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 use the rejection region to test the claim that for the modified component the mean time between failures is greater than 520 hours.
List the null and alternative hypothesis
Find the value of the test statistic
Rejection region (s)
Reject/fair to reject H0:
sketch
State the final conclusion that addresses the original claim
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
However, I will start you out.
Ho: mean = 520
Ha: mean > 520
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Z = (sample mean-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Is it ≤ .05?
To test whether the mean time between failures for the modified component is greater than 520 hours, we need to set up the null and alternative hypothesis.
Null Hypothesis (H0): The mean time between failures for the modified component is 520 hours.
Alternative Hypothesis (Ha): The mean time between failures for the modified component is greater than 520 hours.
To find the value of the test statistic, we need to calculate the sample mean and the sample standard deviation.
Sample mean (X̄) = (518 + 548 + 561 + 523 + 536 + 499 + 538 + 557 + 528 + 563) / 10 = 541.4
Sample standard deviation (s) = sqrt(((518-541.4)^2 + (548-541.4)^2 + (561-541.4)^2 + (523-541.4)^2 + (536-541.4)^2 + (499-541.4)^2 + (538-541.4)^2 + (557-541.4)^2 + (528-541.4)^2 + (563-541.4)^2) / 9) = 20.33
Next, we calculate the test statistic:
t = (X̄ - μ) / (s / sqrt(n))
where μ is the population mean (520), s is the sample standard deviation (20.33), and n is the sample size (10).
t = (541.4 - 520) / (20.33 / sqrt(10)) = 3.71
To determine the rejection region (s), we look up the critical value in the t-distribution table using the significance level of 0.05 and degrees of freedom (df) of n-1 (10-1 = 9). In this case, the critical value is approximately 1.833.
Therefore, the rejection region (s) is t > 1.833.
Since the test statistic (t = 3.71) falls in the rejection region, we reject the null hypothesis (H0) and conclude that there is sufficient evidence to support the claim that for the modified component, the mean time between failures is greater than 520 hours.
The final conclusion is that there is significant evidence to suggest that the modification made to the computer component has increased its mean time between failures.