I am not sure what info you need so I gave you the entire worksheet with answers
ID: 3263826 • Letter: I
Question
I am not sure what info you need so I gave you the entire worksheet with answers. I just need #8 done. I have all the other answers except #8. I need the work shown with #8
Boeing is teamed with Lockheed Martin, Pratt & Whitney and the U.S. Air Force to produce the F-22 Raptor air dominance fighter. The Raptor's unique combination of stealth, speed, agility, precision and situational awareness make it overwhelmingly effective in its combined air-to-air and air-to-ground mission capability. This aircraft is the next generation of precision strike aircraft replacing the F-117A.
Boeing has a contract with an outside company, Opus Enterprise Inc., to supply rivets. These rivets are used to fasten the wings to the fuselage (body) of the plane, etc.
The contract specification requires the rivets have a mean (µ) diameter of 1.48 cm.
To ensure quality, you have an assistant measure the diameters of 10 randomly selected rivets:
1.364 1.189 0.743 1.711 1.119
1.814 1.589 1.646 0.998 2.067
1.
As the rivet will work or not work, p = .5. Assuming the data is normally distributed, construct a 95% confidence interval, using the sample of 10, to estimate the population proportion p. You may use the TI or formula card. Round to the hundredths. (Section 7.2)
.19 <.p< 81
2.
Is it safe to assume that 50% of the rivets can be used? Why or why not?
Yes, because 50% falls between the 19% and 81% of the population proportion
3.
The tolerance, allowable error, of useable rivets is .15. Using p = .5 and the allowable error, find the sample size needed to construct a 95% confidence interval for the population proportion. (Section 7.2)
P= .5
42.68 rounded to 43. You need 43 samples to obtain a 95% CI for the proportion which marginal error 0.15
4. Below are 90 diameters. Find a x and Sx. Round to the nearest hundredth.
1.189
1.364
1.104
1.325
1.502
0.923
2.216
2.309
0.657
0.929
2.142
1.375
1.631
1.949
1.481
1.076
0.743
2.076
1.977
1.416
1.34
1.83
1.466
1.711
2.214
1.736
1.815
1.273
1.661
1.595
1.135
1.399
1.814
1.374
1.34
0.911
1.259
1.198
0.289
1.911
0.813
1.472
1.484
0.346
0.843
1.036
1.793
2.284
1.589
0.983
1.825
1.387
1.632
2.125
1.297
1.113
1.405
1.805
1.975
0.569
1.893
1.557
1.22
1.209
1.646
1.814
1.74
1.307
0.977
1.344
2.003
1.832
1.343
0.591
2.195
2.007
0.966
1.655
1.68
1.268
0.998
2.48
1.938
1.119
1.218
1.662
1.372
0.854
0.858
0.995
x = 1.45
Sx =.46
5.
Assuming the sample data is essentially normal, using your Sx and x, construct a 95% confidence interval to estimate the population mean µ. You may use the TI or formula card. Round to the hundredths. Use df = 90. (Section 7.3)
x= 1.45
Sx= .46
CI= 95%
1.36 < < 1.55
6.
The contract has specified a population standard deviation of .23 cm. Using the sample mean and the given population standard deviation of .23, construct a 95% confidence interval to estimate µ assuming the data is normal You may use the TI or formula card. Round to the hundredths. (Section 7.3)
x= 1.45
Sx= .23
CI= 95%
1.40 < < 1.49 / 1.50
7.
Using the n = 90 sample, find, with 95% confidence, the estimate of the population standard deviation. Use df = 90. (Section 7.4)
** Remember:
DF= 90
CI=95%
N=90
Sx=.23
.2678 < < .1996
8.
The contract asserts:
50% of the rivets must be usable
The mean rivet diameter should be 1.48 cm
The standard deviation for ALL rivets should be .23 cm
Based upon the contract's specifications for the rivets, does Boeing accept or reject the shipment of 200,000 rivets from Opus Enterprise, Inc.? Comment on whether the contract specifications were satisfied by the confidence intervals. No math = no points!
1.189
1.364
1.104
1.325
1.502
0.923
2.216
2.309
0.657
0.929
2.142
1.375
1.631
1.949
1.481
1.076
0.743
2.076
1.977
1.416
1.34
1.83
1.466
1.711
2.214
1.736
1.815
1.273
1.661
1.595
1.135
1.399
1.814
1.374
1.34
0.911
1.259
1.198
0.289
1.911
0.813
1.472
1.484
0.346
0.843
1.036
1.793
2.284
1.589
0.983
1.825
1.387
1.632
2.125
1.297
1.113
1.405
1.805
1.975
0.569
1.893
1.557
1.22
1.209
1.646
1.814
1.74
1.307
0.977
1.344
2.003
1.832
1.343
0.591
2.195
2.007
0.966
1.655
1.68
1.268
0.998
2.48
1.938
1.119
1.218
1.662
1.372
0.854
0.858
0.995
Explanation / Answer
Solution
As specified in the post, only Q8 is solved.
For solving Q8, results on CI for µ and are necessary. These are directly taken from answers of Q5 and Q7 taking them to be correct.
Now to work out the solution,
Since, as per the answer of Q5, the CI for µ is: (1.36, 1.556) and this CI contains 1.45, it can be concluded that the rivets satisfy the contractual stipulation of µ.
Similarly, as per the answer of Q7, the CI for is: (0.19968, 0.2678) and this CI contains 0.23, it can be concluded that the rivets satisfy the contractual stipulation of .
Now, the initial conditions stipulate that the mean diameter of rivets must be 1.48 and Q3 specifies that the tolerance is 0.15. These imply that the diameters must lie between 1.33 and 1.63.
If X = diameter of the rivets, the above implies that the rivet is usable only if 1.33 < X < 1.63.
Given X is Normally distributed, Z = {(X – 1.45)/0.46} is Standard Normal. Hence,
P(1.33 < X < 1.63) = P[{(1.33 – 1.45)/0.46} < Z < {(1.63 – 1.45)/0.46}]
= P(- 0.2609 < Z < 0.3913) = P(Z < 0.3913) - P(Z < - 0.2609) = 0.6522 – 0.3971 = 0.2551
=> only 25.51% of rivets are usable.
Hence, although contractual stipulations on µ and are fully met, usability percentage is well below the acceptable level. => the lot cannot be accepted.
DONE
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