Question

in tests of a computer component, it is found that the mean time between failures is 891 hours. a modification is made which is supposed to increase reliability by increasing the time between failures. tests on a sample of 47 modified components produce a mean time between failures of 922 hours, with a standard deviation of 74 hours. test the claim that on average, the modified components last longer than 891 hours between failures, use a level of significance of 1%.
a. what type of test will be used in this problem? select an answer
b. identify the null and alternative hypotheses?
$h_0$: select an answer?
$h_a$: select an answer?
c. is the original claim located in the null or alternative hypothesis? select an answer
d. calculate your test statistic. write the result below, and be sure to round your final answer to two decimal places.

Explanation:

Step1: Determine test type

Since the population standard - deviation is unknown and we are testing a claim about the population mean with a sample, we use a one - sample t - test.

Step2: State hypotheses

The null hypothesis \(H_0\) is a statement of no effect. The claim is that the modified components last longer than 891 hours. So, \(H_0:\mu\leq891\) and \(H_a:\mu > 891\).

Step3: Locate original claim

The original claim "on average, the modified components last longer than 891 hours" is \(H_a:\mu > 891\), so the original claim is in the alternative hypothesis.

Step4: Calculate test statistic

The formula for the one - sample t - test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x} = 922\), \(\mu = 891\), \(s = 74\), and \(n = 47\).
\[t=\frac{922 - 891}{74/\sqrt{47}}\]
\[t=\frac{31}{74/\sqrt{47}}\]
\[t=\frac{31}{74/6.8557}\]
\[t=\frac{31}{10.794}\]
\[t\approx2.87\]

Answer:

a. One - sample t - test
b. \(H_0:\mu\leq891\), \(H_a:\mu > 891\)
c. Alternative hypothesis
d. \(2.87\)