1- An equipment manufacturer claims his equipment will produce product with fewe
ID: 3227288 • Letter: 1
Question
1- An equipment manufacturer claims his equipment will produce product with fewer defectives than your current equipment. You have been monitoring your process for several months and the process percent defective has been stable with a mean of .025. A test run of 200 of your parts is performed with the new equipment and the number defective in the lot is 3. At 95% confidence, can you conclude that the new equipment is better?
The null and alternative hypothesis for this problem is?
2- A medical study is performed on the impact of a new blood pressure medication. An assessment was made of 24 study participants. For each participant, blood pressure readings were taken upon first awakening prior to beginning the study and then four weeks after the initial dose of medication. Which of the following types of hypothesis test would be most appropriate to test the effectiveness of the new drug?
3- An hypothesis test is performed with H0: µ=600 and Ha: µ 600 using a 95% confidence. The x1 and x2 values that define the upper and lower bounds of the test are identified in the figure. The actual process has shifted to a mean of 610. The sample size is 40 and s=20. What is the power of the hypothesis test to detect the shift?
4- In establishing a control chart, the reason for considering initially developed control limits as "trial" limits is?
5- The value D3 used in establishing control charts is used to?
6- An chart with n=5 has been established for a process with = 34.1 and = 4.8. During production, a sample was taken and the pieces measured 33, 31, 38, 42, and 35. For this sample?
7- A controlled process has a mean of 50.1 and variance of 0.04. The specification limits for a part on the process are 50±0.5. Cpk for this process is?
8- A process which is in statistical control?
Explanation / Answer
1)
Below are the null and alternate hypothesis
H0: p >= 0.025
H1: p < 0.025 (claim)
pcap = 3/200 = 0.015
SE = sqrt(0.025*(1-0.025)/200) = 0.011
test statistics, z = (pcap - p)/SE = (0.015 - 0.025)/0.011 = -0.9091
p-value = 0.1817
As p-value is greater than significance level of 0.05, we fail to reject the null hypothesis. This means we can not accept the claim of equipment manufacturer.
2)
Here the best method to use is paired t-test.
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