SOLUTION: Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the tradition method or P-value method. Identify (a

Question 1198706: Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the tradition method or P-value method. Identify (a) the null and alternative hypotheses, (b) test statistic, (c) critical value(s) or P-value (or range of P-value) as appropriate, and (d) state the final conclusion that addresses the final claim.
In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours) between failures.
518 548 561 523 536
499 538 557 528 563
At the 0.05 significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours? You may assume the times between failures are normally distributed. Use the P-value method of testing hypotheses.
Note: The value of sample mean (x̄≈537.1hrs) and sample standard deviation (s≈20.701hrs) are calculated.
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**a) Hypotheses:**
* **Null Hypothesis (H₀):** μ ≤ 520 hours
* This states that the mean time between failures for the modified components is less than or equal to 520 hours.
* **Alternative Hypothesis (H₁):** μ > 520 hours
* This states that the mean time between failures for the modified components is greater than 520 hours.
**b) Test Statistic**
* **Calculate the t-statistic:**
* t = (x̄ - μ) / (s / √n)
* where:
* x̄ = sample mean = 537.1 hours
* μ = hypothesized population mean = 520 hours
* s = sample standard deviation = 20.701 hours
* n = sample size = 10
* t = (537.1 - 520) / (20.701 / √10)
* t ≈ 2.632
**c) P-value**
* **Degrees of Freedom (df):** df = n - 1 = 10 - 1 = 9
* **Using a t-distribution table or statistical software:** Find the P-value associated with t = 2.632 and df = 9 for a one-tailed test.
* **P-value ≈ 0.013**
**d) Conclusion about the Null Hypothesis**
* **Compare P-value to significance level:**
* P-value (0.013) < α (0.05)
* **Decision:** Since the P-value is less than the significance level, we **reject the null hypothesis**.
**e) Conclusion about the Claim**
* **There is sufficient evidence at the 0.05 significance level to support the claim that the mean time between failures for the modified components is greater than 520 hours.**
**In Summary:**
* The calculated t-statistic falls in the rejection region.
* The P-value is less than the significance level, indicating strong evidence against the null hypothesis.
* We can conclude that the modification likely increases the mean time between failures for the computer components.

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