1.) A random sample of 25 college graduates found 18 were proficient with Microsoft Access. Determine a 95% confidence interval for the percent college graduate proficient with Microsoft Access.
Select one:
A. [0.854, 0.934]
B. 0.910, 0.978]
C. 0.544, 0.896]
D. 0.904, 0.984]
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2.)
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students out of total of 4100 university students and finds that 118 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid.
Select one:
A. 0.59 + 0.0572
B. 0.59 + 0.0553
C. 0.59 + 0.0682
D. 0.59 + 0.0659
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3.) A confidence interval was used to estimate the proportion of statistics students that are females. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Which of the following interpretations is correct?
Select one:
A. 90% of the sampled students are female.
B. We are 90% confidence that proportion of all statistics female students falls in the interval 0.438 to 0.642
C. 90% of all statistic students are female.
D. We are 90% confident that the sample proportion of statistics female students falls in interval 0.438 to 0.642.
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4.) The Friday after Thanksgiving is the biggest shopping day of the year. You are interested in the number of people who claim to have finished their Christmas shopping at the end of this weekend. On Monday, you take a random sample of people by standing at a toll booth at 7:00 a.m. and asking every third commuter if he or she has finished Christmas shopping. Of the 67 commuters you poll, 14 claim to have finished Christmas shopping. A 90% confidence interval of the people in the surrounding communities who have finished Christmas shopping is
Select one:
A. [0.1116, 0.3064]
B. [0.1273, 0.2907]
C. [0.0616, 0.2566]
D. None of the above is the correct answer.
In: Statistics and Probability
Sleep – College Students (Raw Data, Software Required): Suppose you perform a study about the hours of sleep that college students get. You know that for all people, the average is about 7.0 hours per night. You randomly select 35 college students and survey them on the number of hours of sleep they get per night. The data is found in the table below. You want to construct a 99% confidence interval for the mean hours of sleep for all college students. You will need software to answer these questions. You should be able to copy the data directly from the table into your software program.
| Student | Sleep |
| 1 | 5.0 |
| 2 | 6.6 |
| 3 | 7.3 |
| 4 | 6.4 |
| 5 | 7.3 |
| 6 | 4.6 |
| 7 | 7.8 |
| 8 | 5.7 |
| 9 | 8.7 |
| 10 | 4.8 |
| 11 | 3.6 |
| 12 | 4.8 |
| 13 | 5.3 |
| 14 | 9.3 |
| 15 | 9.9 |
| 16 | 6.5 |
| 17 | 8.8 |
| 18 | 5.1 |
| 19 | 4.7 |
| 20 | 7.0 |
| 21 | 5.0 |
| 22 | 7.4 |
| 23 | 4.4 |
| 24 | 7.5 |
| 25 | 5.7 |
| 26 | 5.8 |
| 27 | 7.3 |
| 28 | 6.4 |
| 29 | 6.1 |
| 30 | 7.6 |
| 31 | 7.1 |
| 32 | 9.7 |
| 33 | 7.3 |
| 34 | 7.1 |
| 35 | 5.2 |
(a) What is the point estimate for the mean nightly hours of sleep
for all college students? Round your answer to 2 decimal
places.
(b) Construct the 99% confidence interval for the mean nightly
hours of sleep for all college students. Round your answers
to 2 decimal places.
Are you 99% confident that the mean nightly hours of sleep for all college students is below the average for all people of 7.0 hours per night? Why or why not?
Yes, because 7.0 is above the upper limit of the confidence interval for college students.
No, because 7.0 is below the upper limit of the confidence interval for college students.
Yes, because 7.0 is below the upper limit of the confidence interval for college students.
No, because 7.0 is above the upper limit of the confidence interval for college students.
In: Math
Review the following project proposal. Respond with the following clearly labeled
Gotham Textbook Exchange
The problem: Buying and selling books from the bookstore is tedious and expensive. Direct sales from students to students can “cut out the middleman” and result in both higher revenue for the seller and a cheaper price for the buyer. The issue, is there is not a good forum to connect interested sellers to interested buyers.
The solution: Create a web application that enables students to login into the site and indicate books they would like to sell. Sellers must supply the ISBN number (i.e. the unique ID) for each book, the asking price, the condition of the book (poor, fair, good or excellent, and comments about the book’s condition. Potential buyers can log into the system, search for books and check a book to indicate interest. When they check a book, the seller will receive the buyer’s email address so they can communicate. After a sale, the seller can mark the book ‘sold’ and can update the sales price. A seller can also ‘uncheck’ a sold book to be available again should there be an issue with the sale. Book’s listed more than 3 months, should be marked ‘stale’ and require the seller (if still interested in selling the book), to log in and reset the state.
In addition, buyers and sellers can rate each other, enabling users to build up confidence in each other over time. Admins can ban users for inappropriate behavior and list all the book sales in the most recent time period.
In: Computer Science
QUESTION 9
1. A 1,000-turn coil has a cross section of 7.0 cm ^ 2 and a length of 25 cm. Determine how much energy is stored in the coil's magnetic field when it has a current of 10.0 A.
a. 0.18 J
b. 0.36 J
c. 0.10 J
d. 28 J
e. 2.8 J
QUESTION 10
1. Determine which of the following types of waves is intrinsically different from the other four.
a. ultraviolet radiation
b. gamma rays
c. radio waves
d. visible light
e. sound waves
QUESTION 11
1. A circular copper cable is located perpendicular to a uniform magnetic field of 0.50 T. Due to external forces the cable area decreases at a ratio of 1.26 x 10 ^ -3 m ^ 2 / s. Determine the "emf" induced in the circular cable.
a. 7.9 x 10 ^ -3 V
b. 3.1 V
c. 1.2 x 10 ^ -3 V
d. 6.3 x 10 ^ -4 V
e. 3.1 x 10 ^ -4 V
QUESTION 12
1. A cable consists of 240 circular turns, each radius 0.044 m, and the current is 2.2 A. Determine the magnetic moment of the cable.
a. 23 Am ^ 2
b. 0.65 Am ^ 2
c. 15 am ^ 2
d. 0.21 Am ^ 2
e. 3.2 Am ^ 2
In: Physics
c. A protein called Merlin (Nf2) has been identified as a tumor suppressor. An explanation for its role as a tumor suppressor is likely found in Merlin’s influence on vesicular trafficking! Researchers labeled a v-SNARE with green florescent protein to monitor vesicular trafficking in cells lacking merlin protein. Why would monitoring a v-SNARE allow researchers to track vesicular trafficking? Rubric (2): explanation shows understanding of where vSNARES are present in the cell.
Answer: Once a vesicle is formed with its cargo inside, there is a v-snare protruding from the vesicle. By labeling the v-snare, the researchers could monitor the path of the vesicle from when the vesicle is released by coat assembly to when the v-snare and t-snare interact and fuse the vesicle and target membrane. This occurs during vesicular transport from the donor membrane to the target membrane.
e. The researchers found that anterograde vesicular transport was slower in cells with no Merlin protein present. This suggests that Merlin regulates motor protein function. Which type of motor protein might be affected by Merlin? Rubric (2): correct motor protein.
f. Merlin slowed vesicular transport by influencing motor protein function. Hypothesize as to how Merlin might slow the walking of a motor protein. Rubric (2): all plausible answers get full points!
In: Biology
Let X1,X2,...,Xn be a random sample from a uniform distribution
on the interval (0,a). Recall that the maximum likelihood estimator
(MLE) of a is ˆ a = max(Xi).
a) Let Y = max(Xi). Use the fact that Y ≤ y if and only if each Xi
≤ y to derive the cumulative distribution function of Y.
b) Find the probability density function of Y from cdf.
c) Use the obtained pdf to show that MLE for a (ˆ a = max(Xi)) is
biased.
d) Say I would like to consider another estimator for a, I will
call it ˆ b = 2 ¯ X. Is it unbiased estimator of a (show)? How you
can explain someone without calculations why ˆ b = 2 ¯ X is a
reasonable estimator of a?
e) Based on the result in (c), I will propose to use unbiased
estimator for a instead of ˆ a = max(Xi), say ˆ c = n+1 n max(Xi).
Given that the relative efficiency of any two unbiased estimators ˆ
b,ˆ c is the ratio of their variances
V ar(ˆ b) V ar(ˆ c)
,
explain which of these two unbiased estimators is more efficient. You
can obtain the V ar(ˆ c) = V ar(n+1 n max(Xi)) from V ar(ˆ a) =
Var(Y ). The variance of the Y = max(Xi) is
Var(Y ) =
n/( (n + 1)^2(n + 2))*a^2
In: Statistics and Probability
Consider the utility functions of three individuals: u(x) = x1/2, v(x) = ln x, and h(x) = x – 0.01 x2, where x represents wealth.
Consider also the following lotteries: X = (w0 + x1, w0 + x2, w0 + x3; ¼, ½, ¼ ) = (4, 16, 25; ¼, ½, ¼), where w0 = $2, and lottery Y in which w1 = $10, so that Y = (12, 24, 33; ¼, ½, ¼). Note that Y = X + (10; 1)
1. Tell whether u(x), v(x) and h(x) are risk averse, risk neutral or risk lovers individuals.
2. Compare individuals u(x) and v(x) with respect to their degree of risk aversion. Tell who is more risk averse.
3. Calculate the risk premium of individuals u(x) and v(x) with respect to lottery X. Did you obtain that the risk premium of v((x) is larger than that of u(x)? Is that result expected? Why?
4. Compare the risk premium of individual u(x) with respect to lotteries X and Y. Did you obtain that the risk premium with respect to lottery X is larger than that with respect to Y? Is that result expected? Why?
5. Compare the risk premium of individual h(x) with respect to lotteries X and Y. Did you obtain that the risk premium with respect to lottery X is larger than that with respect to Y? Is that result expected? Why?
In: Finance
(Vr = VLRcos(α) – VR and r = Vr/I).
| f (Hz) | R (ohms) | Vsource (V) | Vc (V) | VLr (V) | VLR (V) | VR (V) |
| 0.5 | 10 | 10 | 1.0 | 10 | 10 | 3.5 |
| 1.5 | 10 | 10 | 0.1 | 10 | 10 | 1. |
In: Electrical Engineering
Assume you wish to evaluate the risk and return behaviors
associated with various combinations of assets V and W under three
assumed degrees of correlation: perfect positive, uncorrelated,
and perfect negative. The following average return and risk values
were calculated for these assets:
Asset Average Return, {r}} Risk (Standard
Deviation), s
V 7.6% 4.7%
W 13.2% 9.4%
a. If the returns of assets V and W are perfectly positively
correlated (correlation coefficient equals plus 1 ), describe the
range of (1) return and (2) risk associated with all possible
portfolio combinations.
(1) Range of expected return: between % and % (Round to one
decimal place.)
(2) Range of the risk: between % and % (Round to one decimal
place.)
b. If the returns of assets V and W are uncorrelated (correlation
coefficient equals 0 ), describe the approximate range of (1)
return and (2) risk associated with all possible portfolio
combinations.
(1) Range of expected return: between % and % (Round to one decimal place.)
(2) Range of the risk: between % and % (Round to one decimal
place.)
c. If the returns of assets V and W are perfectly negatively
correlated (correlation coefficient equals negative 1 ), describe
the range of (1) return and (2) risk associated with all possible
portfolio combinations.
(1) Range of expected return: between % and % (Round to one decimal place.)
(2) Range of the risk: between % and % (Round to one decimal place.)
In: Finance
Scenario
The enrolment process starts when a student submits all the required forms and documentation to the online enrolment system. As a first step, the academic qualifications of the student are verified with an external agency, and the student is given an academic score. Then, at the closing date of applications, a selection is made of the students based on their academic merits. Then, the payment is performed followed by a visa application for international students, while in the meantime the advanced standing requests are processed. After all these steps have been completed, the student is registered for the course, and for the chosen units. At any moment students have been asking for updates on the status of their enrolment applications. At any time during the process, a student can withdraw the enrolment application for a full refund, or inquire about the status of the enrolment application. Once the payment has been processed, a visa sponsorship application is made with the immigration department of the Australian government. If there is no reply from the immigration department within 20 days, a manual inquiry is made by the professional staff1 . Once the approval has been received, the student is notified by professional staff and encouraged to lodge their visa application. After 30 days, the student is notified by an automatic system to inform UH of any status changes in the visa processing. Once the student has notified UH of the outcome, UH verifies the granted visa with the immigration department. Unfortunately, in the current situation, it has occurred that already granted visa were cancelled by the immigration department, in which case UH was notified, and provided the student with the option to withdraw the enrolment application with a certificate of acceptance, or to continue with a reduced list of units that could be taken online. If at any stage the student withdraws, UH notifies the immigration department and recommends cancellation of the granted visa. Professional staff assigns students an academic score based on several criteria, partly assisted by input of academic staff members. After the closing date of enrolment applications, the highestranking students are admitted, as many as there are places available. Payments are performed by an automated system that interacts with UH’s standard payment provider. Sometimes, transactions are successfully disputed well after being approved, which voids the payment retroactively, and consequently the application or enrolment. In their initial application, students can include requests for advanced standing of particular units. For each such request, professional staff identifies the unit coordinator and forwards the request to the unit coordinator for assessment. The unit coordinator might get in touch with the student to obtain more information regarding the advanced standing request; if the student does not respond within 10 working days, the decision will be negative. On a positive decision of the unit coordinator, the decision is registered in the student progress information system. Hint: unit coordinators are academic staff.
Model the given scenario (Use Sub Processes). • . • Model the control flow, the data and the resource perspective in BPMN; • Make sure that it adheres as much as possible to the as-is description in Scenario. However, as with any real-life modelling effort, not every aspect of the process is described, and not all described behaviour can be modelled. • Make sure your model is structurally correct; • In the report, the model should be readable on A4 paper without zooming in;
In: Accounting