Question 1179172: Question 3 (20 marks)
3(a)The Sales manager at Hybrid Industries a high-tech farming agency
claims that there has been a drop in the sales from the target of $2800/- per
week. Data was collected for the last 48 weeks and the sample mean was
computed as $ 2435/- with a SD $ 351. Verify at 5% los if the manager’s
claim can be accepted based on the sample observations.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to conduct a hypothesis test to verify the sales manager's claim:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The population mean sales are equal to or greater than $2800 (μ ≥ 2800).
* **Alternative Hypothesis (H1):** The population mean sales are less than $2800 (μ < 2800).
This is a left-tailed test.
**2. Set the Significance Level:**
* α = 0.05 (5%)
**3. Choose the Test Statistic:**
* Since the population standard deviation is unknown and the sample size is large (n = 48 > 30), we will use a z-test.
**4. Calculate the Test Statistic (z-score):**
* Sample mean (x̄): $2435
* Population mean (μ): $2800
* Sample standard deviation (s): $351
* Sample size (n): 48
z = (x̄ - μ) / (s / √n)
z = (2435 - 2800) / (351 / √48)
z = -365 / (351 / 6.928)
z = -365 / 50.66
z ≈ -7.20
**5. Find the Critical Value:**
* For a left-tailed test with α = 0.05, the critical z-value is -1.645. (You can find this using a z-table or calculator).
**6. Make a Decision:**
* Compare the calculated z-score (-7.20) to the critical z-value (-1.645).
* Since -7.20 < -1.645, the calculated z-score falls in the rejection region.
* Therefore, we reject the null hypothesis.
**7. Draw a Conclusion:**
* There is sufficient evidence at the 5% significance level to support the sales manager's claim that there has been a drop in sales from the target of $2800 per week.
**In summary:**
The calculated z-score of -7.20 is far into the rejection region of the left tailed test. Therefore we can confidently reject the null hypothesis and accept the sales managers claim.
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