Applied statistics assignment help sample

Statistics is used in many disciplines today for effective analysis of data. Among these fields include engineering where it is used for gauge quantification. For instance, it can be used to quantify uncertainties in a gauge study. In such a study, engineers gather the appropriate data based on the specific variable (s) being observed and estimate standard deviations for the various sources that contribute to uncertainty. They apply powerful statistical techniques to arrive at the most accurate solutions.

Assuming a centered process what are the process control limits for the part

given USL = 0.95 in, LSL = 0.7 in

  [email protected] = 0.0252 [email protected] = 0.0231

As we know for a centered process Xbar = (USL + LSL)/2 = 0.825 in

P_(pk ) = min( [USL - Xbar]/3*σ , [Xbar - LSL] / 3*σ ) = [USL - LSL] /6*σ

therefore σ = [0.95 - 0.7]/6*0.8

which gives σ = 0.05208 and total variance = σ = 0.05208

 also d = [[email protected] + [email protected]] /2

therefore d = 0.02415

 Now covering 99% of the variance

    GRR = d/(5.15*TV)

which gives GRR = 0.02415/(5.15*0.05208)

                = 0.09004 or 9%

process control limits

UCL =Xbar + 3*σ = 0.825 + 3*0.05208 = 0.98 in

LCL =Xbar - 3*σ = 0.825 - 3*0.05208 = 0.67 in

Assuming a centered process what are the process control limits for the part

given USL = 0.95 in, LSL = 0.7 in

  [email protected] = 0.0252 [email protected] = 0.0231

As we know for a centered process Xbar = (USL + LSL)/2 = 0.825 in

P_(pk ) = min( [USL - Xbar]/3*σ , [Xbar - LSL] / 3*σ ) = [USL - LSL] /6*σ

therefore σ = [0.95 - 0.7]/6*1.3

which gives σ = 0.03205 and total variance = σ = 0.03205

 also d = [[email protected] + [email protected]] /2

therefore d = 0.02415

 Now covering 99% of the variance

    GRR = d/(5.15*TV)

which gives GRR = 0.02415/(5.15*0.03205)

                = 0.1463 or 14.63%

process control limits

UCL =Xbar + 3*σ = 0.825 + 3*0.03205 = 0.921 in

LCL =Xbar - 3*σ = 0.825 - 3*0.03205 = 0.73 in

Since the part is gaged by the ring gage for the maximum material condition, therefore,it's a go gage.

The diameter of the shaft =〖2.000〗_(-.005)^(+.008)in , range of diameter = 1.995 - 2.008 in

total tolerance = 0.013 in

generally gage tolerance is 10% of total tolerance therefore gage tolerance = 0.0013 in

for bilateral tolerance

we will divide gage tolerance by 2

we get 0.0013/2 = 0.00065 in

therefore size of go ring gage - 〖2.008〗_(-.00065)^(+.00065)

and the mean size of the ring gauge would be 2.008 in.

No, more number of data would provide more accurate results.

repeatability = variation due to measurement environment

therefore repeatability = 0.001921

  To determine if the bias is significantly different from 0, we need to do a t test to test that

  Null Hypothesis: Mean of bias is zero

  Alternate Hypothesis: Mean of bias is not zero

  We need to perform a two-tailed t test to test our hypothesis

Mean of bias= .00655

Sample SD = 0.001921

  Degrees of freedom = n-1 = 9

  Test value of t-statistic = (Mean of bias - 0)/(Sample SD/SQRT(10)) = 10.78

  At a level of significance = 0.01 and df = 9, the critical values of t are 土3.2498

  As the test value of t falls outside the range of critical values of t, we reject the null hypothesis

  In conclusion, at 99% confidence, we can say that bias is not zero

To plot a gage performance curve we require following inputs

  USL = 2.008, LSL = 1.995

  Accuracy ( = -bias) = -0.00655 GRR(SD) = 0.001921

To get quality assistance with this topic, hire our hypothesis testing assignment help providers.

given USL = 0.56 , LSL = 0.44

 P_pk = 0.6

gage tolerance = 10% of part tolerance

Assuming a centered process

P_(pk ) = min( [USL - Xbar]/3*σ , [Xbar - LSL] / 3*σ ) = [USL - LSL] /6*σ

therefore σ = [0.56 - 0.44]/6*0.6

which gives σ = 0.0333

mean = 0.50

 Agreement calculation between A and B

(heret represent agreement and f represent disagreement.)

part	Agreement (A/B)[1 or 0 by A]	Agreement (A/C)	Agreement (B/C)	Pass or fail (based on reference dimension (0.44-0.56)	part	Agreement (A/B) [1 or 0 by A]	Agreement (A/C)	Agreement (B/C)	Pass or fail (based on reference dimension (0.44-0.56)
1	t[1]	t	t	1	8	t[1]	t	t	1
1	t[1]	t	t	1	8	t[1]	t	t	1
1	t[1]	t	t	1	8	t[1]	t	t	1
2	t[0]	t	t	0	9	t[1]	t	t	1
2	t[0]	t	t	0	9	t[1]	t	t	1
2	t[0]	t	t	0	9	t[1]	t	t	1
3	t[0]	t	t	0	10	t[0]	t	t	1
3	t[0]	t	t	0	10	t[0]	f	f	1
3	t[0]	t	t	0	10	t[0]	t	t	1
4	t[0]	t	t	0	11	t[1]	t	t	1
4	t[0]	t	t	0	11	t[1]	t	t	1
4	t[0]	t	t	0	11	t[1]	t	t	1
5	t[1]	t	t	1	12	t[1]	t	t	1
5	t[1]	f	f	1	12	t[1]	f	f	1
5	t[0]	t	t	1	12	f[0]	t	f	1
6	t[1]	t	t	1	13	t[1]	t	t	1
6	t[1]	f	f	1	13	t[1]	t	t	1
6	t[1]	t	t	1	13	t[1]	t	t	1
7	t[1]	t	t	1	14	t[1]	t	t	1
7	t[1]	t	t	1	14	t[1]	t	t	1
7	t[1]	t	t	1	14	t[1]	t	t	1

part	Agreement (A/B) [1 or 0 by A]	Agreement (A/C)	Agreement (B/C)	Pass or fail (based on reference dimension (0.44-0.56)	part	Agreement (A/B) [1 or 0 by A]	Agreement (A/C)	Agreement (B/C)	Pass or fail (based on reference dimension (0.44-0.56)
15	t[1]	t	t	1	22	t[1]	t	t	1
15	t[1]	t	t	1	22	t[1]	t	t	1
15	t[1]	t	t	1	22	t[1]	t	t	1
16	t[1]	t	t	1	23	t[0]	t	t	1
16	t[1]	t	t	1	23	f[1]	f	t	1
16	t[1]	t	t	1	23	t[0]	f	f	1
17	t[1]	t	t	1	24	t[1]	t	t	1
17	t[1]	t	t	1	24	t[1]	t	t	1
17	t[1]	t	t	1	24	t[1]	t	t	1
18	t[1]	t	t	1	25	t[1]	t	t	1
18	t[1]	t	t	1	25	t[1]	t	t	1
18	t[1]	t	t	1	25	t[1]	t	t	1
19	t[1]	f	f	1	26	t[1]	t	t	1
19	f[1]	t	f	1	26	t[1]	t	t	1
19	f[0]	t	f	1	26	t[1]	t	t	1
20	t[0]	f	f	1	27	t[0]	t	t	0
20	f[0]	f	t	1	27	t[0]	t	t	0
20	f[1]	f	t	1	27	f[0]	t	f	0
21	t[1]	t	t	1	28	t[1]	t	t	1
21	t[1]	t	t	1	28	t[1]	t	t	1
21	t[1]	t	t	1	28	t[1]	t	t	1

part	Agreement (A/B)	Agreement (A/C)	Agreement (B/C)	Pass or fail (based on reference dimension (0.44-0.56)	part	Agreement (A/B)	Agreement (A/C)	Agreement (B/C)	Pass or fail (based on reference dimension (0.44-0.56)
43	t[1]	t	t	1	45	t[0]	t	t	1
43	t[1]	t	t	1	45	t[0]	t	t	1
43	t[1]	t	t	1	45	t[0]	t	t	1
44	t[0]	t	t	0
44	t[0]	t	t	0
44	t[0]	t	t	0

We will further create a cross-tabulation table for A/B, A/C, and B/C Agreement table for A/B kappa = (po -pe)/(1 - pe) where po = the sum of the actual counts in the diagonal cells/overall total pe = the sum of the expected counts in the diagonal cells/over total so                     k(A/B)   =  (126/135 -  79.08/135)/(1- 79.08/135)   =  83.91% for A/C

operator			C	C	C	C
			1	1	0	0	total
			actual count	expected	actual count	expected
A	1	actual count	86		8		94
A	1	expected		64.76		29.24	94
A	0	actual count	7		34		41
A	0	expected		28.24		12.76	41
	total		93	93	42	42	135

Kappa(A/C)                        =   (120 - 77.52)/(135 - 77.52)   =  73.90% for B/C

operator			C	C	C	C
			1	1	0	0	total
			actual count	expected	actual count	expected
B	1	actual count	88		9		97
B	1	expected		66.82		30.18	97
B	0	actual count	5		33		38
B	0	expected		26.18		11.82	38
	total		93	93	42	42	135

Kappa(B/C)                         =   (121 - 78.64)/(135 - 78.64)   =  75.16% A general rule of thumb is that values of kappa greater than 0.75 indicate good to the excellent agreement (with a maximum kappa = 1); values less than 0.40 indicate poor agreement." Thus we can say that A/B and B/C has good agreement while A/C agreement is slightly poorer.

Sampling

Miss rate of an operator = total of accept the bad/(total sample that are bad in study*number of trials per operator) miss rate of A  =   0/(8*3)    =    0% false alarm rate  = total of reject the good/(total sample that are good in study*number of trials per operator) false alarm rate  =  17/ (37*3) =  15.32% (5) effectiveness of operator A  for agreement with reference value =  [ number of good correct  + number of bad correct]/number of samples =   [  27 + 8] /45   =  35/45 = 77.78% If you would like academic support on this topic, take our sampling assignment help.

Probability

probability of non conforming part being called out as non conforming =  1 -  probability of  non conforming part being called out as conforming =  1  - probability( miss ) =   1  - P(miss by A) - P(miss by B) - P(miss by C ) =   1  -   0/24 - 1/24 - 0/24 =   23/24