Mid Term 1
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School
New England College *
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Course
5330
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
docx
Pages
9
Uploaded by DeanBraveryKouprey26
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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