Question

Quality Specifications for the bottle filling process = 355 ± 1.5 ml

The sample measurements for Process A & Process B can be found in the attached Excel file Other relevant information for the analysis:

Process A = 11.5 million bottles

Process B = 6.9 million bottles

Estimated cost of overfilling = $0.071 per bottle

Estimated cost of underfilling = $0.134 per bottle

Process A			Process B
bottle nbr	fill volume, ml		bottle nbr	fill volume, ml
1	353.8716		1	356.4036
2	356.4629		2	354.8854
3	354.3566		3	356.2884
4	354.9326		4	355.9886
5	354.1558		5	355.3441
6	354.6894		6	355.141
7	353.1613		7	355.5605
8	354.492		8	354.7924
9	353.2064		9	355.7594
10	355.0353		10	356.2499
11	354.1497		11	356.7416
12	355.3837		12	355.6718
13	354.8073		13	355.8648
14	354.52		14	354.8881
15	354.127		15	355.7184
16	354.0073		16	355.5325
17	354.5865		17	355.5129
18	355.2267		18	356.0567
19	354.1048		19	355.9526
20	354.7092		20	356.2481
21	354.8251		21	355.5009
22	355.1323		22	355.8066
23	355.6323		23	355.6735
24	355.0618		24	355.5141
25	355.6289		25	355.6569
26	354.5315		26	355.0254
27	354.6454		27	356.4625
28	354.1473		28	356.3046
29	355.5054		29	356.2273
30	354.6658		30	355.4353
31	354.777		31	356.0174
32	354.6489		32	356.3742
33	354.9304		33	355.3607
34	356.0081		34	355.6758
35	353.5191		35	355.2479
36	354.089		36	356.5349
37	355.4005		37	356.1038
38	354.7968		38	356.1314
39	354.5803		39	356.1499
40	354.5402		40	356.9204
41	353.9612		41	356.0494
42	355.3751		42	355.8082
43	355.2035		43	355.7958
44	354.7033		44	356.4498
45	355.5842		45	355.0805
46	355.2069		46	355.6821
47	355.291		47	355.7853
48	355.5132		48	356.0396
49	354.4062		49	354.5163
50	354.8773		50	355.1708
51	354.0812		51	356.8699
52	355.5711		52	355.8047
53	356.8612		53	356.1663
54	354.6389		54	356.2781
55	355.4831		55	355.6501
56	354.4165		56	355.1498
57	354.4106		57	356.1733
58	353.8034		58	355.3848
59	355.779		59	355.4425
60	354.3574		60	355.7853
61	354.2061		61	355.6234
62	355.3		62	355.2701
63	353.8064		63	355.3693
64	355.0172		64	356.0998
65	355.2049		65	354.3443
66	356.0506		66	355.2375
67	355.5254		67	356.0556
68	355.9298		68	355.6644
69	354.6942		69	355.9695
70	354.879		70	356.0207
71	354.876		71	355.8412
72	353.2011		72	356.013
73	355.69		73	356.0578
74	355.2879		74	355.0693
75	354.881		75	356.2371
76	353.4271		76	356.4531
77	354.3281		77	355.8708
78	355.4182		78	355.7516
79	354.7104		79	356.0311
80	354.5383		80	355.336
81	354.2397		81	356.6274
82	355.7615		82	355.5591
83	355.7941		83	355.577
84	353.7047		84	356.3873
85	355.3057		85	355.4378
			86	355.4152
			87	355.7074
			88	354.8495
			89	356.3219
			90	355.2006
			91	356.162
			92	356.7196
			93	354.69
			94	354.7049
			95	355.2266
			96	355.7611
			97	356.1532
			98	355.2149
			99	354.9555
			100	356.2889
			101	356.1144
			102	355.8599
			103	356.2266
			104	356.2091
			105	355.8744
			106	355.8112
			107	355.1513
			108	355.2167
			109	355.2743
			110	355.2112
			111	355.8082
			112	355.8028

a. Calculate numerical measures for Central Tendency and for Dispersion on both processes.

b. Construct confidence interval for the true Mean (µ) of each of the processes. Comment on the results. Is there an issue with the mean of any of the processes? Explain the results.

c. Construct confidence interval for the true Standard Deviation (σ) of each of the processes. Comment on the results. Is there an issue with the dispersion of any of the processes? Explain the results.

d. Run a hypothesis test (a t-test) for the Mean of each process being equal to the target value of 355 ml. Comment on the results e. Draw a histogram (with normal-distribution fit) for both processes; interpret the results (also, draw the two theoretical normal distributions overlapping in the same graph to facilitate interpretation)

f. Construct a Normal Probability Plot for each of the process. What can you conclude?

g. Estimate the expected number of bottles overfilled per year from each of the processes

h. Estimate the expected number of bottles overfilled per year from each of the processes

i. Calculate and compare the annual cost of overfilling AND underfilling per process. Comment on the results.

j. Make your Final Conclusions and Recommendations

minitab plz

Answer 1

A) Calculate numerical measures for Central Tendency and for Dispersion on both processes.

STEP 1: Delete row with naming bottle nbr and fill volumn ml.

STEP 2: copy all four columns of dataset and paste in minitab.

STEP 3:

1) Go to tab Stat>Basic statistics>Display descriptive statistics.

2) Select variable process A and process B

3) Go to statistics tab to select all the required statistics function.(mean, variance, standard deviation, median, maximum etc)

4) Click ok.

Descriptive Statistics: Process A, Process B

Statistics

Variable	N	N*	Mean	SE Mean	StDev	Variance	Minimum	Q1	Median	Q3
Process A	85	0	354.80	0.0796	0.734	0.538	353.16	354.34	354.78	355.34
Process B	112	0	355.75	0.0494	0.522	0.273	354.34	355.36	355.79	356.15

Variable	Maximum
Process A	356.86
Process B	356.92

B)Construct confidence interval for the true Mean (µ) of each of the processes. Comment on the results. Is there an issue with the mean of any of the processes? Explain the results.

STEPS:

1) Go to tab Stat>Basic statistics>1-Sample t and click.

2) Select >one or more samples each in a column

3) Select>Process A

4) go to options and write as follows.

5) click ok.

One-Sample T: Process A

Descriptive Statistics

N	Mean	StDev	SE Mean	95% CI for μ
85	354.804	0.734	0.080	(354.646, 354.962)

μ: mean of Process A

REPEAT THE SAME PROCEDURE FOR PROCESS B

One-Sample T: Process B

Descriptive Statistics

N	Mean	StDev	SE Mean	95% CI for μ
112	355.747	0.522	0.049	(355.649, 355.845)

μ: mean of Process B

C) Construct confidence interval for the true Standard Deviation (σ) of each of the processes. Comment on the results. Is there an issue with the dispersion of any of the processes? Explain the results.

STEPS:

1) Go to tab Stat>Basic statistics>1variance and click.

2) Select >one or more samples each in a column

3) Select>Process A

4) go to options and write as follows.

5) click ok.

Test and CI for One Variance: Process A

Method

σ: standard deviation of Process A

The Bonett method is valid for any continuous distribution.

The chi-square method is valid only for the normal distribution.

Descriptive Statistics

N	StDev	Variance	95% CI for σ using Bonett	95% CI for σ using Chi-Square
85	0.734	0.538	(0.635, 0.868)	(0.638, 0.864)

REPEAT THE SAME PROCEDURE FOR PROCESS B

Test and CI for One Variance: Process B

Method

σ: standard deviation of Process B

The Bonett method is valid for any continuous distribution.

The chi-square method is valid only for the normal distribution.

Descriptive Statistics

N	StDev	Variance	95% CI for σ using Bonett	95% CI for σ using Chi-Square
112	0.522	0.273	(0.465, 0.597)	(0.462, 0.601)

d. Run a hypothesis test (a t-test) for the Mean of each process being equal to the target value of 355 ml. Comment on the results STEPS: 1) Go to tab Stat>Basic statistics>1-Sample t and click. 2) Select >one or more samples each in a column 3) Select>Process A 4) check box perform hypothesis test (hypothesized mean=355) 5) click ok. One-Sample T: Process A Descriptive Statistics N Mean StDev SE Mean 95% CI for μ 85 354.804 0.734 0.080 (354.646, 354.962) μ: mean of Process A Test Null hypothesis H₀: μ = 355 Alternative hypothesis H₁: μ ≠ 355 T-Value P-Value -2.46 0.016 Conclusion: p<α , hence reject Null hypothesis at 0.05 level of significance. Perform similarly for Process B: One-Sample T: Process B Descriptive Statistics N Mean StDev SE Mean 95% CI for μ 112 355.747 0.522 0.049 (355.649, 355.845) μ: mean of Process B Test Null hypothesis H₀: μ = 355 Alternative hypothesis H₁: μ ≠ 355 T-Value P-Value 15.13 0.000 Conclusion: p<α , hence reject Null hypothesis at 0.05 level of significance. e. Draw a histogram (with normal-distribution fit) for both processes; interpret the results (also, draw the two theoretical normal distributions overlapping in the same graph to facilitate interpretation)

STEPS:

1) Go to tab Graph>Histogram>With fit

2) Select graph variable>Process A

3) click ok.

Perform similarly for process B

f. Construct a Normal Probability Plot for each of the process. What can you conclude?

STEPS:

1) Go to tab Graph>Probability plot>Single

2) Select graph variable>Process A

3) click on Distribution and do as follows

4)click ok

g. Estimate the expected number of bottles overfilled per year from each of the processes

Quality Specifications for the bottle filling process = 355 ± 1.5ml=(353.5,356.5)

Process A = 11.5 million bottles

Process B = 6.9 million bottles

Estimated cost of overfilling = $0.071 per bottle

Estimated cost of underfilling = $0.134 per bottle

For process A :

1)calculate number of bottles >356.5

Go to data>recode>to numeric

Follow steps as follows:

Summary

Lower End	Upper End	Recoded Value	Number of Rows
0	356.5	0	84
356.51	400	1	1

Source data column	Process A
Recoded data column	Recoded Process A_3

Each interval includes its lower end.

So number of overfilled bottles in sample of size 85 is 1.

There are 365 days in a year hence 1*365=365 bottles overfilled per year.do similarly for process B.

i. Calculate and compare the annual cost of overfilling AND underfilling per process. Comment on the results.

So number of overfilled bottles in sample of size 85 is 1.

There are 365 days in a year hence 1*365=365 bottles overfilled per year.do similarly for process B.

Estimated cost of overfilling = $0.071 per bottle

365*0.071=25.915 dollar is estimated cost for overfilling per year for process A.

similarly for underfilling and both process.