Essay about Flyrock Tires

3) Compute limits for the sample mean X around μ=12 such that, as long as a new sample mean is within those limits, the process will be considered to be operating satifactorily. If X exceeds the upper limit or if X is below the lower limit, corrective action will be taken. These limits are refferred to as upper and lower control limits fro quality control purposes.

The area under the curve to the left of the unknown quantity must be 0.7 (70%). So, we must first find the z value that cuts off an area of 0.7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0.53.

You work in marketing for a company that produces work boots. Quality control has sent you a memo detailing the length of time before the boots wear out under heavy use. They find that the boots wear out in an average of 208 days, but the exact amount of time varies, following a normal distribution with a standard

(a) Then mean of the sample and the value of Z with an area of 10% in right tail.

First we look for the area of both by doing “2nd ,Vars, NORMALCDF” and inputting “-1000, “Z,” 0, 1 then find the difference between both.

Decision Rule: The calculated test statistic of -3.024 does fall in the rejection region of z<-1.645, therefore I would reject the null and say there is sufficient evidence to indicate mu<50.

σB = 0.3 × (0.14)2 + 0.4 × (0.04)2 + 0.3 × (0.08)2 − (0.05)2 = 0.00594.

18. Suppose that the scores of architects on a particular creativity test are normally distributed. Using a normal curve table, what percentage of architects have Z scores:

There are only 25 numbers in the sample collected. For purposes of analysis, it will be considered that the process itself is normally distributed, and its standard deviation is unknown. Under these conditions, the formula that gives the confidence interval is:

C. An unknown, rectangular substance measures 3.6 cm high, 4.21 cm long, and 1.17 cm wide.

Quality Associates, Inc. is a consulting firm that advises its clients about sampling and statistical procedures that can be used to control manufacturing processes. In one case, a client provided Quality Associates with a sample of 800 observations that were taken during a time when the client's process was operating satisfactorily. The sample standard deviation for these data was .21, hence, the population standard deviation was assumed to be .21. Quality Associates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples, the client could quickly learn whether the process was operating satisfactorily. When the process was not operating

Quality Associates, Inc., a consulting firm advises its clients about sampling and statistical procedures that can be used to control their manufacturing processes. In one particular application a client gave Quality Associates a sample of 800 observations taken during a time in which the client’s process was operating satisfactorily. The sample standard deviation of this data was 0.21; hence with so much data, the population standard deviation was assumed to be 0.21. Quality Associates then suggested that random samples of size 30 be taken periodically to monitor the process on an ongoing basis. By analyzing the new samples, the client could quickly learn whether the process was operating satisfactorily. When the process was not operating satisfactorily, corrective action could be taken to eliminate the problem. The design specification indicated the mean for the process should be 12. The hypothesis test suggested by Quality Associates follows.

5.  Assuming that operators will continue to take samples of 10 sheets each hour to check if the process is in control, what control limits should Douglas set for the case when extrusion is a Six Sigma process?

From the calculation (See Appendix I), we get the 3-sigma control limits for the process, i.e. UCL=0.091, LCL=0.014. These control limits indicate that if the error proportion is within the range of [0.014, 0.091], the process is under control; if not, the process is out of control.

Since the expected demand is 2000, thus, the mean µ is 2000. Through Excel, we get the z value given a 95% probability is 1.96. Thus, we have: z= (x-µ)/ σ=(30000-20000)/