Question 1176438: 3. A principal states that in his school, the average number of days per year missed by a student due to illness is less than 10. The ff. data show the number of days missed by 40 students a year. Is there sufficient evidence to believe the principal’s claim at alpha=.05?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! To test the principal's claim, we'll perform a one-tailed t-test. Since you haven't provided the actual data, I'll outline the steps and provide a general solution. You'll need to plug in the data to complete the calculations.
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** μ ≥ 10 (The average number of days missed is greater than or equal to 10)
* **Alternative Hypothesis (H1):** μ < 10 (The average number of days missed is less than 10)
**2. Determine the Significance Level:**
* α = 0.05
**3. Calculate the Sample Statistics:**
* **Sample Size (n):** 40
* **Calculate the Sample Mean (x̄):** Sum all the data points and divide by 40.
* **Calculate the Sample Standard Deviation (s):** Use the formula:
* s = √[Σ(xᵢ - x̄)² / (n - 1)]
**4. Calculate the Test Statistic (t):**
* t = (x̄ - μ) / (s / √n)
* Where:
* x̄ = sample mean
* μ = population mean (10)
* s = sample standard deviation
* n = sample size
**5. Determine the Critical Value:**
* Since we are performing a left-tailed test at α = 0.05 with df = n - 1 = 39, we need to find the critical t-value from the t-distribution table.
* For df = 39 and α = 0.05 (one-tailed), the critical t-value is approximately -1.684.
**6. Make a Decision:**
* **If the calculated t-statistic is less than the critical t-value (-1.684), reject the null hypothesis.** This means there is sufficient evidence to support the principal's claim.
* **If the calculated t-statistic is greater than or equal to the critical t-value, fail to reject the null hypothesis.** This means there is not sufficient evidence to support the principal's claim.
**Example (Illustrative - You need to plug in your data):**
Let's say, after calculating from your data, you find:
* x̄ = 8.5
* s = 4.0
Then:
* t = (8.5 - 10) / (4.0 / √40)
* t = -1.5 / (4.0 / 6.3245)
* t = -1.5 / 0.63245
* t = -2.372
In this example:
* -2.372 < -1.684
Therefore, you would reject the null hypothesis.
**Conclusion:**
* **If your calculated t-statistic is less than -1.684, there is sufficient evidence to support the principal's claim that the average number of days missed is less than 10 at α = 0.05.**
* **If your calculated t-statistic is greater than or equal to -1.684, there is not sufficient evidence to support the principal's claim at α = 0.05.**
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