In tests of a computer component, it is found that the mean time between failures is 911 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 25 modified components produce a mean time between failures of 957 hours. Using a 1% level of significance, perform a hypothesis test to determine whether the mean time between failures for the modified components is greater than 911 hours. Assume that the population standard deviation is s2 hours. Ho : µ> 957 hours H : µ= 957 hours Test statistic : z=87.6 Critical value : z=1.65 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 957 hours. Ho : µ=911 hours H : µ>911 hours Test statistic : z=4.42 Critical value : z= 2.33 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 911 hours. Ho : µ> 957 hours H : µ= 957 hours O Test statistic : z=4.42 Critical value : z=1.65 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 957 hours. H, : µ= 957 hours H :µ> 957 hours O Test statistic : z=4.42 Critical value : z=2.33 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 957 hours.
In tests of a computer component, it is found that the mean time between failures is 911 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 25 modified components produce a mean time between failures of 957 hours. Using a 1% level of significance, perform a hypothesis test to determine whether the mean time between failures for the modified components is greater than 911 hours. Assume that the population standard deviation is s2 hours. Ho : µ> 957 hours H : µ= 957 hours Test statistic : z=87.6 Critical value : z=1.65 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 957 hours. Ho : µ=911 hours H : µ>911 hours Test statistic : z=4.42 Critical value : z= 2.33 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 911 hours. Ho : µ> 957 hours H : µ= 957 hours O Test statistic : z=4.42 Critical value : z=1.65 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 957 hours. H, : µ= 957 hours H :µ> 957 hours O Test statistic : z=4.42 Critical value : z=2.33 Reject the null hypothesis. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 957 hours.
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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![### Hypothesis Testing on Mean Time Between Failures of Modified Computer Components
In tests of a computer component, it is found that the mean time between failures is **911 hours**. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of **25 modified components** produce a mean time between failures of **957 hours**. Using a **1% level of significance**, perform a hypothesis test to determine whether the mean time between failures for the modified components is greater than **911 hours**. Assume that the population standard deviation is **52 hours**.
#### Hypothesis Testing Steps:
1. **Set Up Hypotheses:**
- Null hypothesis (\(H_0\)): \(\mu = 911\) hours
- Alternative hypothesis (\(H_1\)): \(\mu > 911\) hours
2. **Test Statistic Calculation:**
- Use the Z-test statistic formula for population mean:
\[
z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
- Where:
- \(\bar{X}\) = Sample mean = 957 hours
- \(\mu\) = Population mean under null hypothesis = 911 hours
- \(\sigma\) = Population standard deviation = 52 hours
- \(n\) = Sample size = 25
- Plugging in the values:
\[
z = \frac{957 - 911}{\frac{52}{\sqrt{25}}} = \frac{46}{10.4} \approx 4.42
\]
3. **Critical Value:**
- From Z-tables, the critical value \(z_{\alpha}\) for a one-tailed test at the 1% significance level is 2.33.
4. **Decision Rule:**
- If the calculated test statistic exceeds the critical value (4.42 > 2.33), reject the null hypothesis.
#### Conclusion:
Since the calculated z-value (4.42) is greater than the critical value (2.33), we **reject the null hypothesis**. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 911 hours.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea0820f3-64e4-4dfd-b401-c1fa00677880%2F010e0879-ee4b-49ed-b520-9326bd50f000%2Fczcfuui_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hypothesis Testing on Mean Time Between Failures of Modified Computer Components
In tests of a computer component, it is found that the mean time between failures is **911 hours**. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of **25 modified components** produce a mean time between failures of **957 hours**. Using a **1% level of significance**, perform a hypothesis test to determine whether the mean time between failures for the modified components is greater than **911 hours**. Assume that the population standard deviation is **52 hours**.
#### Hypothesis Testing Steps:
1. **Set Up Hypotheses:**
- Null hypothesis (\(H_0\)): \(\mu = 911\) hours
- Alternative hypothesis (\(H_1\)): \(\mu > 911\) hours
2. **Test Statistic Calculation:**
- Use the Z-test statistic formula for population mean:
\[
z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
- Where:
- \(\bar{X}\) = Sample mean = 957 hours
- \(\mu\) = Population mean under null hypothesis = 911 hours
- \(\sigma\) = Population standard deviation = 52 hours
- \(n\) = Sample size = 25
- Plugging in the values:
\[
z = \frac{957 - 911}{\frac{52}{\sqrt{25}}} = \frac{46}{10.4} \approx 4.42
\]
3. **Critical Value:**
- From Z-tables, the critical value \(z_{\alpha}\) for a one-tailed test at the 1% significance level is 2.33.
4. **Decision Rule:**
- If the calculated test statistic exceeds the critical value (4.42 > 2.33), reject the null hypothesis.
#### Conclusion:
Since the calculated z-value (4.42) is greater than the critical value (2.33), we **reject the null hypothesis**. There is sufficient evidence to support the claim that, for the modified components, the mean time between failures is greater than 911 hours.
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