SOLUTION: 6. A school administrator claims that less than 50% of the students of the school are dissatisfied by the community cafeteria service. Test this claim by using sample data obtained

Question 1186605: 6. A school administrator claims that less than 50% of the students of the school are dissatisfied by the community cafeteria service. Test this claim by using sample data obtained from a survey of 500 students of the school where 54% indicated their dissatisfaction of the community cafeteria service. Use alpha = 0.05
answer the problem using the steps in hypothesis testing.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to conduct a hypothesis test for the school administrator's claim:
**1. State the Hypotheses:**
* **Null Hypothesis (H₀):** p ≥ 0.50 (The proportion of dissatisfied students is greater than or equal to 50%)
* **Alternative Hypothesis (H₁ or Ha):** p < 0.50 (The proportion of dissatisfied students is less than 50%) This is a *left-tailed test*.
**2. Determine the Level of Significance (alpha):**
* α = 0.05 (This is given in the problem.)
**3. Calculate the Test Statistic:**
* **Sample Proportion (p̂):** p̂ = 0.54 (54% of the 500 students surveyed)
* **Hypothesized Proportion (p₀):** p₀ = 0.50 (The proportion under the null hypothesis)
* **Standard Error (SE):** SE = sqrt[(p₀ * (1 - p₀)) / n] = sqrt[(0.50 * (1 - 0.50)) / 500] = sqrt(0.0005) ≈ 0.0224
* **Z-score:** z = (p̂ - p₀) / SE = (0.54 - 0.50) / 0.0224 ≈ 1.79
**4. Determine the Critical Value or the P-value:**
* **Critical Value (Left-tailed test):** For α = 0.05, the critical z-value is -1.645. (You can find this using a z-table or a calculator.)
* **P-value:** Since this is a left-tailed test, the p-value is the probability of observing a z-score as extreme as, or more extreme than, 1.79. Using a z-table or calculator, we find the area to the *left* of z = 1.79. Because this is the opposite of the tail we are interested in, we subtract this from 1. So p-value = 1 - 0.9633 = 0.0367.
**5. Make a Decision:**
* **Using Critical Values:** Our calculated z-score (1.79) is *greater* than the critical z-value (-1.645). Therefore, we fail to reject the null hypothesis.
* **Using P-value:** The p-value (0.0367) is *less* than the level of significance (0.05). Because this is a left tailed test, we are interested in the values to the left of the critical value and therefore, reject the null hypothesis.
**6. Conclusion:**
There *is* sufficient evidence at the 0.05 level of significance to support the school administrator's claim that less than 50% of students are dissatisfied with the cafeteria service.

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