10. Consider the simplified scenario for genetic determinants of height in humans, where there are three genes (A/B/C) with varying numbers of alleles (3/3/2) affecting height, and with different effects in males and females. (Assume additive contributions, thus the effect of having genotype A1A2 is +0.1” + +0.2” = +0.3”). Average height for men and women is 69” and 64”.
|
Gene |
Allele |
Male(effect |
Female(effect |
|
A |
1 |
+0.1 |
+0.1 |
|
2 |
+0.2 |
+0.2 |
|
|
3 |
41.3 |
41.5 |
|
|
B |
1 |
+0.5 |
+.3 |
|
2 |
40.4 |
40.2 |
|
|
3 |
40.1 |
40.1 |
|
|
C |
1 |
42 |
+1.5 |
|
2 |
+0.1 |
0 |
|
What is the expected height for a male with genotype A2A2B1B3C1C2 (3 points)?
11. Consider a cross between two heterozygotes.
What is the probability that their first offspring has recessive
phenotype? (2 points)
What is the probability that their first offspring has recessive phenotype and the second offspring also has the recessive phenotype? (2 points)
What is the probability that out of their first offspring, one has dominant phenotype and one has recessive phenotype? (2 points)
15. The gene for petal color in a flower has incomplete dominance, so that individuals with two A1 alleles (A1A1) are black, individuals with two A2 alleles (A2A2) are white, and individuals with one of each allele (A1A2) are mottled.
In a cross between two black flowers, what is the probability of getting a mottled offspring? (2 points)
3" 4"
5" 6" 7" 8" 9" 10" 11" 12"
1" 2"
In a cross between a black flower and a mottled flower, what is the probability of getting a black offspring? (2 points)
In a cross between two mottled flowers, if there are two offspring, what is the probability of getting one black offspring and one mottled offspring? (2 points)
In a cross between two mottled flowers, if there are two offspring, what is the probability of getting one white offspring and one mottled offspring? (2 points)
In a cross between two mottled flowers, if there are nine offspring, what is the probability of getting exactly three mottled offspring? (2 points)
16. Two individuals that are heterozygous for a recessive autosomal trait have an offspring with dominant phenotype. What is the probability that that offspring is a carrier (heterozygote?) (3 points)
If that offspring has an offspring with an individual with the recessive condition, what is the chance their offspring has the condition? (2 points)
17. Most randomly occurring mutations that occur in humans do not have an effect on phenotype. Why is this? (4 points)
18. Imagine that coronavirus has a 0.002% incidence in the population. A test for the virus has a 0.001% false positive rate and no false negative rate (false positive rate means the chance that if an uninfected individual takes the test the test will falsely identify them as infected). If a random person takes the test and gets a positive result, what is the chance that they are infected? (Show your work to earn partial credit) (4 points)
Now consider the case in the future, where the incidence of the virus has increased to 1%. Now if a random person takes the test and gets a positive result, what is the chance that they are infected? (2 points)
19. Your colleague is studying long toes in the California vole (Microtus californicus). She proposes that this trait is due to to an X-linked dominant allele.
You go for a hike in Oakland, and notice that very few of the California voles you see have the long toe trait. Does this affect your colleague’s hypothesis? How? Why? (2 points)
You go for a walk in Golden Gate Park, and notice that, among California voles in the Golden Gate population, females are much less likely than males to have the long toe trait. Does this affect your colleague’s hypothesis? How? Why? (2 points)
20. You are a genetic counseler. A mother and father with a son and a daughter come to see you. The mother and the father both have a very rare condition that no one has ever studied, but neither their son or their daugther does. Karyotype analysis shows that the mother and the daugther are XX and the father and the son are XY. You think about it and realize that this pattern cannot be due to a number of simple inheritance patterns. Explain why:
Why can’t it be an autosomal dominant condition? (2 points) Why
can’t it be Y-linked condition? (2 points)
Why can’t it be an X-linked dominant condition? (2 points) Why
can’t it be an X-linked recessive condition? (2 points) Why can’t
it be a mitochondrial condition? (2 points)
In: Biology
A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 31 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was 53.6. Another random sample of 23 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was 33.7. Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a 5% level of significance. How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems?
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ12 = σ22; H1: σ12 > σ22
Ho: σ12 > σ22; H1: σ12 = σ22
Ho: σ22 = σ12; H1: σ22 > σ12
Ho: σ12 = σ22; H1: σ12 ≠ σ22
(b) Find the value of the sample F statistic. (Round your
answer to two decimal places.)
What are the degrees of freedom?
| dfN | |
| dfD |
What assumptions are you making about the original
distribution?
The populations follow independent normal distributions.
The populations follow independent chi-square distributions. We have random samples from each population.
The populations follow dependent normal distributions. We have random samples from each population.
The populations follow independent normal distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.200
0.100 < P-value < 0.200
0.050 < P-value < 0.100
0.020 < P-value < 0.050
0.002 < P-value < 0.020
P-value < 0.002
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the
application.
Fail to reject the null hypothesis, there is sufficient evidence that the variance in consumption of gasoline is greater in the new fuel injection systems.
Reject the null hypothesis, there is insufficient evidence that the variance in consumption of gasoline is greater in the new fuel injection systems.
Reject the null hypothesis, there is sufficient evidence that the variance in consumption of gasoline is different in both fuel injection systems.
Fail to reject the null hypothesis, there is insufficient evidence that the variance in consumption of gasoline is different in both fuel injection systems.
In: Statistics and Probability
A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 41 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was 54.8. Another random sample of 27 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was 36.4. Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a 5% level of significance.
How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems? (a) What is the level of significance? 0.05
State the null and alternate hypotheses. Ho: σ12 = σ22; H1: σ12 > σ22 Ho: σ12 > σ22; H1: σ12 = σ22 Ho: σ22 = σ12; H1: σ22 > σ12 Ho: σ12 = σ22; H1: σ12 ≠ σ22
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom? dfN dfD
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.
The populations follow dependent normal distributions. We have random samples from each population.
The populations follow independent normal distributions.
The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic.
P-value > 0.200 0.100 < P-value < 0.200 0.050 < P-value < 0.100 0.020 < P-value < 0.050 0.002 < P-value < 0.020 P-value < 0.002
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Fail to reject the null hypothesis, there is sufficient evidence that the variance in consumption of gasoline is greater in the new fuel injection systems.
Reject the null hypothesis, there is insufficient evidence that the variance in consumption of gasoline is greater in the new fuel injection systems.
Reject the null hypothesis, there is sufficient evidence that the variance in consumption of gasoline is different in both fuel injection systems.
Fail to reject the null hypothesis, there is insufficient evidence that the variance in consumption of gasoline is different in both fuel injection systems.
In: Statistics and Probability
A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 41 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was 53. Another random sample of 27 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was 34.6. Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a 5% level of significance. How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems?
(a) What is the level of significance?
State the null and alternate hypotheses.
Ho: σ12 = σ22; H1: σ12 > σ22
Ho: σ12 > σ22; H1: σ12 = σ22
Ho: σ22 = σ12; H1: σ22 > σ12
Ho: σ12 = σ22; H1: σ12 ≠ σ22
(b) Find the value of the sample F statistic. (Round your
answer to two decimal places.)
What are the degrees of freedom?
| dfN | |
| dfD |
What assumptions are you making about the original
distribution?
The populations follow dependent normal distributions. We have random samples from each population.
The populations follow independent normal distributions.
The populations follow independent chi-square distributions. We have random samples from each population.
The populations follow independent normal distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test
statistic.
P-value > 0.200
0.100 < P-value < 0.200
0.050 < P-value < 0.100
0.020 < P-value < 0.050
0.002 < P-value < 0.020
P-value < 0.002
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the
application.
Fail to reject the null hypothesis, there is sufficient evidence that the variance in consumption of gasoline is greater in the new fuel injection systems.
Reject the null hypothesis, there is insufficient evidence that the variance in consumption of gasoline is greater in the new fuel injection systems.
Reject the null hypothesis, there is sufficient evidence that the variance in consumption of gasoline is different in both fuel injection systems.
Fail to reject the null hypothesis, there is insufficient evidence that the variance in consumption of gasoline is different in both fuel injection systems.
In: Statistics and Probability
*** It flagged the word support so job sub = job support
Gallatin Carpet Cleaning is a small, family-owned business operating out of Bozeman, Montana. For its services, the company has always charged a flat fee per hundred square feet of carpet cleaned. The current fee is $22.60 per hundred square feet. However, there is some question about whether the company is actually making any money on jobs for some customers—particularly those located on remote ranches that require considerable travel time. The owner’s daughter, home for the summer from college, has suggested investigating this question using activity-based costing. After some discussion, she designed a simple system consisting of four activity cost pools. The activity cost pools and their activity measures appear below:
| Activity Cost Pool | Activity Measure | Activity for the Year | |
| Cleaning carpets | Square feet cleaned (00s) | 12,500 | hundred square feet |
| Travel to jobs | Miles driven | 315,500 | miles |
| Job sup. | Number of jobs | 1,700 | jobs |
| Other (organization-sustaining costs and idle capacity costs) | None | Not applicable | |
The total cost of operating the company for the year is $379,000 which includes the following costs:
| Wages | $ | 150,000 |
| Cleaning supplies | 35,000 | |
| Cleaning equipment depreciation | 19,000 | |
| Vehicle expenses | 36,000 | |
| Office expenses | 61,000 | |
| President’s compensation | 78,000 | |
| Total cost | $ | 379,000 |
Resource consumption is distributed across the activities as follows:
| Distribution of Resource Consumption Across Activities | ||||||||||
| Cleaning Carpets | Travel to Jobs | Job Sup | Other | Total | ||||||
| Wages | 77 | % | 15 | % | 0 | % | 8 | % | 100 | % |
| Cleaning supplies | 100 | % | 0 | % | 0 | % | 0 | % | 100 | % |
| Cleaning equipment depreciation | 69 | % | 0 | % | 0 | % | 31 | % | 100 | % |
| Vehicle expenses | 0 | % | 84 | % | 0 | % | 16 | % | 100 | % |
| Office expenses | 0 | % | 0 | % | 57 | % | 43 | % | 100 | % |
| President’s compensation | 0 | % | 0 | % | 33 | % | 67 | % | 100 | % |
Job sup consists of receiving calls from potential customers at the home office, scheduling jobs, billing, resolving issues, and so on.
Required:
1.Prepare the first-stage allocation of costs to the activity cost pools.
|
2. Compute the activity rates for the activity cost pools. (Round your answers to 2 decimal places.)
|
3. The company recently completed a 200 square foot carpet-cleaning job at the Flying N Ranch—a 52-mile round-trip journey from the company’s offices in Bozeman. Compute the cost of this job using the activity-based costing system.
Cost of job =
4. The revenue from the Flying Ranch was $45.20 (200 square feet @ $22.60 per hundred square feet). Calculate the customer margin earned on this job.
Customer Margin =
In: Accounting
Cardiorespiratory fitness is widely recognized as a major component of overall physical well-being. Direct measurement of maximal oxygen uptake (VO2max) is the single best measure of such fitness, but direct measurement is time-consuming and expensive. It is therefore desirable to have a prediction equation for VO2max in terms of easily obtained quantities. A sample is taken and variables measured are age (years), time necessary to walk 1 mile (mins), and heart rate at the end of the walk (bpm) in addition to the VO2 max uptake. The equation from a multiple regression is (V02) = 0.017*(age) - 0.028*(HR) + 0.017*(time) + 3.483. If a person is 29 years old, blood pressure of 130 bpm, and a walking time of 13 minutes, what is his/her expected maximum oxygen uptake?
part A options:
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Suppose that a researcher studying the weight of female college athletes wants to predict the weights based on height, measured in inches, and the percentage of body fat of an athlete. The researcher calculates the regression equation as (weight) = 3.979*(height) + 0.85*(percent body fat) - 87.814. If a female athlete is 67 inches tall and has a 20 percentage of body fat, what is her expected weigh
Part B options:
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A trucking company considered a multiple regression model for relating the dependent variable of total daily travel time for one of its drivers (hours) to the predictors distance traveled (miles) and the number of deliveries of made. After taking a random sample, a multiple regression was performed and the equation is (time) = 0.076*(distance) + 0.579*(deliveries) - 0.466. Suppose for a given driver's day, he is scheduled to drive 58.908 miles and make 10.97 stops. Suppose it took him 14.302 hours to complete the trip. What is the residual based on the regression model?
Part C options:
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Suppose that a researcher studying the weight of female college athletes wants to predict the weights based on height, measured in inches, the percentage of body fat of an athlete, and age. The researcher calculates the regression equation as (weight) = 4.08*(height) + 1.103*(percent body fat) - 0.995*(age) - 82.074. If a female athlete is 68.575 inches tall, has a 23.901 percentage of body fat, is 22.696 years old, and has a weight of 167.763, the residual is -33.7293. Choose the correct interpretation of the residual.
Part D options:
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In: Statistics and Probability
In both your professional and personal life, you will make a variety of decisions. You should consider the financial and nonfinancial aspects while making decisions. To evaluate scenarios, you will also use some tools such as net present value (NPV), internal rate of return (IRR), modified internal rate of return (MIRR), payback, discounted payback, and return on investment (ROI) that you have discussed so far.
In this assignment, you will evaluate the financial aspects of making decisions.
Tasks:
Investigate any two of the following four financial decisions:
1) Using net present value calculations, determine which has a higher ROI. Assume the average mileage under both options is 15,000 miles. The car will be sold for its Kelly Blue Book value at the end of ownership or it will be returned to the leasing dealership for no additional lease/return/mileage cost. The automobile being considered is a 4-cyl, 2.5 liter, two-wheel drive, Nissan Rogue sport utility. At the end of the 6 years, the automobile is in very good condition. Tax, title, or license fees are not considered under either option.
Buying a Nissan Rogue today for $32,000, putting $10,000 down and taking a six-year loan for the rest at 4%
or
Leasing the Rogue for 6 years at $360 a month with a down payment of $3,500 due at delivery. The car must be returned at the end of the lease. 15,000 miles per year are allowed under this lease plan.
Show your work and explain your rationale.
2) Commit to buy a vacation home in the climate of your choice,
rent the home out when you are not using it, or sign a five-year
lease for the home for the two months a year you plan on using it.
You will need to make up the numbers for your home for this
exercise.
3) A different buy-or-lease alternative—you could buy a home for
$300,000, putting 20% down and renting it out at $1,700 a month.
Which would make more financial sense? Buying or leasing the home?
Explain your rationale.
4)Lease your home for the next three years or sell it with the intent to return to the same geographic area after you complete a three-year expatriate assignment in the country of your choice. Given the facts above, should you lease the house or sell it? The current market value is approximately $320,000. Explain your rationale, and show your work.
For the two options that you selected (and using the figures given above for those options), investigate the realistic assumptions for your location and include the information you found in the analysis. Create a paper in about 1,000–1,200 words, including the following:
Initial information/approach: purchase price, rebate, down
payment, amount to finance, etc.
Payments formulas and calculations
Explanation of the financial factors that you are employing in the
selected decisions
Conclusion containing the "best answer" for your personal life on
the basis of these financial factors
Probability of following the recommended "best answer" (assuming
that these are the only decision options)
In: Finance
One for the Road—Anyone?
“Florence Yozefu is a brilliant scientist who heads a robotics
research laboratory at one of the top
ten research universities. Florence has been developing wearable
robotics gear that can take over
the driving functions of a vehicle from a human operator when it is
worn by the driver. In
laboratory tests, the robot, nicknamed Catchmenot, has performed
successfully whenever Florence
and her assistants have worn the robot. However, no real-life
experiment has ever been conducted
outside the lab. Florence has been planning to try it out in her
project plan but has not yet had a
chance to do so. For New Year’s Eve, Florence has plans to visit
her mother and sister, about 100
miles away. This was a good opportunity to show her mother and her
sister what she has been
working on in the last few months. So, she decides to take
Catchmenot with her. She packs her car
the evening before and on the morning of the trip, she passes by
the lab to get her robot and put it
in the car. She drives the 100 miles in a little under her usual
time and arrives at her mother’s house
earlier than usual”.
“In the evening, Florence bids her mother good-bye and passes by
her sister’s apartment as
promised. At her sister’s apartment, she finds a few of her teen
friends and they get right into a
party mode. Florence drinks and dances and forgets about the time.
There are many stories to tell
and to listen to. About 1:00 a.m., after the midnight champagne
toast, she decides to leave and
drive back to her apartment. She had promised to accompany her
friends to a pre-planned
engagement. Although she is very drunk, and against her friend’s
advice and insistence that she
should not drive, Florence puts on Catchmenot and in a few minutes
she is off. Thirty minutes later,
she is cruising at 70 mph and she is also sound asleep. She is
awakened by a squirrel running all
over her car at about 5:00 a.m. She is parked by the roadside in
front of her apartment complex.
She has made it home safely. She has no idea when and where she
passed out and what happened
along the way. She will never know. Although she is surprised,
confused, and feels guilty, she is
happy how well Catchmenot has worked. She decides to market it. How
much should she charge
for it, she wonders”.
[Source: Kizza J.M. History of Computing. In: Ethical and Social
Issues in the Information Age, 2010,
Texts in Computer Science. Springer, London]
Please answer the following questions:
1. As AI applications increase, such as in the use of robotics,
will the wider use of these
“manlike” machines compromise our moral values system? Why or why
not? [2 Marks]
2. Discus the future of computer ethics in the integrated
environment of Artificial Intelligence
(A I), Virtual Reality (VR), and cyberspace. [3 Marks]
3. If anything went wrong during the ride home, would Florence be
responsible? Who should
be? What are the consequences? [3 Marks]
4. Discuss the ethical implications of Artificial Intelligence. [2
Marks]\
In: Computer Science
Gallatin Carpet Cleaning is a small, family-owned business operating out of Bozeman, Montana. For its services, the company has always charged a flat fee per hundred square feet of carpet cleaned. The current fee is $22.40 per hundred square feet. However, there is some question about whether the company is actually making any money on jobs for some customers—particularly those located on remote ranches that require considerable travel time. The owner’s daughter, home for the summer from college, has suggested investigating this question using activity-based costing. After some discussion, she designed a simple system consisting of four activity cost pools. The activity cost pools and their activity measures appear below:
| Activity Cost Pool | Activity Measure | Activity for the Year | |
| Cleaning carpets | Square feet cleaned (00s) | 13,000 | hundred square feet |
| Travel to jobs | Miles driven | 218,000 | miles |
| Job support | Number of jobs | 1,600 | jobs |
| Other (organization-sustaining costs and idle capacity costs) | None | Not applicable | |
The total cost of operating the company for the year is $354,000 which includes the following costs:
| Wages | $ | 148,000 |
| Cleaning supplies | 21,000 | |
| Cleaning equipment depreciation | 15,000 | |
| Vehicle expenses | 33,000 | |
| Office expenses | 64,000 | |
| President’s compensation | 73,000 | |
| Total cost | $ | 354,000 |
Resource consumption is distributed across the activities as follows:
| Distribution of Resource Consumption Across Activities | ||||||||||
| Cleaning Carpets | Travel to Jobs | Job Support | Other | Total | ||||||
| Wages | 74 | % | 16 | % | 0 | % | 10 | % | 100 | % |
| Cleaning supplies | 100 | % | 0 | % | 0 | % | 0 | % | 100 | % |
| Cleaning equipment depreciation | 72 | % | 0 | % | 0 | % | 28 | % | 100 | % |
| Vehicle expenses | 0 | % | 80 | % | 0 | % | 20 | % | 100 | % |
| Office expenses | 0 | % | 0 | % | 64 | % | 36 | % | 100 | % |
| President’s compensation | 0 | % | 0 | % | 35 | % | 65 | % | 100 | % |
Job support consists of receiving calls from potential customers at the home office, scheduling jobs, billing, resolving issues, and so on.
Required:
1. Prepare the first-stage allocation of costs to the activity cost pools.
|
2. Compute the activity rates for the activity cost pools.
|
3. The company recently completed a 200 square foot carpet-cleaning job at the Flying N Ranch—a 51-mile round-trip journey from the company’s offices in Bozeman. Compute the cost of this job using the activity-based costing system.
4. The revenue from the Flying N Ranch was $44.80 (200 square feet @ $22.40 per hundred square feet). Calculate the customer margin earned on this job.
In: Accounting
As part of a study designed to compare hybrid and similarly equipped conventional vehicles, Consumer Reports tested a variety of classes of hybrid and all-gas model cars and sport utility vehicles (SUVs). The following data show the miles-per-gallon rating Consumer Reports obtained for two hybrid small cars, two hybrid midsize cars, two hybrid small SUVs, and two hybrid midsize SUVs; also shown are the miles per gallon obtained for eight similarly equipped conventional models.
| Make/Model | Class | Type | MPG |
|---|---|---|---|
| Honda Civic | Small Car | Hybrid | 37 |
| Honda Civic | Small Car | Conventional | 28 |
| Toyota Prius | Small Car | Hybrid | 44 |
| Toyota Corolla | Small Car | Conventional | 32 |
| Chevrolet Malibu | Midsize Car | Hybrid | 27 |
| Chevrolet Malibu | Midsize Car | Conventional | 23 |
| Nissan Altima | Midsize Car | Hybrid | 32 |
| Nissan Altima | Midsize Car | Conventional | 25 |
| Ford Escape | Small SUV | Hybrid | 27 |
| Ford Escape | Small SUV | Conventional | 21 |
| Saturn Vue | Small SUV | Hybrid | 28 |
| Saturn Vue | Small SUV | Conventional | 22 |
| Lexus RX | Midsize SUV | Hybrid | 23 |
| Lexus RX | Midsize SUV | Conventional | 19 |
| Toyota Highlander | Midsize SUV | Hybrid | 24 |
| Toyota Highlander | Midsize SUV | Conventional | 18 |
At the α = 0.05 level of significance, test for significant effects due to class, type, and interaction.
Find the value of the test statistic for class. (Round your answer to two decimal places.)
Find the p-value for class. (Round your answer to three decimal places.)
State your conclusion about class.
A) Because the p-value > α = 0.05, class is not significant.
B) Because the p-value ≤ α = 0.05, class is significant.
C) Because the p-value ≤ α = 0.05, class is not significant.
D) Because the p-value > α = 0.05, class is significant.
Find the value of the test statistic for type. (Round your answer to two decimal places.)
Find the p-value for type. (Round your answer to three decimal places.)
State your conclusion about type.
A) Because the p-value ≤ α = 0.05, type is not significant.
B) Because the p-value > α = 0.05, type is significant.
C) Because the p-value > α = 0.05, type is not significant.
D) Because the p-value ≤ α = 0.05, type is significant.
Find the value of the test statistic for interaction between class and type. (Round your answer to two decimal places.)
Find the p-value for interaction between class and type. (Round your answer to three decimal places.)
State your conclusion about interaction between class and type.
A) Because the p-value > α = 0.05, interaction between class and type is significant.
B) Because the p-value ≤ α = 0.05, interaction between class and type is significant.
C) Because the p-value ≤ α = 0.05, interaction between class and type is not significant.
D) Because the p-value > α = 0.05, interaction between class and type is not significant.
In: Statistics and Probability