Mid Term 1

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New England College *

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5330

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Industrial Engineering

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Dec 6, 2023

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Name – Suyash Satish Kochrekar uml id – 01745432 PART 1 A snowmobile company makes an average of 9 engines/day, with engine average output of 10 horsepower (hp) based on historical record. No other information is available. Q.1) During the day of Monday, 9 engines were made and tested at with average = 11 hp and sample standard deviation = 5.0. Did the factory experience a problem in materials or methods for the day that was significant? Solution – During Monday, number of engines tested = n = 9 Sample average = Xbar = 11hp Sample Standard Deviation=s=5 Here, on Monday, same number of engines were tested but with an average of 11 hp. Thus, we use t test to compare the daily sample mean of 10 hp to the sample mean on Monday which is 11 hp with Sample Standard deviation for Monday known. We have population mean = 10hp Thus, t = (Xbar – myu)/s/sq.rt(n) = (11-10)/5/sq.rt.(9) = (1x3)/5 T calculated = 0.6 Nyu = DOF = n-1 = 8
Table value of t for DOF = 8 and for value of significance alpha = 0.025 ie two sided, we have t(alpha/2,n-1) = 2.306 Since, t calculated is less than t table value, factory did not experience a problem in materials or methods for the day was not significant. Confidence interval = Xbar +- (t table (alpha/2,n-1) x s/sq.rt n) = 11 +- (2.306 x 5/3) = 11+- (3.843) Upper limit for Confidence Interval = 11 + 3.843 = 14.843 Lower Limit for Confidence Interval = 11 – 3.843 = 7.157 Q.2) What is the maximum acceptable average engine horsepower that can be produced based on Monday production data (9 engines were made and tested with average = 11 hp and standard deviation = 5.0)? Solution – The upper limit for Confidence Interval is the maximum acceptable average engine horsepower which can be produced based on Monday’s production data. It is calculated in the above step and which is equal to 14.843 hp. Confidence interval = Xbar +- (t calculated x s/sq.rt n) = 14.843 +- (2.306 x 5/3) = 14.843 +- (3.843) Upper limit for confidence interval = 14.843+3.843=18.686 hp Lower limit for Confidence Interval = 14.843-3.843=11 hp
Q.3) A check of the company records indicated that over the last week, 30 engines were made, with and average or 10 hp and standard deviation of 3. Given this new information, does the answer in question 1 change? Solution - Number of engines made = n = 30 Sample average = Xbar = 10hp Sample Standard Deviation = 3 Using same t test to check for significance since it is used for sample mean of n = 30 or less than 30, we have, T calculated = (Xbar-myu)/s/sq.rt n = (10-10) / 3/3 = 0 T calculated = 0 since the new sample mean and the daily sample mean are equal. Degrees of Freedom = n-1 =29 and alpha = 0.025, t table = 2.045 which is found out by interpolation. Thus, since t calculated is within the value of T table, the value of the experiment is not significant. Q.4) . For the situation described in problem 3, (a check of the company records indicated that over the last week, 30 engines were made, with average = 10 and standard deviation of 3. What is the maximum acceptable average engine horsepower that can be produced in this factory for a typical day production of 9 units? SOLUTION –
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n =30, Average = myu=10, Standard Deviation = s=3 Confidence Interval = myu +- t(alpha/2,29) x s/sq.rt (n) = 10 +- 2.045 x 3/sq.rt(30) = 10 +- 6.135/5.477 = 10+-1.120 Thus, Upper value of Confidence Interval = 10+1.120=1.120 hp Lower Value of the Confidence Interval =10-1.120=8.880 hp Thus maximum acceptable average engine horsepower that can be produced in this factory for a typical day production of 9 units is the Upper limit of the the Confidence Interval which is equal to 11.120 hp Q.5) Is the variability of the 9 engines made on Monday (with average = 11 hp and standard deviation = 5.0) significantly different from last week’s engine Standard deviation (=3) of 30 engines? Solution – Since it is asked to calculate the variability of the engines, thus we have to check for significance based on the Standard Deviations of the two samples. Thus, we use F test to check for the significance based on the sample variances. The comparison is between sample standard deviations of 9 engines made on a particular day which is the sample for that day and last weeks total engine standard deviation which is the population standard deviation. For Sample 1, n=9, DOF = 9-1 = 8
For Sample 2, n=30, DOF = 30-1 =29 Thus, sample standard deviation=s=5 Population Standard Deviation = 3 Variance V1 = S1^2 = 5^2 = 25 Variance V2 = S2^2 = 3^2 = 9 Thus value of F test = 25/9 = 2.778 From table for F test, we have degrees of freedom = n-1 = 8 And for alpha = 0.05 We have from F table = 2.2783 On comparing with calculated value of F test, we see that variability of the 9 engines made on Monday (with average = 11 hp and standard deviation = 5.0) is significantly different from last week’s engine Standard deviation (=3) of 30 engines. Q.6) Management decided to check the engines made on Tuesday. Seven other engines were made with average = 9 hp and standard deviation = 2.0. Ignoring the historical record and the data from the previous week, is the 7 engines average output = 9 on Tuesday significantly different than the 9 engines on Monday (11 average and sigma = 5.0)? Please extrapolate from the t table for the answer Solution – On Monday, sample size = 9 Average = 11 hp Sample Standard Deviation = 5
On Tuesday, sample size = 7 Average output = 9 hp Sample Standard Deviation = 2 Difference in means of the two samples = 11-9 = 2 hp = X Diff Number of observations in sample 1 = 9 DOF for sample 1 = n-1 = 8 Number of observations in sample 2 = 7 DOF for sample 2 = n-1 =6 Pooled Standard Deviation = Sp Sp = sq.rt [((n1-1) x (SD1)^2 +(n2-1)x(SD2)^2) / (n1+n2-2)] = sq. rt [((8)x(5)^2 +(6)x(2)^2) / (9+7-2) = sq.rt [(200+24) / 14 ] = sq.rt (16) = 4 Thus value of t = (X diff) / Sp x sq.rt (1/n1 + 1/n2) = (2) / 4 x sq.rt (1/9 + 1/7) = 2 / 4 x 0.503 = 2 / 2.012 = 0.994 Comparing with t table value ie for Degrees of Freedom = n1+n2-2 = 9+7-2 = 14 and alpha = 0.05, we have t table = 1.2075 which is calculated by interpolation. Ie table value for DOF 10 = 1.812
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Table value of t for DOF 20 = 1.725 Calculating for table value 14, we have by interpolation formula, Y = (1.725-1.812)/(20-10) x (14-10) + 1.812 = (-0.087x4) / 10 + 1.812 = -0.0348 + 1.812 = 1.7772 Ie table value at DOF = 14 is 1.7772 Since calculated value of t lies within the table value of t, we can say that the experiment is not significant ie range is within the table range of the experiment. CONFIDENCE INTERVAL- CI = (X1bar – X2bar) +- t = (2) +- t at 0.025 = 2 +- 2.1712 Upper value of Confidence Limit = 2+2.1712 = 4.1712 Lower Value of confidence Limit = 2-2.1712 = -0.1712 T value from table for alpha =0.025 and DOF = 14 has been calculated by using interpolation formula. Q.7) Can the management test for the assumption that the engine variances (9 engines on Monday with 11 average and sigma = 5.0 and 7 engines on Tuesday average 9 hp and standard deviation = 2.0 are significant based on the Monday and Tuesday variance outputs? Solution –
We have to check for significance based on Standard Deviations of the two samples with sigma known. We use F test to check for the significance of Sample variances F = S1^2/S2^2 = 5^2 / 2^2 = 25/4 = 6.25 DOF 1 = n1-1 =9-1 =8 DOF 2 = n2-1 =7-1 =6 Thus, from table of F test, and by interpolation, Y={(4.28-4.06)x(8-10) / (6-10)} + 4.74 = 4.85 Ie for table value of F at Numerator = 8 and Denominator = 6, we have F table at 0.05 significance = 4.85 Since calculated value of F ie 6.25 is beyond the table value of F, we can say that the experiment is significantly different based on Variance outputs. 8) Historically the engines yields a Mean µ = 10 hp and s = 3.0. A new engine is being developed. How many engines are needed to test to insure (95% confidence) that new mean is within .5 hp of the old engine? Solution – Mean of Sample 1 = 10hp
Standard Deviation of sample 1 =3 New mean should be 0.5 times that of mean of old engine. Confidence Interval for old sample = myu +- z(alpha/2) x (sigma/sq.rt. (n)) = 10 +- 1.960 x (3/sq.rt (n)) = 10+-(5.89/sq.rt(n)) ….(1) But Confidence Interval need to be within +-0.5 times Ie CI = 10 +-0.5 …(2) Equating (1) and (2), 10 + 5.89/sq.rt.(n) = 10+0.5 5.89/sq.rt(n) = 0.5 Sq.rt(n) = 11.78 N = 139 engines
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