Question 1171066: The performance of 50 Psychology students showed a mean of 98 with a standard deviation of 9.2. A sample of 40 students showed an average performance of 95 with a standard deviation of 7.3. Is there a difference in performance between the two samples using 0.05 level, two-tailed?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to conduct the hypothesis test to determine if there's a difference in performance between the two samples.
**1. State the Hypotheses**
* **Null Hypothesis (H₀):** There is no difference in the mean performance between the two samples.
* H₀: μ₁ = μ₂
* **Alternative Hypothesis (H₁):** There is a difference in the mean performance between the two samples.
* H₁: μ₁ ≠ μ₂ (two-tailed test)
**2. Determine the Test Statistic**
Since we are comparing the means of two independent samples and we have sample standard deviations, we will use a two-sample z-test. (We can use a z-test because the sample sizes are reasonably large.)
The formula for the z-statistic is:
z = (x̄₁ - x̄₂) / √( (σ₁²/n₁) + (σ₂²/n₂) )
Where:
* x̄₁ = mean of sample 1
* x̄₂ = mean of sample 2
* σ₁ = standard deviation of sample 1
* σ₂ = standard deviation of sample 2
* n₁ = sample size of sample 1
* n₂ = sample size of sample 2
**3. Calculate the Test Statistic**
* x̄₁ = 98
* σ₁ = 9.2
* n₁ = 50
* x̄₂ = 95
* σ₂ = 7.3
* n₂ = 40
z = (98 - 95) / √((9.2²/50) + (7.3²/40))
z = 3 / √((84.64/50) + (53.29/40))
z = 3 / √(1.6928 + 1.33225)
z = 3 / √3.02505
z = 3 / 1.739
z ≈ 1.725
**4. Determine the Critical Value or P-value**
* Significance level (α) = 0.05
* Type of test: Two-tailed
* **Critical Value Approach:**
* For a two-tailed test with α = 0.05, the critical z-values are ±zα/2 = ±z0.025.
* Using a standard normal distribution table or a calculator, we find that z0.025 ≈ 1.96.
* **P-value Approach:**
* Using a z-table or calculator, we find the p-value associated with z = 1.725 for a two-tailed test.
* P(Z > 1.725) ≈ 0.0423
* P(Z < -1.725) ≈ 0.0423
* p-value = 2 \* 0.0423 ≈ 0.0846
**5. Make a Decision**
* **Critical Value Approach:**
* The calculated z-statistic is 1.725.
* The critical z-values are ±1.96.
* Since |1.725| < 1.96, we fail to reject the null hypothesis.
* **P-value Approach:**
* The p-value is approximately 0.0846.
* The significance level is 0.05.
* Since the p-value (0.0846) is greater than the significance level (0.05), we fail to reject the null hypothesis.
**6. Conclusion**
There is not sufficient evidence at the 0.05 significance level to conclude that there is a difference in performance between the two samples.
|
|
|
