Question 1170788: A large nation wide chain of home improvement centres was having an end of season clearance of lawn mowers. the number of lawn mowers sold during this sale from a sample of 10 stores was as follows;
77;12;21;50;70;44;67;7;11;73
at the 0.05 level of significance, is there evidence that an average of more than 25 lawn mowers per store have been sold during this sale?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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.
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