Question 1170788
To determine if there is evidence that an average of more than 25 lawn mowers per store have been sold during the sale, we'll perform a one-sample t-test. Here's how:

**1. State the Hypotheses**

* **Null Hypothesis (H₀):** The average number of lawn mowers sold per store is 25 or less. (μ ≤ 25)
* **Alternative Hypothesis (H₁):** The average number of lawn mowers sold per store is more than 25. (μ > 25)

This is a right-tailed test.

**2. Calculate the Sample Mean and Standard Deviation**

* **Sample Data:** 77, 12, 21, 50, 70, 44, 67, 7, 11, 73
* **Sample Size (n):** 10

Let's calculate the sample mean and standard deviation:

* **Sample Mean (x̄):**
    * x̄ = (77 + 12 + 21 + 50 + 70 + 44 + 67 + 7 + 11 + 73) / 10
    * x̄ = 432 / 10
    * x̄ = 43.2

* **Sample Standard Deviation (s):**
    * To calculate this, we first find the variance (s²):
        * s² = \[Σ(xᵢ - x̄)²\] / (n - 1)
        * s² = \[(77-43.2)² + (12-43.2)² + (21-43.2)² + (50-43.2)² + (70-43.2)² + (44-43.2)² + (67-43.2)² + (7-43.2)² + (11-43.2)² + (73-43.2)²\] / 9
        * s² = \[1142.44 + 973.44 + 492.84 + 46.24 + 718.24 + 0.64 + 566.44 + 1303.24 + 1036.84 + 888.04\] / 9
        * s² = 7168 / 9
        * s² ≈ 796.44
    * s = √s²
    * s = √796.44
    * s ≈ 28.22

**3. Calculate the Test Statistic (t-value)**

* The formula for the t-test statistic is:
    * t = (x̄ - μ) / (s / √n)
    * Where:
        * x̄ is the sample mean
        * μ is the population mean (from the null hypothesis)
        * s is the sample standard deviation
        * n is the sample size

* t = (43.2 - 25) / (28.22 / √10)
* t = 18.2 / (28.22 / 3.16)
* t = 18.2 / 8.93
* t ≈ 2.04

**4. Determine the Degrees of Freedom and Critical Value**

* Degrees of Freedom (df) = n - 1 = 10 - 1 = 9
* Significance Level (α) = 0.05
* Since this is a right-tailed test, we look up the critical t-value in a t-table or use a calculator with df = 9 and α = 0.05.
* The critical t-value is approximately 1.833.

**5. Make a Decision**

* Calculated t-value (2.04) > Critical t-value (1.833)

Since the calculated t-value is greater than the critical t-value, we reject the null hypothesis.

**6. Conclusion**

At the 0.05 level of significance, there is sufficient evidence to conclude that the average number of lawn mowers sold per store is more than 25.