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Student Content Assessment

Instrument 28: Pre-electronic ITS Evaluation System

Project: Case-Based Reasoning for Engineering Statistics
Arizona State University

Funding Source: NSF - Division of Undergraduate Education (DUE)

Purpose: To evaluate the effectiveness of practice-problems for an upcoming quiz and different methods of preparation (using conditions as close to the actual test situation as possible) for an undergraduate engineering statistical course.

Administered To: Undergraduates who attended an optional study session for the engineering statistics course.

Topics Covered:

Content Specific Assessment: Engineer-related statistics

Format/Length: 8 open-ended items

Summary of Pre-electronic ITS Evaluation System
Fall, 2002

Preparation
An announcement was made to inform the students of the date and time of the experiment. This announcement can be found in A0.

In preparation for the experiment three primary packets were prepared. One packet consisted of the 6 problems with an example for each problem. The next packet consisted of only the 6 problems. Stapling the six exercises together in a random fashion made each packet. The final products can be found in A 1 and A2. An instruction/debriefing page was then added as the cover. The instruction page for the group with exercises can be found in A 11 and the group without can be found in A21. The problems with examples were labeled an even number and the problems without were labeled an odd number. These two packets were then put together to form one group in numeric order. So when students received their packet they were unaware that there were two different groups.

The third packet consisted of two post problems. Stapling the two exercises in random order made each packet. The final product can be seen in A3. An instruction and debriefing page was then added as the cover, A31. These packets were than randomly put together to form one group.

Solutions to the exercises and post problems along with a debriefing were generated to give to students and the end of the experiment. The debriefing can be found in A4, the solutions to the exercises are A5 and the solutions to the post test are A6.

The Experiment Day
Graduate students along with Dr. Hubele and Dr. Runger proctored the day of the experiment. There were always two individuals proctoring the examples throughout the day. One was in charge of distributing and recording information and the other one was in charge of answering questions after an individual completed the experiment.

When a student entered the room they were given a formula sheet, A7, a numbered packet along with a blue book to complete their work inside of. The proctor recorded their packet number onto the blue book and also recorded the time they distributed the material onto an excel spreadsheet. this excel spreadsheet consisted of the number, time out and in of the exercises, and the time out and in of the post problems. When they completed the exercises, they gave the proctor their exercises and blue book. The proctor recorded the time in and gave them the post problem exercises with their number on it. When they finished the post problem exercises they gave it to the proctor, the proctor then supplied them with the solutions and informed them that there was a graduate student outside available to answer any questions.

ECE 380 ANNOUNCEMENT

There will be a study session for Engineering Statistics on Monday, September 30, 9:00 AM to 9:00 PM in Goldwater Center Room 487. To make it most worthwhile, please anticipate being in the session for about 1.5 hours.

During this study session, you will have an opportunity to work some practice quic problems and get immediate feedback. At the same time, we will be collecting information to help us help you learn statistics.

The instructors are interested in how these sorts of practice sessions affect student performance on quizzes and tests. We will keep track of your performance, and use this information to make changes to the course, the text, and the assignments. We believe that this will help you to learn statistics more effectively, and that it will help other students in the future.

The study session is optional, and open to all students; your grade in the class will not be affected by your attendance, and you can stay as long as you like. If you have any questions about the study session or the research we are doing, please fee free to ask us.

We hope to see you there!

ECE 380 Team
Fall, 2002

Figure A1: Exercises with examples

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1)

If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: YARN

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean elongation force of 14 kg. The company wishes to test the hypothesis that the mean elongatin force is less than 14 kg. The company collects the following sample data: 13.87, 13.85, 13.72, 14.32 which has a sample mean 13.92 and sample standard deviation 0.24. Assuming normality of the data, use = 0.05 to perform the test.

EXAMPLE: THREAD
A fabric producer is researching a new curtain thread, which the company asserts has a mean breaking strength of 800 g. The manufacturer wants to claim that the mean force is under 800 g. The company records the following data: 795, 790, 775, 825, 780, which has a sample mean 793 and sample standard deviation 19.56. Assuming normality of the data, use = 0.05 to perform the test.

Solution

The parameter of interest is, the mean breaking strength.

H₀: = 800 g

H₁: < 800 g

= 0.05

The test statistic is
Reject H₀ if t ₀ < -t_{0.05, 4} = -2.132.

Computation: Since = 793, s = 19.56, = 800, and n = 5, we have

Conclusion: Since t₀=-0.80 > - 2.132, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to show that the mean breaking strength is less than 800 g.

Figure A11: Debriefing for group with exercises and examples

In this activity, we are going to present you with 6 statistics problems to solve. You will notice that we have given you several aids to help you solve them. At the top of the page are the equations for the t and z test statistics. Also, below each problem to solve, there is a worked example.

This worked exampled is the same kind of problem as the one you are being asked to solve. If you can figure out how the worked example was done, you will be able to reach the correct solution on the problem you are solving. Read both problems carefully, and try to make use of the worked example as much as possible.

Put all of your work in the blue books you have been provided; please do not make any marks in this packet. Show as much of your work as possible, and don't worry about making mistakes, or us seeing your mistakes. If you make an error, don't bother erasing, just draw an X through it and keep going. This is practice; what matters is that you learn how to solve the problems, and we learn how to help you solve the problems. Knowing what you are thinking when you are solving the problems is more important than a neat blue book or hiding your mistakes.

This activity is not graded; you'll notice that we don't even know who you are, you are identified only by a number. So you don't have to worry about us seeing your false starts and mistakes.

Please work through the problems in the order they are given. Do your best with each, but don't get stuck on any one problem. Once you have done all you can on a problem, turn the page and move on. Please do not return to any earlier problems once you have turned the page.

If you have any questions, please feel free to ask us. Otherwise, you may begin. When you are finished, please bring your packet and blue book back up to us.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1)

If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: CABLE HARNESS

The diameter of holes for cable harness is known to have a standard deviation of 0.02 in. A random sample of size 10 yields the following data: 1.76, 1.69, 1.74, 1.73, 1.76, 1.77, 1.75, 1.78, and 1.76, with sample mean 1.749. Is the true mean diameter less than 1.75? Use = 0.01.

EXAMPLE: METAL CORD

The diameter of slots for metal cord tackle is known to have a standard deviation of 0.02 in. A random sample of amount 10 renders the subsequent data: 2.49, 2.44, 2.49, 2.48, 2.51, 2.52, 2.50, 2.53, 2.50, and 2.51, with sample mean 2.497. Is the true mean diameter under 2.50? Use = 0.01.

Solution

The parameter of interest is, the mean diameter.

H₀: = 2.50 in

H₁: < 2.50 in

= 0.01

The test statistic is

Reject H₀ if Z₀ < Z_0.01 = -2.32

Computation: Since = 2.497, = 0.02, = 2.50, and n = 10, we have
Conclusion: Since Z₀ = -0.474 > -2.32, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to determine that the mean diameter is less than 2.50 in.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1)

If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: ROCKWELL HARDNESS

The mean Rockwell hardness of a specimen of material is thought to be greater than 60. Five specimens are tested and found to have an average 60.2. The population standard deviation is known to be 0.12. Test the appropriate hyptheses with a significance level = 0.10.

EXAMPLE: COUPON OF METAL

The mean hardness measurement of a coupon of metal is claimed to be greater than 50. Five coupons are examined and produce an average of 50.3. The population standard deviation is known to be 0.80. Test the appropriate hypotheses with a significance level of 0.10.

Solution

The parameter of interest is, the mean hardness.

H₀: = 50

H₁: > 50

= 0.10

The test statistic is

Reject H₀ if Z₀ > Z_0.10 = 1.28

Computation: Since = 50.3, = 0.08, = 50, and n = 5, we have
Conclusion: Since z₀=8.39 > 1.28, we reject H₀ and conclude that the mean hardness is greater than 50.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1)

If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: CEREAL MANUFACTURER
A cereal manufacturer requires that the mean weight of a box be more than 20 ounces of cereal. The content of a sample of ten boxes has an average weight 20.15 and standard deviation 0.15 ounces of cereal. Assuming that the weight of the cereal is a normally distributed random variable, test if the mean weight is greater than 20 ounces with a significance level of = 0.05.

EXAMPLE: GRAIN PRPODUCER

A grain producer demands that the mean weight of a bag be above 32 ounces of grain. The weight of a sample of ten bags has an average weight 32.1 and standard deviation 0.12 ounces of grain. Assuming that the weight of the train is a normally distributed random variable, test if the mean weight is above 32 ounces with a significance level = 0.05.

Solution

The parameter of interest is, the mean weight.

H₀: = 32 ounces

H₁: > 32 ounces

= 0.05

The test statistic is

Reject H₀ if t₀ > - t_{0.05, 9} = 1.833

Computation: Since = 32.1, s = 0.12, = 32, and n = 10, 23 have

Conclusion: Since t₀ = 2.64 > 1.833, we reject H₀ and conclude that the mean weight is more than 32 ounces.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1)

If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: BEVERAGE MACHINE

A post-mix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of 1.098 fluid ounces and a standard deviation of 0.016 fluid ounces. Do the data presented support the claim that the mean amount of syrup dispensed is not 1.0 fluid ounce? Assuming normality of the data, use = 0.01.

EXAMPLE: SOFT DRINK MACHINE

A soft drink mixing device is modified to discharge a particular quantity of syrup into a container where it is combined with carbonated water. A random sample of 25 servings was determined to have a mean syrup composition of 0.068 fluid ounces and a standard deviation of 0.008 fluid ounces. Is there evidence ot claim the mean quantity of syrup dispensed is different from 0.050 fluid ounces? Assuming normality of the data, use = 0.01.

Solution

The parameter of interest is, the mean quantity of syrup dispensed.

H₀: = 0.050 fluid ounces

H₁: 0.050

= 0.01

The test statistic is

Reject H₀ if t₀ < = t_{0.005, 24} = -2.797 or t₀ > t_{0.005, 24} = 2.797

Computation: Since = 0.068, s = 0.008, = 0.05, and n = 25, we have
Conclusion: Since t₀ = 11.25 > 2.797, we reject H₀, we reject H₀ and conclude that the mean quantity of syrup is different from 0.050 fluid ounces.

Figure A21: Instructions for group without examples

In this activity, we are going to present you with 6 statistics problems to solve. You will notice that we have given you aids to help you solve them. At the top of the page are the quations for the t and the z tests.

This activity is not graded; you'll notice that we don't even have who you are, you are identified only by a number. so you don't have to worry about us seeing yoru false starts and mistakes.

If you have any questions, please feel free to ask us. Otherwise, you may begin. When you are finished, please bring your packet and blue book back up to us.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: YIELD

The yield of a chemical process is being studied. From previous experience with this process the standard deviation of teh yield is known to be 3. The past 5 days of plant operation have resulted in the following yields: 91.6%, 88.75%, 90.8%, 89.95%, and 91.3%. The sample mean is calculated as 90.48%. Is there evidence that the mean yield is not 90%? Use = 0.10.

EXAMPLE: PRODUCT

The product of a chemical procedure is being analyzed. From prior studies with this procedure the standard deviation of the product is known to be 2.5. The last 5 days of factory production have generated the subsequent products: 99.7%, 96.75%, 98.80%, 97.90%, and 99.30%. The sample mean is calculated as 98.49%. Can the company conclude that the mean is different from 98%? Use = 0.10.

Solution

The parameter of interest is, the mean product.

H₀: = 98%

H₁: 98%

= 0.10

The test statistic is

Reject H₀ if Z₀ > Z_0.05 = 1.65 or if Z₀ < -Z_0.05 = -1.65

Computation: Since = 98.49%, = 2.5%, = 98%, and n = 5, we have
Conclusion: Since Z₀ = 0.438 < 1.65 and Z₀ = 0.438 > -1.65, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to determine that the mean product is not equal to 98%.

Figure A2: Exercises without examples

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1)

If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: YARN

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean elongation force of 14 kg. The company wishes to test the hypothesis that the mean elongatin force is less than 14 kg. The company collects the following sample data: 13.87, 13.85, 13.72, 14.32, 13.92 which has a sample mean 13.92 and sample standard deviation 0.24. Assuming normality of the data, use = 0.05 to perform the test.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: BEVERAGE MACHINE

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: CEREAL MANUFACTURER

A cereal manufacturer requires that the mean weight of a box be more than 20 ounces of cereal. The content of a sample of ten boxes has an average weight 20.15 and standard deviation 0.15 ounces of cereal. Assuming that the weight of the cereal is a normally distributed random variable, test if the mean weight is greater than 20 ounces with a significant level = 0.05.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: ROCKWELL HARDNESS

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: CABLE HARNESS

The diamter of holes for cable harness is known to have a standard deviation of 0.02 in. a random sample of size 10 yields the following data: 1.76, 1.69, 1.74, 1.73, 1.76, 1.77, 1.75, 1.78, 1.75, and 1.76, with sample mean 1.749. Is the true mean diameter less than 1.75? Use = 0.01.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: YIELD

The yield of a chemical process is being studied. From previous experience with this process the standard deviation of the yield is known to be 3. The past 5 days of plant operation have resulted in the following yields: 91.6%, 88.75%, 90.8%, 89.95%, and 91.3%. The sample mean is calculated as 90.48%. Is there evidence that the mean yield is not 90%? Use = 0.10.

Figure A3: Post Test Problem

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: VOLTAGE

A sample of 10 breakdown voltages, u_d, of coaxial insulation are recorded and found to have an average of 148 kV and a standard deviation of 2.5 kV. The power company would like to determine if the mean breakdown voltage is 150 kV. use a significance level of 0.05.

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: VISCOSITY

According to the engineering specification, a chemical process is designed to generate a fluid with a viscosity greater than 10.2 Pa s (Pascal seconds). The standard deviation of viscosity is known to be 0.5 Pa s. A sample of 20 batches of the fluid yields an average viscosity measurement of 10.23. Based on the sample data, does the process conform to the engineering specification? Use = 0.10.

A31: Debriefing for Post Problems

Now that you have had a chance to practice the problems, we'd like you to do some post-tutorial problems. This will provide you with a "dress rehearsal" for some of the quiz problems that you wil take in class this week.

There are just 2 problems in this packet. Put all of your work in the blue books you have been provided; please do not make any marks in the packet. show as much of your work as possible, and don't worry about making mistakes, or us seeing your mistakes. If you make an error, don't bother erasing, just draw an X through it and keep going.

Once you have finished this exercise, you will be given a packet with complete step-by-step solutions. Also, we will return a copy of your post-tutorial exercises to you fully graded just as it would be in class. This will help you understand how we evaluate the problems. IMPORTANT: Remember your id number--we don't know who is who, so to pick up your results, you have to have your id number!

Please work through the problems in the order they are given. do your best with each, but don't get stuck on any one problem. Once you have done all you can on a problem, turn the page adn go on to the second.

If you have any questions, please feel free to ask us. Otherwise, you may begin. When you are finished, please bring your packet and blue book back up to us.

Figure A3: Post Problems

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: VOLTAGE

If the approrpriate assumptions are satisfied,

has a standard normal distribution N(0,1) If the appropriate assumptions are satisfied,

has a t distribution with n-1 degrees of freedom

EXERCISE: VISCOSITY

Figure A4: Solution Debriefing

Debriefing

Important: We hope that these exercises have been helpful in partial preparation for your upcoming Quiz 2. However, please be informed that you can expect to see other types of questions and topics on the quiz.

This booklet contains complete solutions to all of the problems, both the practice exercises and the practice post-tutorial exercises. This is yours to take home and study from.

Don't forget to write your id number on here as a reminder! Otherwise, you won't be able to get your graded post-tutorial exercises back!

The purpose of this study session was two-fold: First, we wanted to give you a chance to prepare for the upcoming quiz by working problems in a situation that is close to the actual test conditions. Psychologists have shown that learning is very depedent on context; the closer the practice conditions are to the test conditions, the better peopl do on the test. So, we hope this will help you when you take the real quiz!

Second, we also wanted to know how different kinds of practice work. That is why different people received types of practice problems. From the perspective of how our minds process information, we really don't know how these different types of practice work. They may even be exactly the same. We do know that all practice is helpful, so everyone who came should benefit from attending this session.

Because we are trying to do research on how practice methods work, we need your help. Please do not discuss the study session with students who have not yet attended. At the end of the day, all of the study sessions will be over, and you can share these packets and your thoughts all you like. But until then, please keep this packet and your ideas about the session to yourself. Thank you for understanding.

If you have any questions, just ask, or you get in touch with us. Good luck on the quiz! We hope this helped you!

Figure A5: Solutions to Exercises

EXERCISE: YARN

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean elongation force of 14 kg. The company wishes to test the hypothesis that the mean elongatin force is less than 14 kg. The company collects the following sample data: 13.87, 13.85, 13.72, 14.32, 13.92 which has a sample mean 13.92 and sample standard deviation 0.24. Assuming normality of the data, use = 0.05 to perform the test.

Solution

The parameter of interest is, the mean elongation force.

H₀: = 14 kg

H₁: < 14 kg

= 0.05

The test statistic is

Reject H₀ if t₀ < -t_{0.05, 4} = -2.132.

Computation: Since = 13.92, s = 0.24, = 14, and n = 5, we have
Conclusion: Since t₀ = -0.75 > -2.132, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to show that the mean breaking strength is less than 14 kg.

EXAMPLE: THREAD

A fabric producer is researching a new curtain thread, which the company asserts has a mean breaking strength of 800 g. The manufacturer wants to claim that the mean force is under 800 g. The company records the following data: 795, 790, 775, 825, 780, which has a sample mean 793 and sample standard deviation 19.56. Assuming normality of the data, use to = 0.05 perform the test.

Solution

The parameter of interest is of interest is, the mean breaking strength.

H₀: = 800 g

H₁: < 800 g

= 0.05

The test statistic is

Reject H₀ if t₀ < -t_{0.05, 4} = -2.132.

Computation: Since = 793, s = 19.56, = 800, and n = 5, we have
Conclusion: Since t₀= -0.80 > - 2.132, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to show that the mean breaking strength is less than 800 g.

EXERCISE: BEVERAGE MACHINE

Solution

The parameter of interest is, the mean quantity of syrup dispensed.

H₀: = 1.0 fluid ounces

H₁: 1.0

= 0.01

The test statistic is

Reject H₀ if t₀ < -t_{0.005, 24} = -2.797 or t₀ > t_{0.005, 24} = 2.797

Computation: Since = 1.098, s = 0.016, = 0.05, and n = 25, we have
Conclusion: Since t₀ = 30.63 > 2.797, we reject H₀ and conclude that the mean quantity of syrup is different from 1.0 fluid ounces.

EXAMPLE: SOFT DRINK MACHINE

A soft drink mixing device is modified to discharge a particular quantity of syrup into a container where it is combined with carbonated water. A random sample of 25 servings was determined to have a mean syrup composition of 0.068 fluid ounces and a standard deviation of 0.008 fluid ounces. Is there evidence to claim the mean quantity of syrup dispensed is different from 0.050 fluid ounces? Assuming normality of the data, use = 0.01.

Solution

The parameter of interest is, the mean quantity of syrup dispensed.

H₀: = 0.050 fluid ounces

H₁: 0.050

= 0.01

The test statistic is
Reject H₀ if t₀ < - t_{0.005, 24} = -2.797 or t₀ > t_{0.005, 24} = 2.797

Computation: Since = 0.068, s = 0.008, = 0.05, and n = 25, we have

Conclusion: Since t₀ = 11.25 > 2.797, we reject H₀ and conclude that the mean quantity of syrup is different from 0.050 fluid ounces.

EXERCISE CEREAL MANUFACTURER

A cereal manufacturer requires that the mean weight of a box be more than 20 ounces of cereal. The content of a sample of ten boxes has an average weight 20.15 and standard deviation 0.15 ounces of cereal. Assuming that the weight of the cereal is a normally distributed random variable, test if the mean weight is greater than 20 ounces with a significance level of = 0.05.

Solution

The parameter of interest is , the mean weight.

H_0: = 20 ounces

H₁: > 20 ounces

= 0.05

The test statistic is
Reject H₀ if t₀ > t_{0.05, 9} = 1.833

Computation: Since = 20.15, s = 0.15, = 20, and n = 10, we have
Conclusion: Since t₀ = 3.16 > 1.833, we reject H₀ and conclude that the mean weight is more than 20 ounces.

EXAMPLE: GRAIN PRODUCER

A grain producer demands that the mean weight of a bag be above 32 ounces of grain. The weight of a sample of ten bags has an average weight 32.1 and standard deviation 0.12 ounces of grain. Assuming that the weight of the grain is a normally distributed random variable, test if the mean weight is above 32 ounces with a significance level = 0.05.

Solution

The parameter of interest is, the mean weight.

H₀: = 32 ounces

H₁: > 32 ounces

= 0.05

The test statistic is

Reject H₀ if t₀ > t_{0.05, 9} = 1.833

Computation: Since = 32.1, s = 0.12, = 32, and n = 10, we have

Conclusion: Since t₀ = 2.64 > 1.833, we reject H₀ and conclude that the mean weight is more than 32 ounces.

EXERCISE: ROCKWELL HARDNESS

Solution

The parameter of interest is, the mean Rockwell hardness.

H₀: = 60

H₁: > 60

= 0.10

The test statistic is

Reject H₀ if Z₀ > Z_0.10 = 1.28

Computation: Since = 60.2, = 0.12, = 60, and n = 5, we have
Conclusion: since z₀ = 3.73 > 1.28, we reject H₀ and conclude that the mean Rockwell hardness is greater than 60.

EXAMPLE: COUPON OF METAL

Solution

The parameter of interest is, the mean hardness.

H₀: = 50

H₁: > 50

= 0.10

The test statistic is

Reject H₀ if Z₀ > Z_0.10 = 1.28

Computation: Since = 50.3, = 0.08, = 50, and n = 5, we have

Conclusion: Since z₀=8.39 > 1.28, we reject H₀ and conclude that the mean hardness is greater than 50.

EXERCISE: CABLE HARNESS

The diamter of holes for cable harness is known to have a standard deviation of 0.02 in. A random sample of size 10 yields the following data: 1.76, 1.69, 1.74, 1.73, 1.76, 1.77, 1.75, 1.78, 1.75, and 1.76, with sample mean 1.749. Is the true mean diameter less than 1.75? Use = 0.01.

Solution

The parameter of interest is , the mean diameter.

H₀: = 1.75 in

H₁: > 1.75 in

= 0.01

The test statistic is

Reject H₀ if Z₀ < -Z_0.01 = -2.32

Computation: Since 1.749, = 0.02, = 1.75, and n = 10, we have

Conclusion: Since z₀ = -0.16 > -2.57, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to determine that the mean diameter is than 1.75 in.

EXAMPLE: METAL CORD

The diameter of slots for metal cord tackle is known to have a standard deviation of 0.02 in. A random sample of amount 10 renders the subsequent data: 2.49, 2.44, 2.49, 2.48, 2.51, 2.52, 2.50, 2.53, and 2.51, with sample mean 2.497. Is the true mean diameter under 2.50? Use = 0.01.

Solution

The parameter of interest is, the mean diameter.

H₀: = 2.50 in

H₁: < 2.50 in

= 0.01

The test statistic is

Reject H₀ if Z₀ < -Z_0.01 = -2.32

Computation: Since = 2.497, = 0.02, = 2.50, and n = 10, we have
Conclusion: since Z₀ = -0.474 > - 2.57, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to determine that the mean diameter is less than 2.50 in.

EXERCISE: YIELD

Solution

The parameter of interest is, the mean yield.

H₀: = 90%

H₁: = 90%

= 0.10

The test statistic is

Reject H₀ if Z₀ > Z_0.05 = 1.65 or if Z₀ < -Z_0.05 = -1.65

Computation: Since = 90.48%, = 3%, = 90%, and n = 5, we have

Conclusion: Since z₀ = 0.358 <1.65 and z₀ = 0.358 > - 1.65, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to determine that the mean yield is not equal to 90%.

EXAMPLE: PRODUCT

The product of a chemcial procedure is being analyzed. From prior studies wtih this procedure the standard deviation of the product is known to be 2.5. The last 5 days of factory production have generated the subsequent products: 99.7%, 96.75%, 98.80%, 97.90%, and 99.30%. the sample mean is calculated as 98.49%. Can the company conclude that the mean is different from 98%? Use = 0.10.

Solution

The parameter of interest is, the mean product.

H₀: = 98%

H₀: = 98%

= 0.10

The test statistic is

Reject H₀ if Z₀ > Z_0.05 = 1.65 or if Z₀ < -Z_0.05 = -1.65

Computation: Since = 98.49%, = 2.5%, = 98%, and n = 5, we have

Conclusion: Since z₀ = 0.438 < 1.65 and z₀ = 0.438 > - 1.65, we fail to reject H₀ and conclude that the data do not provide sufficient evidence to determine that the mean product is not 98%.

Figure A6: Solutions to Post Problems

EXERCISE: VOLTAGE

A sample of 10 breakdown voltages u_d, of coaxial insulation are recorded and found to have an average of 148 kV and a standard deviation of 2.5 kV. The power company would like to determine if the mean breakdown voltage is 150 kV. use a significance level of 0.05.

Solution

The parameter of interest is , the mean breakdown voltage

H₀: = 150 kV

H₁: = 150 kV

= 0.05

The test statistic is

Reject H₀ if t₀ < t_{0.025, 9} = -2.262 or if t₀ > t_{0.025, 9} = 2.262

Computation: Since = 148, s = 2.5, = 150, and n = 10, we have

Conclusion: Since t₀ = -2.52 < -2.262, we reject H₀ and conclude that the mean breakdown voltage is less than 150 kV, i.e. different from 150 kV.

EXERCISE: VISCOSITY

Solution

The parameter of interest is , the mean viscosity.

H₀: = 10.2 Pa s

H₁: > 10.2 Pa s

= 0.10

The test statistic is
Reject H₀ if Z₀ > Z_0.10 = 1.282

Computation: Since = 10.23, = 0.5, = 10.2, and n = 20, we have

Conclusion: Since z₀ = 0.27 < 1.282, we do not reject H₀ and conclude that the mean viscosity is not more than 10.2 Pa s at a significance level of 0.1, that is, the process does not conform to specifications.