Statistical Decision Making and Quality Control Homework Assignment

This assignment solution addresses statistical decision making and quality control, focusing on calculating control limits (UCL and LCL) based on sample size and confidence intervals. It explores the impact of different sample sizes and confidence levels on the width of the control limits. The solution includes calculations for various scenarios, comparing the effectiveness of different procedures to maintain control over labor time. Furthermore, the assignment covers hypothesis testing, including setting up null and alternative hypotheses and calculating the test statistic (Z-score) to validate a company's claim. The solution provides detailed steps and reasoning, making it a valuable resource for students studying statistical quality control and hypothesis testing.

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QUESTION 3 Statistical Decision Making and Quality Control
The following formulas are used to find the upper control limit UCL, and lower control limit LCL,
(Montgomery, 2003)
UCL=mean-Z*StDev/sqrt(n)
LCL=mean+Z*StDev/sqrt(n)
Where mean is the sample mean (20),
StdDev is the population standard deviation (10),
n is the given size of the sample which will be changing for different parts of the problem, and
Z represents the critical value (as obtained from Z table) according to the given level of confidence
which will change in different parts of the problem
In what follows, we substitute in the above formula
1. 95% of confidence with Samples of 64 observations :
LCL=20-1.96*10/ sqrt (64)= 17.55
UCL=20+1.96*10/sqrt(64)= 22.45
2. 95% of confidence with 16 observations
LCL= 20-1.96*10/ sqrt (16)= 15.1
UCL= 20+1.96*10/ sqrt (16)= 24.9
3. In this part we investigate how the level of confidence and sample size will affect the
calculation of the confidence interval, more specifically the effect on the interval width.
 Increasing the sample size to 16 observations and setting confidence level to 90%
L= 20-1.645*10/ sqrt( 16)= 15.8875
U= 20+1.645*10/ sqrt(16)= 24.1125
 Keeping the confidence level at 95% and upraising the sample size to 64 observations
L = 20-1.96*10/ sqrt(64)= 17.55
U=20+1.96*10/ sqrt(64)= 22.45
 Holding the confidence level at 95% while reducing the sample sizes of to 36 observations.
L= 20-1.96*10/ sqrt(36)= 16.73333
U= 20+1.96*10/ sqrt(36)= 23.26667
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As seen from the result above the second confidence interval has narrower limits (17.55, 22.45)
(b) This is left tail Z test for mean ( Johnson, R.A., Bhattacharyya, G.K., 2014)
Null hypothesis: μ ≥ 1 .5
Alternative Hypothesis: μ < 1.5
The test statistic is Z is computed by the formula
z=(mean-1.5)*sqrt(n)/StDev=(1.3-1.5)*sqrt(100)/0.3= -6.6667
-1.6449 is the Critical z is obtained from the Z table corresponding to left tail 0.05
The following graph gives idea about the rejection region of size 0.05 to the left of the critical value (-
1.645). Since the observed Z=-6.7 is very far below the critical value, then we reject the null
hypothesis and declare the validity of the company claim at 0.05 level of significance.
References
Johnson, R.A., Bhattacharyya, G.K. (2014). Statistics: Principles and Methods, 7 edition. Wiley,
Hoboken, NJ.
Montgomery, D. C. (2003). Introduction to statistical quality control. Hoboken, NJ: Wiley.
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