Questions: In tests of a computer component, it is found that the mean time between failures is 610 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 43 modified components produce a mean time between failures of 631 hours, with a standard deviation of 52 hours. Test the claim that on average, the modified components last longer than 610 hours between failures, use a level of significance of 1%. a. What type of test will be used in this problem? b. What evidence justifies the use of this test? Check all that apply: - The population standard deviation is known - The original population is approximately normal - There are two different samples being compared - The sample standard deviation is not known - n p>5 and n q>5 - The sample size is larger than 30 - The population standard deviation is not known c. Enter the null hypothesis for this test. H0: ? v ? d. Enter the alternative hypothesis for this test. H1: ? v ? e. Is the original claim located in the null or alternative hypothesis?

In tests of a computer component, it is found that the mean time between failures is 610 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 43 modified components produce a mean time between failures of 631 hours, with a standard deviation of 52 hours. Test the claim that on average, the modified components last longer than 610 hours between failures, use a level of significance of 1%. a. What type of test will be used in this problem? b. What evidence justifies the use of this test? Check all that apply: - The population standard deviation is known - The original population is approximately normal - There are two different samples being compared - The sample standard deviation is not known - n p>5 and n q>5 - The sample size is larger than 30 - The population standard deviation is not known c. Enter the null hypothesis for this test. H0: ? v ? d. Enter the alternative hypothesis for this test. H1: ? v ? e. Is the original claim located in the null or alternative hypothesis?
Transcript text: In tests of a computer component, it is found that the mean time between failures is 610 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 43 modified components produce a mean time between failures of 631 hours, with a standard deviation of 52 hours. Test the claim that on average, the modified components last longer than 610 hours between failures, use a level of significance of $1 \%$. a. What type of test will be used in this problem? b. What evidence justifies the use of this test? Check all that apply: The population standard deviation is known The original population is approximately normal There are two different samples being compared The sample standard deviation is not known $n p>5$ and $n q>5$ The sample size is larger than 30 The population standard deviation is not known c. Enter the null hypothesis for this test. $H_{0}$ $\square$ ? $v$ $\square$ d. Enter the alternative hypothesis for this test. $H_{1}$ : $\square$ ? v $\square$ e. Is the original claim located in the null or alternative hypothesis? $\square$
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Solution

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Solution Steps

Step 1: Hypothesis Formulation

We are testing the claim that the modified components last longer than the original mean time between failures of 610 hours. Thus, we set up our hypotheses as follows:

  • Null Hypothesis: \[ H_0: \mu = 610 \]

  • Alternative Hypothesis: \[ H_1: \mu > 610 \]

Step 2: Calculate Standard Error

The standard error \(SE\) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{52}{\sqrt{43}} \approx 7.9299 \]

Step 3: Calculate Test Statistic

The test statistic \(Z_{test}\) is calculated using the formula: \[ Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{631 - 610}{7.9299} \approx 2.6482 \]

Step 4: Calculate P-value

For a right-tailed test, the p-value is calculated as: \[ P = 1 - T(z) \approx 0.004 \]

Step 5: Decision Rule

We compare the p-value to the significance level \(\alpha = 0.01\):

  • Since \(P \approx 0.004 < 0.01\), we reject the null hypothesis.

Final Answer

The evidence suggests that the modified components last longer than 610 hours on average. Therefore, the answer is:

\(\boxed{H_0 \text{ is rejected, indicating that } \mu > 610}\)

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