Question

A modification is made to a computer component to try to increase its mean time between failures, which was originally 520 hours. Tests on a simple random sample of 10 modified components resulted in the following times (in hours) between failures. Assume that the sample comes from a normally distributed population.
518 548 561 523 536
499 538 557 528 563
At the 0.05 significance level, test the claim that, with the modified component, the mean time between failures is greater than 520 hours. Show all work. You may use your calculator to input data and find sample mean and standard deviation to use below. Answer all parts, including the sketch.

Proportion or mean???
Do we have STD deviation or not? As a result, do we use z- or t-distribution?
From the sample data, state the mean and standard deviation:???
List the null and alternative hypotheses: Left, right, or two-tail test?
Find the value of the test statistic:???
Critical value(s) / Draw Rejection region(s): Sketch:???
Reject/fail to reject H0:???
(Is the test statistic in the rejection region?)
State the final conclusion that addresses the original claim:???
(Use your Hypothesis Test Conclusion Flowchart.)

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 520
Alternative hypothesis: u > 520

Note that these hypotheses constitute a one-tailed test.  

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 6.546

z = (x - u) / SE

z = 2.61

z_critical = 1.645

Rejection regioon is z > 1.645

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of 2.61.

Thus the P-value in this analysis is 0.005

Interpret results. Since the P-value (0.005) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the mean time between failures is greater than 520 hours.