Question 1174678: The manager of the credit department for an oil company would like to determine whether the average monthly balance of credit card holders is higher than P3,000. An auditor randomly samples 150 accounts and finds that the average owed is P4170 with a standard deviation of P1,182.50. Using the 0.05 level of significance, can the auditor conclude that there is evidence that the average monthly balance is really higher than P3,000?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this hypothesis test problem:
**1. Set up Hypotheses**
* **Null Hypothesis (H0):** μ ≤ 3000 (The average monthly balance is less than or equal to P3,000)
* **Alternative Hypothesis (HA):** μ > 3000 (The average monthly balance is higher than P3,000)
**2. Determine the Test Statistic**
* Since the sample size (n = 150) is large, we can use the t-test.
* The formula for the t-statistic is:
* t = (x̄ - μ) / (s / √n)
* Where:
* x̄ = sample mean (4170)
* μ = population mean (3000)
* s = sample standard deviation (1182.50)
* n = sample size (150)
**3. Calculate the Test Statistic**
* t = (4170 - 3000) / (1182.50 / √150)
* t = 1170 / (1182.50 / 12.247)
* t = 1170 / 96.55
* t = 12.12
**4. Determine the Critical Value**
* Significance level (α) = 0.05
* Degrees of freedom (df) = n - 1 = 150 - 1 = 149
* Because this is a right-tailed test, we need to find the critical t-value that corresponds to an area of 0.05 in the right tail of the t-distribution with 149 degrees of freedom.
* Using a t-table or calculator, the critical t-value is approximately 1.656.
**5. Make a Decision**
* Compare the calculated t-statistic (12.12) with the critical t-value (1.656).
* Since 12.12 > 1.656, we reject the null hypothesis.
**6. State the Conclusion**
* There is sufficient evidence at the 0.05 level of significance to conclude that the average monthly balance of credit card holders is higher than P3,000.
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