Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. Choose the correct option based on the problem information. 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. Use the P-value method of testing hypotheses. Choose the correct P-value and conclusion based on the null hypothesis. 0 0.01

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Statistics Question

**Statistical Hypothesis Testing: A Case Study**

This exercise involves testing a claim regarding the mean time between failures of a modified computer component, using a simple random sample from a normally distributed population.

### Problem Statement

A test determines that the mean time between failures for a computer component is initially 520 hours. A modification to the component is made, intended to increase this time. A random sample of 10 modified components yields the following results (in hours between failures):

- Sample Times: 518, 548, 561, 523, 536, 499, 538, 557, 528, 563

**Objective**: At a 0.05 significance level, test the claim that the mean time between failures for the modified components is greater than 520 hours using the P-value approach.

### Methodology

1. **Hypotheses Setup**:
   - Null Hypothesis (\( H_0 \)): The mean time between failures is 520 hours or less.
   - Alternative Hypothesis (\( H_1 \)): The mean time between failures is greater than 520 hours.

2. **Significance Level**: \( \alpha = 0.05 \)

3. **Test the Claim**:
   - Use the sample data and an appropriate statistical test to find the P-value.
   - Compare the P-value with the significance level to decide on rejecting the null hypothesis.

### Conclusion Options

Based on the P-value, choose one of the following conclusions:

- \( 0.01 < \text{P-value} < 0.025 \). Reject \( H_0 \).
- \( 0.05 < \text{P-value} < 0.1 \). Reject \( H_0 \).
- \( 0.01 < \text{P-value} < 0.025 \). Fail to reject \( H_0 \).
- \( 0.1 < \text{P-value} < 0.2 \). Reject \( H_0 \). 

This exercise emphasizes the application of the P-value method in hypothesis testing, focusing on interpreting the statistical output and drawing conclusions from it.
Transcribed Image Text:**Statistical Hypothesis Testing: A Case Study** This exercise involves testing a claim regarding the mean time between failures of a modified computer component, using a simple random sample from a normally distributed population. ### Problem Statement A test determines that the mean time between failures for a computer component is initially 520 hours. A modification to the component is made, intended to increase this time. A random sample of 10 modified components yields the following results (in hours between failures): - Sample Times: 518, 548, 561, 523, 536, 499, 538, 557, 528, 563 **Objective**: At a 0.05 significance level, test the claim that the mean time between failures for the modified components is greater than 520 hours using the P-value approach. ### Methodology 1. **Hypotheses Setup**: - Null Hypothesis (\( H_0 \)): The mean time between failures is 520 hours or less. - Alternative Hypothesis (\( H_1 \)): The mean time between failures is greater than 520 hours. 2. **Significance Level**: \( \alpha = 0.05 \) 3. **Test the Claim**: - Use the sample data and an appropriate statistical test to find the P-value. - Compare the P-value with the significance level to decide on rejecting the null hypothesis. ### Conclusion Options Based on the P-value, choose one of the following conclusions: - \( 0.01 < \text{P-value} < 0.025 \). Reject \( H_0 \). - \( 0.05 < \text{P-value} < 0.1 \). Reject \( H_0 \). - \( 0.01 < \text{P-value} < 0.025 \). Fail to reject \( H_0 \). - \( 0.1 < \text{P-value} < 0.2 \). Reject \( H_0 \). This exercise emphasizes the application of the P-value method in hypothesis testing, focusing on interpreting the statistical output and drawing conclusions from it.
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