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

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

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 () 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

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

Explanation / Answer

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

85

354.804

0.734

0.080

(354.646, 354.962)

: 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

112

355.747

0.522

0.049

(355.649, 355.845)

: mean of Process B

C) Construct confidence interval for the true Standard Deviation () 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

: 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
using Bonett

95% CI for
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

: 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
using Bonett

95% CI for
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

85

354.804

0.734

0.080

(354.646, 354.962)

: mean of Process A

Test

Null hypothesis

H: = 355

Alternative hypothesis

H: 355

T-Value

P-Value

-2.46

0.016

Conclusion: p< , 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

112

355.747

0.522

0.049

(355.649, 355.845)

: mean of Process B

Test

Null hypothesis

H: = 355

Alternative hypothesis

H: 355

T-Value

P-Value

15.13

0.000

Conclusion: p< , 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.

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

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