Q.Explain the flaws in the following analysis or conclusion. (Note that this is not about chi-squared test per se, perhaps, it is about what a statistical analysis (hypothesis testing in this case) can tell us and what it cannot) Background - The Scholarship Committee comprises 3 faculty members, one of whom is Professor X. Professor X also wrote recommendation letter for 5 students who took his class earlier and four of them were awarded scholarships (out of a total of six scholarships awarded to graduate students). A student who applied but was not awarded a scholarship accused Professor X of favoritism, claiming that he/she was denied a scholarship despite having a very high GPA because he/she did not take class under Professor X and had declined him/her a recommendation letter. The student then did following statistical analysis as an evidence of favoritism . The student runs a Chisq-test and his result shows not independent between students who took class under Professor X and Students got scholarship. H0: students who get scholarship is independent with whether they have taken class under Professor X H1: not independent # Let total number of students who applied for scholarship is 70, and only 5 people get the scholarship, conservatively estimate only 3 out of 5 students took Class with ProfessorX, the other two selected did not take Class with ProfessorX # chisquare test shows the p-value <0.05, thus we reject H0 at .05 level and conclude students who got scholarship was affected by whether they have class under Professor X #student in addition tests , how about the total applicants were 60, or 80? and find out either case we reject H0, they are not independent.

(a) Find the Riemann sum for

f(x) = 4 sin(x), 0 ≤ x ≤ 3π/2,

with six terms, taking the sample points to be right endpoints. (Round your answers to six decimal places.)
R₆ =

(b) Repeat part (a) with midpoints as the sample points.
M₆ =

If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval [a, b], then

m(b − a) ≤

≤ M(b − a).

Use this property to estimate the value of the integral.

Q.Explain the flaws in the following analysis or conclusion. (Note that this is not about chi-squared test per se, perhaps, it is about what a statistical analysis (hypothesis testing in this case) can tell us and what it cannot) Background - The Scholarship Committee comprises 3 faculty members, one of whom is Professor X. Professor X also wrote recommendation letter for 5 students who took his class earlier and four of them were awarded scholarships (out of a total of six scholarships awarded to graduate students). A student who applied but was not awarded a scholarship accused Professor X of favoritism, claiming that he/she was denied a scholarship despite having a very high GPA because he/she did not take class under Professor X and had declined him/her a recommendation letter. The student then did following statistical analysis as an evidence of favoritism . The student runs a Chisq-test and his result shows not independent between students who took class under Professor X and Students got scholarship. H0: students who get scholarship is independent with whether they have taken class under Professor X H1: not independent # Let total number of students who applied for scholarship is 70, and only 5 people get the scholarship, conservatively estimate only 3 out of 5 students took Class with ProfessorX, the other two selected did not take Class with ProfessorX # chisquare test shows the p-value <0.05, thus we reject H0 at .05 level and conclude students who got scholarship was affected by whether they have class under Professor X #student in addition tests , how about the total applicants were 60, or 80? and find out either case we reject H0, they are not independent.

Explain why a discrete time system is stable if its eigenvalues are all in the unit circle.

Explain why a symmetric matrix will always have real eigenvalues.

Explain why a continuous time system is stable if the real part of its eigenvalues are negative.

1. Reflect on the concept of function. What concepts (only the names) did you need to accommodate the concept of function in your mind?

2. What is the simplest function you can imagine?

3. In your day to day, is there any occurring fact that can be interpreted as a function?

4. Is it possible to view a function?

5. What strategy are you using to get the graph of a function?

Kindly keep it detailed and informative but do not copy and paste from other sources!

Import the text into Excel. The column delimiter is the minus sign, ‘-‘. Make a table of the total population of each city in each state in excel and post all steps used.

State Code-County Code-City Code-Population
2-117-13-49174
4-134-21-72247
3-137-27-91811
8-140-13-13913
4-133-16-11028
1-119-12-37710
10-101-31-76485
8-138-16-82956
6-125-26-67320
2-108-19-27973
6-119-22-25359
8-122-17-16881
4-140-18-63178
2-111-13-53192
2-112-30-31156
1-127-19-37547
9-135-23-48125
3-108-19-61645
6-113-26-16166
5-112-31-17847
7-135-23-48441
5-124-15-20688
6-106-15-95046
3-105-16-35144
6-120-29-97858
9-117-26-19441
6-120-18-71854
2-114-24-80466
3-138-15-73195
3-104-28-62538
1-133-24-74205
2-126-11-56200
3-129-15-41400
8-121-21-13075
10-138-19-57237
8-112-21-99683
2-114-13-70453
4-113-28-33874
9-112-23-23172
6-105-18-74177
7-140-21-11851
1-115-28-28103
5-134-20-48211
6-114-20-51120
3-111-24-55273
7-119-28-19041
8-122-29-19288
3-134-11-94750
9-119-26-62117
6-138-21-66254
4-117-29-21650
2-135-28-31025
5-135-24-30519
10-118-27-22788
9-102-19-92033
1-138-22-49667
2-101-24-81847
3-122-20-13381
3-124-18-34141
6-140-17-52557
6-131-28-23671
2-126-12-15206
7-124-23-28605
10-101-23-85366
10-106-19-57824
4-132-21-53104
3-119-29-51525
4-107-30-53327
5-108-17-97390
5-128-26-24022
3-117-19-37134
7-135-17-64097
3-140-31-93398
10-129-19-30970
2-112-12-85156
6-136-29-52765
5-121-21-73339
4-113-27-54497
8-113-22-94934
8-138-12-51563

Consider an unweighted, undirected graph G = <V, E>. The neighbourhood of a node v ∈ V in the graph is the set of all nodes that are adjacent (or directly connected) to v. Subsequently, we can define the neighbourhood degree of the node v as the sum of the degrees of all its neighbours (those nodes that are directly connects to v).

(a) Design an algorithm that returns a list containing the neighbourhood degree for each node v ∈ V, assuming that the input is the graph, G, represented using an adjacency list. Each item i in the list that you generate will correspond to the correct value for the neighbourhood degree of node vi. Your algorithm should be presented in unambiguous pseudocode. Your algorithm should have a time complexity value O(V +E).

(b) If an adjacency matrix was used to represent the graph instead of an adjacency list, what is the new value for the time complexity? Justify your answer by explicitly referring to the changes that would be necessary to your algorithm from part (a).

Prove that if a ≡ b(mod d) then a, b have the same remainder when divided by d

Use Euler's method to obtain a four-decimal approximation of the indicated value. Carry out the recursion of (3) in Section 2.6

y_{n + 1} = y_n + hf(x_n, y_n)        (3)

by hand, first using

h = 0.1

and then using

h = 0.05.

y' = 2x − 3y + 1, y(1) = 8;   y(1.2)

Euclidean metric) The metric

d

((

x

1

,x

2

)

,

(

y

1

,y

2

)) =

√

(

x

1

−

x

2

)

2

+ (

y

1

−

y

2

)

2

on

R

2

generates the standard (product) topology on

R

2

.

In an article in the Journal of Marketing, Bayus studied the differences between "early replacement buyers” and "late replacement buyers” in making consumer durable good replacement purchases. Early replacement buyers are consumers who replace a product during the early part of its lifetime, while late replacement buyers make replacement purchases late in the product’s lifetime. In particular, Bayus studied automobile replacement purchases. Consumers who traded in cars with ages of zero to three years and mileages of no more than 35,000 miles were classified as early replacement buyers. Consumers who traded in cars with ages of seven or more years and mileages of more than 73,000 miles were classified as late replacement buyers. Bayus compared the two groups of buyers with respect to demographic variables such as income, education, age, and so forth. He also compared the two groups with respect to the amount of search activity in the replacement purchase process. Variables compared included the number of dealers visited, the time spent gathering information, and the time spent visiting dealers.

(a) Suppose that a random sample of 807 early replacement buyers yields a mean number of dealers visited of x⎯⎯x¯ = 3.3, and assume that σ equals .79. Calculate a 99 percent confidence interval for the population mean number of dealers visited by early replacement buyers. (Round your answers to 3 decimal places.)

The 99 percent confidence interval is            [ , ].

(b) Suppose that a random sample of 493 late replacement buyers yields a mean number of dealers visited of x⎯⎯x¯ = 4.2, and assume that σ equals .66. Calculate a 99 percent confidence interval for the population mean number of dealers visited by late replacement buyers. (Round your answers to 3 decimal places.)

The 99 percent confidence interval is            [ , ].

(c) Use the confidence intervals you computed in parts a and b to compare the mean number of dealers visited by early replacement buyers with the mean number of dealers visited by late replacement buyers. How do the means compare?

Mean number of dealers visited by late replacement buyers appears to be ( lower or higher?)

∀, ∃ what is their meanings and the differences between them. This may take a paragraph or two of some writing, which should at minimum begin with a definition of each in your words.

Consider monthly demand for the ABC Corporation as shown below. Forecast the monthly demand for Year 6 using the 3-period moving average and 4-period moving average. Evaluate the bias, TS, MAD, MAPE and MSE. Evaluate the quality of the forecast

24. Show that (x ^p) − x has p distinct zeros in Zp, for any prime p. Conclude that (x ^p) − x = x(x − 1)(x − 2)· · ·(x − (p − 1)).

(this is not as simple as showing that each element in Z_p is a root -- after all, we've

seen that in Z₆[x], the polynomial x^2-5x has 4 roots, 0, 5, 2, and 3, but x^2-5 is not equal to (x-0)(x-5)(x-2)(x-3))

Project Details

John and Jane Doe are newlyweds with executive track careers at ACME Gadget Company. In five years, the Does would like to have a family, envisioning two young children, Jack and Jill. With an eye for the future, John and Jane are now looking to ensure that their future family has a place to call home, that their future children will have access to all the education they desire, and that they themselves will be able to enjoy retirement when the time comes. As such, they’ve come to your financial planning company for advice for purchasing a house, planning for retirement, setting up a RESP and for your perspective on a side venture. They’ve provided you with the background and questions below.

A side-venture

Jane is an inventor, working for the ACME Gadget Company in research and development. She recently proposed the development of an advanced technology, but it was deemed too risky for R&D at ACME. However, ACME has agreed that if Jane successfully develops the technology on her own, ACME will acquire a license to use the technology for a period of 10 years. To develop the technology will require an initial expenditure of $150,000 and an additional expenditure of $150,000 at the end of each of the next 2 years. When the patent is approved in Year 4, it is expected to be licensed to the ACME Gadget Company for an upfront fee of 100,000 plus an additional fee of $90,000/year for 10 years. At that time the product that uses the technology will be replaced by a new model. What is the rate of return on the Jane’s advanced technology?

Please include calculations and diagrams.