n) 

⇒ t = (7 – 0) / (2.828 / √5)

⇒ t ≈ 5.535

Next, calculate the value of degrees of freedom (df): 

df = n – 1 = 5 – 1 = 4. 

And, define the level of significance(α) as 0.05. 

Now, from the t-distribution table, we find that the critical t-value is 2.776. As the calculated t-value is greater than the critical t-value (5.535 > 2.776), thus, the null hypothesis is rejected. And we conclude that there is a significant difference in the average test scores before and after the implementation of the scheme.

T-Test is a method used in statistics to determine if there is a significant difference between the means of two groups and how they are related. In T-Test statistics, the sample data is a subset of the two groups that we use to draw conclusions about the groups as a whole. 

For example, if we want to know the average weight of mangoes grown on a farm, the population would consist of all the mangoes that grew on the farm. However, it would be time-consuming to weigh each mango. Instead, we could take a sample of mangoes from trees at different locations on the farm and use their weights to make inferences about the average weight of all the mangoes grown on the farm.