Statistics for Engineers and Scientists
4th Edition
ISBN: 9780073401331
Author: William Navidi Prof.
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 5.4, Problem 1E
To study the effect of curing temperature on shear strength of a certain rubber compound, 80 specimens were cured at 150°C and 95 were cured at 130°C. The specimens cured at 150°C had an average shear strength of 620 psi, with a standard deviation of 20 psi. Those cured at 130°C had an average shear strength of 750 psi, with a standard deviation of 30 psi. Find a 95% confidence interval for the difference between the mean shear strengths of specimens cured at the two temperatures.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The tensile strength of paper using 5% hardwood and 10% hardwood are to be compared. A random sample of 41 sheets made from 5% hardwood had an average tensile strength of 1.52 lbs. with a standard deviation of 0.33 lbs. A random sample of 61 sheets made from 10% hardwood had an average tensile strength of 1.88 lbs. with a standard deviation of 0.42 lbs.
Can we conclude the average tensile strength is the same for the two types of paper? Use 5% significance and p-values.
Part 1) Conduct the test using the assumption that the sample standard deviations are the same as the population standard deviations.
Part 2) Conduct the test using the assumption that the population standard deviations are unknown but equal (you also must assume a normal population here).
Part 3) Conduct the test with only the assumption that we have simple random samples from normal populations.
May you please explain which stat test you used on your calculator and why. Thank you very much for your help!
Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials.
An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.4 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide
compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use a = 0.05.)
State the appropriate hypotheses.
O H,: H 48
H:u = 48
O H,: = 48
Hi u 48
OH:= 48
H: u + 48
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
P-value =
What can you conclude?
O The data provides compelling…
Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens
with 2% fiber content, the sample mean tensile strength (MPa) was 51.3 and the sample standard deviation was 1.4. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this
value? (Use α = 0.05.)
State the appropriate hypotheses.
ⒸH₂: μ = 48
Ha: > 48
Ho:μ = 48
H₂H 48
H₂: μ = 48
Ho: μ< 48
H₂₁:μ = 48
USE SALT
Ho: M = 48
H₂:48
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)
t= 7.68
X
P-value =
Chapter 5 Solutions
Statistics for Engineers and Scientists
Ch. 5.1 - Find the value of Z/2 to use in expression (5.1)...Ch. 5.1 - Find the levels of the confidence intervals that...Ch. 5.1 - As the confidence level goes up, the reliability...Ch. 5.1 - The article Modeling Arterial Signal Optimization...Ch. 5.1 - In a sample of 100 steel wires the average...Ch. 5.1 - The article Application of Surgical Navigation to...Ch. 5.1 - The capacities (in ampere-hours) were measured for...Ch. 5.1 - Oven thermostats were tested by setting them to...Ch. 5.1 - In a sample of 80 ten-penny nails, the average...Ch. 5.1 - In a sample of 60 electric motors, the average...
Ch. 5.1 - The sugar content in a one-cup serving of a...Ch. 5.1 - Refer to Exercise 5. a. Find a 95% lower...Ch. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Refer to Exercise 9. a. Find a 90% upper...Ch. 5.1 - Refer to Exercise 10. a. Find a 98% lower...Ch. 5.1 - Refer to Exercise 11. a. Find a 95% upper...Ch. 5.1 - Prob. 19ECh. 5.1 - A 95% confidence interval for a population mean is...Ch. 5.1 - Based on a large sample of capacitors of a certain...Ch. 5.1 - Sixty-four independent measurements were made of...Ch. 5.1 - A large box contains 10,000 ball bearings. A...Ch. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.2 - In a simple random sample of 70 automobiles...Ch. 5.2 - During a recent drought, a water utility in a...Ch. 5.2 - A soft-drink manufacturer purchases aluminum cans...Ch. 5.2 - The article HIV-positive Smokers Considering...Ch. 5.2 - The article The Functional Outcomes of Total Knee...Ch. 5.2 - Refer to Exercise 1. Find a 95% lower confidence...Ch. 5.2 - Refer to Exercise 2. Find a 98% upper confidence...Ch. 5.2 - Refer to Exercise 4. Find a 99% lower confidence...Ch. 5.2 - A random sample of 400 electronic components...Ch. 5.2 - Refer to Exercise 9. A device will be manufactured...Ch. 5.2 - When the light turns yellow, should you stop or go...Ch. 5.2 - In a random sample of 150 customers of a...Ch. 5.2 - A sociologist is interested in surveying workers...Ch. 5.2 - Stainless steels can be susceptible to stress...Ch. 5.2 - The article A Music Key Detection Method Based on...Ch. 5.2 - A stock market analyst notices that in a certain...Ch. 5.3 - Find the value of tn1,/2 needed to construct a...Ch. 5.3 - Find the value of tn1, needed to construct an...Ch. 5.3 - Find the level of a two-sided confidence interval...Ch. 5.3 - True or false: The Students t distribution may be...Ch. 5.3 - The article Wind-Uplift Capacity of Residential...Ch. 5.3 - Prob. 6ECh. 5.3 - The article An Automatic Visual System for Marble...Ch. 5.3 - A chemist made eight independent measurements of...Ch. 5.3 - Six measurements are taken of the thickness of a...Ch. 5.3 - Fission tracks are trails found in uranium-bearing...Ch. 5.3 - The article Effect of Granular Subbase Thickness...Ch. 5.3 - The article Influence of Penetration Rate on...Ch. 5.3 - Prob. 13ECh. 5.3 - Prob. 14ECh. 5.3 - Prob. 15ECh. 5.3 - The concentration of carbon monoxide (CO) in a gas...Ch. 5.3 - The article Filtration Rates of the Zebra Mussel...Ch. 5.4 - To study the effect of curing temperature on shear...Ch. 5.4 - The article Some Parameters of the Population...Ch. 5.4 - The article Inconsistent Health Perceptions for US...Ch. 5.4 - The article Hatching Distribution of Eggs Varying...Ch. 5.4 - The article Automatic Filtering of Outliers in RR...Ch. 5.4 - A group of 78 people enrolled in a weight-loss...Ch. 5.4 - In experiments to determine the effectiveness of...Ch. 5.4 - A stress analysis was conducted on random samples...Ch. 5.4 - In a study to compare two different corrosion...Ch. 5.4 - An electrical engineer wishes to compare the mean...Ch. 5.4 - In a study of the effect of cooling rate on the...Ch. 5.4 - Refer to Exercise 11. Ten more welds will be made...Ch. 5.4 - The article The Prevalence of Daytime Napping and...Ch. 5.4 - The article Occurrence and Distribution of...Ch. 5.5 - In a test of the effect of dampness on electric...Ch. 5.5 - The specification for the pull strength of a wire...Ch. 5.5 - Angioplasty is a medical procedure in which an...Ch. 5.5 - Prob. 4ECh. 5.5 - Prob. 5ECh. 5.5 - Prob. 6ECh. 5.5 - In a study of contamination at landfills...Ch. 5.5 - Prob. 8ECh. 5.5 - A mobile computer network consists of a number of...Ch. 5.5 - The article Evaluation of Criteria for Setting...Ch. 5.5 - In a certain year, there were 80 days with...Ch. 5.6 - In a study comparing various methods of gold...Ch. 5.6 - Prob. 2ECh. 5.6 - In an experiment involving the breaking strength...Ch. 5.6 - A new post-surgical treatment is being compared...Ch. 5.6 - The article Differences in Susceptibilities of...Ch. 5.6 - The article Tibiofemoral Cartilage Thickness...Ch. 5.6 - During the spring of 1999, many fuel storage...Ch. 5.6 - Prob. 8ECh. 5.6 - The article Toward a Lifespan Metric of Reading...Ch. 5.6 - Prob. 10ECh. 5.6 - Measurements of the sodium content in samples of...Ch. 5.6 - Prob. 12ECh. 5.6 - Prob. 13ECh. 5.6 - In the article Bactericidal Properties of Flat...Ch. 5.6 - Prob. 15ECh. 5.7 - The article Simulation of the Hot Carbonate...Ch. 5.7 - The article Effect of Refrigeration on the...Ch. 5.7 - Transepidermal water loss (TEWL) is a measure of...Ch. 5.7 - Breathing rates, in breaths per minute, were...Ch. 5.7 - A group of five individuals with high blood...Ch. 5.7 - A sample of 10 diesel trucks were run both hot and...Ch. 5.7 - For a sample of nine automobiles, the mileage (in...Ch. 5.7 - Prob. 8ECh. 5.7 - Prob. 9ECh. 5.7 - Prob. 10ECh. 5.8 - Find the following values. a. 12,.0252 b. 12,.9752...Ch. 5.8 - Prob. 2ECh. 5.8 - Construct a 99% confidence interval for the...Ch. 5.8 - Prob. 4ECh. 5.8 - Scores on an IQ test are normally distributed. A...Ch. 5.8 - Prob. 6ECh. 5.8 - Boxes of cereal are labeled as containing 14...Ch. 5.8 - Prob. 8ECh. 5.8 - Following are interest rates (annual percentage...Ch. 5.8 - Prob. 10ECh. 5.8 - Prob. 11ECh. 5.8 - Prob. 12ECh. 5.9 - A sample of 25 resistors, each labeled 100, had an...Ch. 5.9 - Prob. 2ECh. 5.9 - The article Ozone for Removal of Acute Toxicity...Ch. 5.9 - Six measurements were made of the concentration...Ch. 5.9 - Five measurements are taken of the octane rating...Ch. 5 - A molecular biologist is studying the...Ch. 5 - Prob. 2SECh. 5 - The article Genetically Based Tolerance to...Ch. 5 - A sample of 87 glass sheets has a mean thickness...Ch. 5 - Prob. 5SECh. 5 - Prob. 6SECh. 5 - Leakage from underground fuel tanks has been a...Ch. 5 - Prob. 8SECh. 5 - Prob. 9SECh. 5 - Prob. 10SECh. 5 - In the article Groundwater Electromagnetic Imaging...Ch. 5 - Prob. 12SECh. 5 - Prob. 13SECh. 5 - Prob. 14SECh. 5 - A metallurgist makes several measurements of the...Ch. 5 - In a study of the lifetimes of electronic...Ch. 5 - The temperature of a certain solution is estimated...Ch. 5 - Prob. 18SECh. 5 - Prob. 19SECh. 5 - The answer to Exercise 19 part (d) is needed for...Ch. 5 - The carbon content (in ppm) was measured for each...Ch. 5 - Diameters, in mm, were measured for eight...Ch. 5 - A sample of eight repair records for a certain...Ch. 5 - Prob. 25SE
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Similar questions
- Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.4 and the sample standard deviation was 1.5. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.) Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value =arrow_forwardPolymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.3. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.) USE SALT State the appropriate hypotheses. ⒸHO: μ> 48 H₂: μ = 48 Ho: μ 48 Ho: μ = 48 H₂: μ = 48 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = What can you conclude? O The data provides compelling…arrow_forwardPolymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.8 and the sample standard deviation was 1.5. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use a = 0.05.) n USE SALT State the appropriate hypotheses. Ho:H = 48 Ha:µ > 48 = 48 Hạ: u + 48 Ho: H = 48 Hai u 48 H: µ = 48 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) t = P-value = What can you conclude? O The data…arrow_forward
- The tensile strengths of two types of materials were investigated by collecting 16 samples of each material (totally 32 samples).The sample mean for material 1 was 423.08 (psi), and for material 2 was 424.7 (psi). The samplestandard deviation of the difference in tensile strengths was sD = 2.8 (psi). Is there sufficientevidence to conclude that two types of materials have the same tensile strength? Formulate andtest the appropriate null H0 and alternative hypotheses H1 to verify the claim at level α = 0.01arrow_forwardTo compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 43 feet. Assume the population standard deviation is 4.7 feet. The mean braking distance for Make B is 44feet. Assume the population standard deviation is 4.5 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e). a) identify the claim and state Ho and Ha b) find the critical values and identify the rejection regions c) Find the standardized test statistic z d) Decide whether to reject or fail to reject the null hypothesis. e) Interpret the decision in the context of the original claim.arrow_forwardTo compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 43 feet. Assume the population standard deviation is 4.6 feet. The mean braking distance for Make B is 46 feet. Assume the population standard deviation is 4.5 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. The critical value(s) is/are Find the standardized test statistic z for μ1−μ2.arrow_forward
- To compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 41 feet. Assume the population standard deviation is 4.6 feet.The mean braking distance for Make B is 42 feet. Assume the population standard deviation is 4.4 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e). (a) Identify the claim and state Ho and Ha. What is the claim? A.The mean braking distance is different for the two makes of automobiles. This is the correct answer. B.The mean braking distance is the same for the two makes of automobiles. C.The mean braking distance is less for Make A automobiles than Make B automobiles. Your answer is…arrow_forwardTo compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 43 feet. Assume the population standard deviation is 4.6 feet. The mean braking distance for Make B is 47 feet. Assume the population standard deviation is 4.2 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e).arrow_forwardTo compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 40 feet. Assume the population standard deviation is 4.8 feet. The mean braking distance for Make B is 43 feet. Assume the population standard deviation is 4.5 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e).arrow_forward
- To compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 44 feet. Assume the population standard deviation is 4.9 feet. The mean braking distance for Make B is 46 feet. Assume the population standard deviation is 4.7 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e).arrow_forwardTo compare the dry braking distances from 30 to 0 miles per hour for two makes of automobiles, a safety engineer conducts braking tests for 35 models of Make A and 35 models of Make B. The mean braking distance for Make A is 42 feet. Assume the population standard deviation is 4.5 feet. The mean braking distance for Make B is 46 feet. Assume the population standard deviation is 4.4 feet. At α=0.10, can the engineer support the claim that the mean braking distances are different for the two makes of automobiles? Assume the samples are random and independent, and the populations are normally distributed. Complete parts (a) through (e).arrow_forwardA synthetic fiber used in manufacturing carpet has tensile strength that is normally distributed with mean 75.5 psi and standard deviation 3.5 psi. A study is actually conducted with n = 6 and the average value observed is 78.7. Is this surprising? Would it be reasonable to conclude the true mean had changed (that is, it is not really 75.5 for this study)?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License