SOLUTION: A standardized test was administered to a thousand pupils with a mean average score of 85 and a standard deviation of 8. A random sample of 50 pupils was given the same test and sh

Question 1171061: A standardized test was administered to a thousand pupils with a mean average score of 85 and a standard deviation of 8. A random sample of 50 pupils was given the same test and showed an average score of 83.20. Is there evidence to show that this group has a lower performance than the ones in general at 0.05 level (two-tailed)?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's conduct the hypothesis test to see if the sample group has a lower performance than the general population.
**1. State the Hypotheses**
* **Null Hypothesis (H₀):** The sample group's mean score is equal to the population mean score.
* H₀: μ = 85
* **Alternative Hypothesis (H₁):** The sample group's mean score is different from the population mean score.
* H₁: μ ≠ 85 (two-tailed test)
**2. Determine the Test Statistic**
* We are given the population standard deviation, so we will use a z-test.
* The formula for the z-statistic is:
z = (x̄ - μ) / (σ / √n)
Where:
* x̄ = sample mean
* μ = population mean
* σ = population standard deviation
* n = sample size
**3. Calculate the Test Statistic**
* x̄ = 83.20
* μ = 85
* σ = 8
* n = 50
z = (83.20 - 85) / (8 / √50)
z = -1.8 / (8 / 7.071)
z = -1.8 / 1.131
z ≈ -1.591
**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 ±z_α/2 = ±z_0.025.
* Using a standard normal distribution table or a calculator, we find that z_0.025 ≈ 1.96.
* **P-value Approach:**
* Using a z-table or calculator, we find the p-value associated with z = -1.591 for a two-tailed test.
* P(Z < -1.591) ≈ 0.0558
* P(Z > 1.591) ≈ 0.0558
* p-value = 2 \* 0.0558 ≈ 0.1116
**5. Make a Decision**
* **Critical Value Approach:**
* The calculated z-statistic is -1.591.
* The critical z-values are ±1.96.
* Since |-1.591| < 1.96, we fail to reject the null hypothesis.
* **P-value Approach:**
* The p-value is approximately 0.1116.
* The significance level is 0.05.
* Since the p-value (0.1116) 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 the sample group has a lower performance than the general population.

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