ISEN 350 - Problem Set M7
pdf
keyboard_arrow_up
School
Texas A&M University *
*We aren’t endorsed by this school
Course
350
Subject
Industrial Engineering
Date
Feb 20, 2024
Type
Pages
5
Uploaded by DoctorRook5940
1. The heat involved in calories per gram of a cement mixture is approximately normally
distributed. The mean is thought to be 95, and the standard deviation 2.4. We wish to test
H0: μ = 95 versus μ ≠ 95 with a sample of size n = 10 specimens.
(a) Type I error.
α = 𝑃(𝑥 < 91) + 𝑃(𝑥 > 96. 4)
SE = 2.4 / √10 ≈ 0.7594
α = 𝑃(𝑧 <− 5. 270462767) + 𝑃(𝑥 > 1. 844661968)
α = 0 + 1 − 0. 9674566368 = 0. 03254336325
So, the Type I error is around 0.0325.
(b) Find β for the case where the true mean heat has evolved to 93.
(91 - 93) / 0.7594 = -2.635231383
(96.4 - 93) / 0.7594 = 4.479893352
α = 𝑃(𝑧 <− 2. 635231383) + 𝑃(𝑧 > 4. 479893352) = 0. 004203997289
So the Type II error is 0.0042.
2. A manufacturer is interested in the power supply’s output voltage used in a PC. Output
voltage is assumed to be normally distributed with a standard deviation 0.26 volts, and
the manufacturer wishes to test the hypothesis H0: μ = 5 against H1: μ > 5 volts using n =
12 units.
(a) Find α, the type I error.
SE = 0.26 / √12 ≈ 0.075
(5.23 - 5) / 0.075 = 3.0667
α = 𝑃(𝑥 > 5. 23) = 𝑃(𝑍 > 3. 0667) = 1 − 0. 9989178202 = 0. 001082179806 α = 0. 0011
(b) Find the power of the test when the true mean voltage output is 4.9 volts.
(4.9 - 5) / 0.075 = -1.33333333333
P(Z > -1.3333) = 1 - NORM.S.DIST(-1.3333, TRUE) = 0.9087887797
Power = 1 -
= 1 - 0.9087887797 = 0.09121122027
β
3. A civil engineer analyzes the compressive strength of concrete. Compressive strength
is normally distributed with σ^2 = 1000 (psi)^2. A random sample of 10 specimens has a
mean compressive strength of 3225 psi.
(a) Construct a 90% two-sided confidence interval on the mean compressive strength.
90% CI = (3225 - 1.645 * 31.623 / √(10), 3225 + 1.645 * 31.623 / √(10)) = (3201.37, 3248.63)
(b) Suppose it is desired to estimate the mean compressive strength with an error that is
less than 10 psi at 95% confidence. What sample size is required?
z * √(s^2 / n) < 10
1.96 * √(1000 / n) < 10
n > 400
4. A manufacturer produces camshafts for an automobile engine. The wear on the
camshafts after 100,000 miles (0.0001 inches) is of interest because it will likely impact
warranty claims. A random sample of 25 shafts is tested, and the sample mean is 2.78. It
is known that σ = 0.6, and the wear is normally distributed.
(a) Test H0: μ = 3 versus H1: μ < 3 using α = 0.01.
SE = σ / √n = 0.6 / √25 = 0.12
Z = (2.78 - 3) / 0.12 = -1.83333333333
Critical Z-value = NORM.S.INV(0.01) ≈ -2.33
-1.8333 > -2.33, so we do not reject the null hypothesis.
(b) Find the type II error if the population mean is 2.80.
Z = (2.80 - 3) / 0.12 ≈ -1.6667
P(Z ≥ -1.6667) = 1 - NORM.S.DIST(-1.6667, TRUE) = 0.9522129635
5. The life in hours of a battery is known to be normally distributed with a standard
deviation of 1.25 hours. A random sample of 8 batteries has a mean life of x = 40.5 hours.
(a) Is there evidence to support the claim that battery life exceeds 40 hours? Use α = 0.01.
SE = 1.25 / √8 ≈ 0.4412
Z = (40.5 - 40) / 0.4412 = 1.13327289211
Critical Z-value = NORM.S.INV(1 - 0.01) ≈ 2.33
1.13327289211 < 2.33, so we reject the null hypothesis
(b) What is the p-value of the test in part (a)?
1 - NORM.S.DIST(1.13327289211, TRUE) = 0.1285498359
(c) What is the probability to FTR if the population mean is 40.9 hours?
(40.9 - 40) / 0.4412 = 2.0398912058
P(Z ≤ 2.0398912058) = NORM.S.DIST(2.0398912058, TRUE) = 0.9793194186
6. Supercavitation is a propulsion technology for undersea vehicles that can greatly
increase their speed. It occurs above approximately 50 meters per second when the
pressure drops sufficiently to allow the water to dissociate into water vapor forming a
gas bubble behind the vehicle. When the gas bubble completely encloses the vehicle,
supercavitation is said to occur. Nine tests were conducted on a scale model of an
undersea vehicle in a towing basin with the average observed speed X = 103.2 meters per
second. Assume that speed is normally distributed with a known standard deviation of σ
= 4 meters per second.
(a) Test the hypothesis H0: μ = 100 versus H1: μ > 100 using α = 0.05.
SE = 4 / √9 ≈ 1.333
Z = (103.2 - 100) / 1.333 = 2.4
Test statistic is higher than the critical value so we reject the null hypothesis.
(b) What sample size would be required to detect a true mean speed of 102 meters per
second if we wanted the power of the test to be at least 0.90? (Use Minitab)
α = 1. 645
β = 1. 28
(
1.645*4+1.28*4
2
)
2
= 34. 25881623
N = 35
7. An HT is performed on the average content of a container. Standards specify a mean
content of 16.0 oz. Nine containers are sampled and checked for over and under-filling.
Content is known to be normally distributed with a standard deviation of 0.1 oz., and the
upper and lower critical values for the test are 15.95 and 16.05, respectively.
(a) Determine the Type I error for the test?
SE = 0.1 / √9 ≈ 0.0333
𝑍
𝐿
= (15. 95 − 16) / 0. 0333 = − 1. 502
𝑃(𝑍 <− 1. 502) = 0. 06680720127
𝑍
𝑈
= (16. 05 − 16) / 0. 0333 = 1. 502
𝑃(𝑍 > 1. 502) = 1 − 0. 9331927987 = 0. 06680720127 Type I error = 0.1336144025
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
(b) If the true mean is 16.08, what is the probability that the test would conclude that
containers were filling correctly?
𝑍
𝐿
= (15. 95 − 16. 08) / 0. 0333 = − 3. 9
𝑃(𝑍 ≤− 3. 9) = 0. 00004809634402
𝑍
𝑈
= (16. 05 − 16. 08) / 0. 0333 = − 0. 9
𝑃(𝑍 ≥− 0. 9) = 1 − 0. 9331927987 = 0. 1840601253 𝑃(− 3. 9 ≤ 𝑍 ≤− 0. 9) = 0. 184
(c) What equation would you plot to obtain the OC curve for the test as a function of μ?
β =
15.95
16.05
∫
𝑁(µ, 0. 0333)𝑑𝑥 = 0. 184
8. Your boss has a presentation and needs information from you on the mouse
production process. Cover-hole diameter is normally distributed with a standard
deviation of 0.0014. A sample of 25 cover holes found the mean diameter to be 0.196. You
have been asked to provide the following information:
(1) A 90% confidence interval for the population mean dimension of cover-hole diameters
produced.
SE = 0.1 / √25 ≈ 0.00028
Z = 1.645
𝐶𝐼 = 0. 196 ± 0. 0005
(2) Given six weeks of mouse production data (690, 692, 680, 675, 684. 686), predict next
week’s total production with 90% confidence (assume a normal distribution can
approximate the production data).
Sample Mean = (690 + 692 + 680 + 675 + 684 + 686) / 6 = 684.5
Sample Stdev = 6.316644679
𝑡
0.05,5
= 2. 015
√(1+
⅙
) = 1.08
2.015*1.08*6.317 = 13.75
Lower = 684.5 - 13.75 = 670.75
Upper = 684.5 + 13.75 = 698.25
Therefore next week’s total production =
670. 75 ≤ 𝑋 ≤ 698. 25
(3) You are 90% confident that 95% of the cover-hole dimensions produced fall within
what range.
K = 2.631
X = 0.196
S = 0.0014
Lower = 0.196 - 2.631*0.0014 = 0.1923
Upper = 0.196 + 2.631*0.0014 = 0.1997
CI = (0.1923, 0.1997)
4.5. The inside diameters of bearings used in an aircraft landing gear assembly are
known to have a standard deviation of u = 0.002 cm. A random sample of 15 bearings has
an average inside diameter of 8.2535 cm.
a. Test the hypothesis that the mean inside bearing diameter is 8.25 cm. Use a two-sided
alternative and a = 0.05.
SE = 0.002 / √15 = 0.00051639777
Z = 8.2535-8.25/SE = 6.78
We do reject the null hypothesis.
b. Find the P-value for this test.
P = 0
c. Construct a 95% two-sided confidence interval on the mean bearing diameter.
Lower = 8.2535 - 1.96*SE = 8.252
Upper = 8.2535 + 1.96*SE = 8.255
CI = (8.252,8.255)
4.9. Ferric chloride is used as a "ux in some types of extraction metallurgy processes.
This material is shipped in containers, and the container weight varies. It is important to
obtain an accurate estimate of mean container weight. Suppose that from long
experience a reliable value for the standard deviation of "ux container weight is
determined to be 4 lb. How large a sample would be required to construct a 95%
two-sided confidence interval on the mean that has a total width of 1 lb?
N = (2 * 1.96 * 4)^2 = 246