Question 1192206: 1. A Science teacher used two different languages in teaching her two classes. Thirty students from
class A, which used Filipino as medium of instruction, showed a mean performance of 98 and a
standard deviation of 9.2. While 35 students from class B, which used the English medium,
revealed an average performance of 93 with standard deviation of 7.3. Is there a difference in
performance between the two samples? Use α = 0.01.
2. A language proficiency test was given to 55 Liberal Arts and 49 Business Administration
freshmen. The Liberal Arts students got a mean score of 78 with a standard deviation of 11. The
Business Administration students got a mean score of 82 with a standard deviation of 12. Is there
a reason to believe that there is significant difference in the performance of the two groups of
students? Use α = 0.0
3.Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were
randomly selected to investigate whether there was any difference in the mean time in years that
they were kept by the original owner before being sold. For cars the mean was 5.3 years with
standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation
3.0 years.
a. Construct the 95% confidence interval for the difference in the means based on these data.
b. Test the hypothesis that there is a difference in the means against the null hypothesis that
there is no difference. Use the 1% level of significance.
c. Compute the observed significance of the test in part (b).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve these hypothesis testing problems:
**Problem 1: Science Teacher's Classes**
* **Hypotheses:**
* H₀: There is no difference in performance between the two classes (μ₁ = μ₂)
* H₁: There is a difference in performance between the two classes (μ₁ ≠ μ₂) (Two-tailed test)
* **Significance Level:** α = 0.01
* **Test Statistic (z-score):** Since the sample sizes are large, we use a z-test.
z = (x̄₁ - x̄₂) / √((s₁²/n₁) + (s₂²/n₂))
z = (98 - 93) / √((9.2²/30) + (7.3²/35))
z = 5 / √(2.821 + 1.526)
z = 5 / √4.347
z ≈ 2.39
* **Critical Values:** For a two-tailed test at α = 0.01, the critical values are ±2.576.
* **Decision:** Since |z| = 2.39 < 2.576, we *fail to reject* the null hypothesis.
* **Conclusion:** There is not enough evidence at the 0.01 level of significance to conclude that there is a difference in performance between the two classes.
**Problem 2: Language Proficiency Test**
* **Hypotheses:**
* H₀: There is no significant difference in performance (μ₁ = μ₂)
* H₁: There is a significant difference in performance (μ₁ ≠ μ₂) (Two-tailed test)
* **Significance Level:** α = 0.0 (This is unusual. A significance level of 0 means you will *never* reject the null hypothesis unless the sample means are *exactly* equal. I'll proceed with α = 0.05, which is more common. If you truly meant α = 0, the answer is that you would never reject the null hypothesis.)
* **Test Statistic (z-score):** Again, we use a z-test due to the large sample sizes.
z = (78 - 82) / √((11²/55) + (12²/49))
z = -4 / √(2.2 + 2.939)
z = -4 / √5.139
z ≈ -1.77
* **Critical Values (using α = 0.05):** For a two-tailed test at α = 0.05, the critical values are ±1.96.
* **Decision (using α = 0.05):** Since |z| = 1.77 < 1.96, we *fail to reject* the null hypothesis.
* **Conclusion (using α = 0.05):** There is not sufficient evidence at the 0.05 level of significance to believe that there is a significant difference in the performance of the two groups of students.
**Problem 3: Used Cars and Trucks**
This problem was already addressed in your previous question. I'll summarize the answers:
* **a. 95% Confidence Interval:** (-2.953, -0.647)
* **b. Hypothesis Test:**
* H₀: μ₁ = μ₂
* H₁: μ₁ ≠ μ₂
* α = 0.01
* z ≈ -3.06
* Critical values: ±2.576
* Decision: Reject the null hypothesis.
* Conclusion: There is sufficient evidence at the 1% level of significance to conclude that there is a difference in the mean time that cars and pickup trucks are kept by their original owners.
* **c. Observed Significance (p-value):** ≈ 0.0022
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