Question 1180643
**1. Hypotheses:**

* **Null Hypothesis (H0):** The population mean voting time using the new computerized method is greater than or equal to the old average (μ ≥ 30).
* **Alternative Hypothesis (H1):** The population mean voting time using the new computerized method is less than the old average (μ < 30).  This is what we want to test.

**2. Level of Significance:**

* You didn't specify a level of significance, so let's use a common one: α = 0.05

**3. Test Statistic:**

Since we have a small sample size (n = 25) and we don't know the population standard deviation, we'll use a t-test.

* **t = (x̄ - μ) / (s / √n)** 
   where:
     * x̄ = sample mean (24 minutes)
     * μ = population mean under the null hypothesis (30 minutes)
     * s = sample standard deviation (2.9 minutes)
     * n = sample size (25)

* **Calculate t:**
   t = (24 - 30) / (2.9 / √25) = -6 / 0.58 ≈ -10.34

**4. Critical Region:**

Since this is a one-tailed test (H1: μ < 30), we need to find the critical t-value for α = 0.05 and degrees of freedom (df) = n - 1 = 24.

* Using a t-table or calculator, the critical t-value is approximately -1.711.

* **Critical Region:**  t < -1.711

**5. Decision and Conclusion:**

* Our calculated t-value (-10.34) falls in the critical region (t < -1.711).

* Therefore, we **reject the null hypothesis (H0)**.

* There is sufficient evidence to support the claim that the population mean voting time using the computerized method is less than the old average of 30 minutes.

**In simpler terms:** The sample data provides strong evidence that the new computerized voting method is faster than the old procedure.