Model Answer
0 min readIntroduction
Hypothesis testing is a fundamental statistical method used to make inferences about population parameters based on sample data. In biological and agricultural research, such as evaluating animal feeds, it provides a robust framework to determine if observed differences in outcomes (like weight gain) are statistically significant or merely due to random chance. This problem specifically delves into the application of t-tests, which are widely used when comparing means of two groups, particularly when sample sizes are small and the population standard deviation is unknown. Understanding the distinction between independent and paired samples is critical for selecting the appropriate t-test and drawing valid conclusions.
(i) Independent Samples t-test (Two-Sample t-test)
This scenario assumes that the two groups of animals (fed Feed A and Feed B) are distinct and unrelated. We will perform a two-sample t-test to compare their mean weight gains.Step 1: State the Hypotheses
- Null Hypothesis (H₀): There is no significant difference in the mean weight gain between Feed A and Feed B. (μ_A = μ_B)
- Alternative Hypothesis (H₁): Feed B is more effective than Feed A, meaning the mean weight gain for Feed B is greater than Feed A. (μ_B > μ_A)
Step 2: Calculate Sample Statistics
We first list the data for each feed:Feed A (n_A = 8): 45, 48, 47, 50, 46, 49, 47, 48
- Mean (x̄_A) = (45+48+47+50+46+49+47+48) / 8 = 380 / 8 = 47.5 kg
- Sum of squared deviations from the mean (Σ(x_i - x̄_A)²)
- (45-47.5)² = 6.25
- (48-47.5)² = 0.25
- (47-47.5)² = 0.25
- (50-47.5)² = 6.25
- (46-47.5)² = 2.25
- (49-47.5)² = 2.25
- (47-47.5)² = 0.25
- (48-47.5)² = 0.25
- Sum = 18
- Sample Variance (s²_A) = 18 / (8-1) = 18 / 7 ≈ 2.5714
Feed B (n_B = 8): 51, 53, 52, 55, 50, 52, 54, 33
- Mean (x̄_B) = (51+53+52+55+50+52+54+33) / 8 = 390 / 8 = 48.75 kg
- Sum of squared deviations from the mean (Σ(x_i - x̄_B)²)
- (51-48.75)² = 5.0625
- (53-48.75)² = 18.0625
- (52-48.75)² = 10.5625
- (55-48.75)² = 39.0625
- (50-48.75)² = 1.5625
- (52-48.75)² = 10.5625
- (54-48.75)² = 27.5625
- (33-48.75)² = 248.0625
- Sum = 360.5
- Sample Variance (s²_B) = 360.5 / (8-1) = 360.5 / 7 ≈ 51.5
Step 3: Calculate the Pooled Standard Deviation (s_p)
Since we assume equal variances (a common assumption for independent t-tests when sample sizes are equal), we calculate the pooled variance: s²_p = [ (n_A - 1)s²_A + (n_B - 1)s²_B ] / [ (n_A - 1) + (n_B - 1) ] s²_p = [ (7 * 2.5714) + (7 * 51.5) ] / [ 7 + 7 ] s²_p = [ 18 + 360.5 ] / 14 = 378.5 / 14 ≈ 27.0357 s_p = √27.0357 ≈ 5.1996 kgStep 4: Calculate the t-statistic
t = (x̄_B - x̄_A) / [ s_p * √(1/n_A + 1/n_B) ] t = (48.75 - 47.5) / [ 5.1996 * √(1/8 + 1/8) ] t = 1.25 / [ 5.1996 * √(2/8) ] t = 1.25 / [ 5.1996 * √0.25 ] t = 1.25 / [ 5.1996 * 0.5 ] t = 1.25 / 2.5998 ≈ 0.4808Step 5: Determine the Critical t-value
- Degrees of Freedom (df) = n_A + n_B - 2 = 8 + 8 - 2 = 14
- Level of Significance (α) = 0.025 (one-tailed)
Step 6: Make a Decision
Since the calculated t-statistic (0.4808) is less than the critical t-value (2.145), we fail to reject the null hypothesis.Conclusion for (i):
At the 0.025 level of significance, we cannot conclude that Feed B is more effective than Feed A when the two samples are considered independent. The observed difference in mean weight gain is not statistically significant and could be due to random chance.(ii) Paired Samples t-test (Dependent Samples t-test)
This scenario assumes the same set of eight animals received both feeds. This means we are looking at the difference in weight gain for each animal, making the samples dependent (paired).Step 1: State the Hypotheses
- Null Hypothesis (H₀): There is no significant mean difference in weight gain between Feed A and Feed B. (μ_d = 0)
- Alternative Hypothesis (H₁): There is a significant positive mean difference in weight gain for Feed B over Feed A. (μ_d > 0)
Step 2: Calculate the Differences (d_i) and their Statistics
For each animal, we calculate d_i = Weight Gain (Feed B) - Weight Gain (Feed A).| Animal | Feed A (x_A) | Feed B (x_B) | Difference (d = x_B - x_A) |
|---|---|---|---|
| 1 | 45 | 51 | 6 |
| 2 | 48 | 53 | 5 |
| 3 | 47 | 52 | 5 |
| 4 | 50 | 55 | 5 |
| 5 | 46 | 50 | 4 |
| 6 | 49 | 52 | 3 |
| 7 | 47 | 54 | 7 |
| 8 | 48 | 33 | -15 |
Now, calculate statistics for the differences (d):
- Number of pairs (n) = 8
- Mean of differences (d̄) = (6+5+5+5+4+3+7-15) / 8 = 20 / 8 = 2.5 kg
- Sum of squared deviations from the mean difference (Σ(d_i - d̄)²)
- (6-2.5)² = 12.25
- (5-2.5)² = 6.25
- (5-2.5)² = 6.25
- (5-2.5)² = 6.25
- (4-2.5)² = 2.25
- (3-2.5)² = 0.25
- (7-2.5)² = 20.25
- (-15-2.5)² = (-17.5)² = 306.25
- Sum = 360.25
- Standard Deviation of differences (s_d) = √[ Σ(d_i - d̄)² / (n-1) ] = √[ 360.25 / 7 ] = √51.4643 ≈ 7.1739 kg
Step 3: Calculate the t-statistic for Paired Samples
t = d̄ / (s_d / √n) t = 2.5 / (7.1739 / √8) t = 2.5 / (7.1739 / 2.8284) t = 2.5 / 2.5364 ≈ 0.9857Step 4: Determine the Critical t-value
- Degrees of Freedom (df) = n - 1 = 8 - 1 = 7
- Level of Significance (α) = 0.01 (one-tailed)
Step 5: Make a Decision
Since the calculated t-statistic (0.9857) is less than the critical t-value (2.998), we fail to reject the null hypothesis.Conclusion for (ii):
At the 0.01 level of significance, even when using a paired t-test, we cannot conclude that Feed B is more effective than Feed A. The mean difference in weight gain is not statistically significant. The substantial negative difference for one animal (-15 kg) significantly impacts the overall mean difference and standard deviation, reducing the t-statistic.Summary of Differences Between Independent and Paired t-tests
The choice between an independent and a paired t-test hinges on the nature of the samples being compared. This distinction is crucial for accurate statistical inference.
| Feature | Independent Samples t-test | Paired Samples t-test |
|---|---|---|
| Sample Relationship | Two distinct, unrelated groups (e.g., Feed A given to one group, Feed B to another). | Two measurements from the same group or matched pairs (e.g., same animals receive both feeds). |
| Data Structure | Two sets of measurements, assumed to be independent. | One set of differences calculated from paired observations. |
| Hypothesis Focus | Compares the means of two independent populations (μ1 vs. μ2). | Compares the mean of the differences (μd) to zero. |
| Degrees of Freedom | n1 + n2 - 2 | n - 1 (where n is the number of pairs) |
| Statistical Power | Generally lower if pairing exists but is ignored. | Higher when pairing is naturally present, as it controls for inter-subject variability. |
| Variance Assumption | Often assumes equal population variances (pooled variance used). | Focuses on the variance of the differences, not individual group variances. |
Conclusion
The analysis reveals that neither the independent samples t-test nor the paired samples t-test supports the conclusion that Feed B is more effective than Feed A at the specified significance levels. In the independent samples scenario (0.025 level), the observed difference was too small relative to the variability within groups to be statistically significant. For the paired samples scenario (0.01 level), despite an overall positive mean difference, the high variability among the differences, particularly influenced by one outlying data point, prevented us from rejecting the null hypothesis. This underscores the importance of statistical rigor and outlier management in research, suggesting that more data or a re-evaluation of experimental conditions might be necessary to definitively assess the feeds' comparative effectiveness.
Answer Length
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