Let S be a set and P be a property of the elements of the set, such that each element either has property P or not. For example, maybe S is the set of your classmates, and P is "likes Japanese food." Then if s ∈ S is a classmate, he/she either likes Japanese food (so s has property P) or does not (so s does not have property P). Suppose Pr(s has property P) = p for a uniformly chosen s ∈ S. Suppose Ofer the Oracle has magic powers that allow him to check property P for any s, but it doesn’t always work since each time he relies on some (independently) generated random numbers to help him. If he says “Yes,” then s has the property. If s has the property, then he says “Yes” with probability at least q.

Suppose we let Ofer check property P on an element s a total of N times, and he responded “No” each time. Find a lower bound (i.e. a smaller number, but a useful one) for Pr(s does not have property P). Suppose p = 99/100 and q = 1/2, how many times should you let Ofer check probability P before you are 99% confident that s does not have property P?

Hypothesis 1: Job stress is associated with absenteeism rates at work.

1. What type of relationship would you predict (positive linear, negative linear, and so on)?

2. Graph and label the relationship you predict. State a prediction based on the relationship

shown by your graph.

3. Which method (experimental or nonexperimental) would you use to test this hypothesis?

Why?

4. Which is the independent or “cause” variable and which is the dependent or “effect”

variable?

5. State an operational definition for each variable. If experimental, describe how the

independent variable might be manipulated and the dependent variable measured. If

nonexperimental, describe how both variables might be measured.

Hypothesis 2: A pregnant woman’s diet affects the birth weight of her baby.

1. What type of relationship would you predict (positive linear, negative linear, and so on)?

2. Graph and label the relationship you predict. State a prediction based on the relationship

shown by your graph.

3. Which method (experimental or nonexperimental) would you use to test this hypothesis?

Why?

4. Which is the independent or “cause” variable and which is the dependent or “effect”

variable?

5. State an operational definition for each variable. If experimental, describe how the

independent variable might be manipulated and the dependent variable measured. If

nonexperimental, describe how both variables might be measured.

Hypothesis 3: The size of a meeting is related to the length of the meeting.

1. What type of relationship would you predict (positive linear, negative linear, and so on)?

2. Graph and label the relationship you predict. State a prediction based on the relationship

shown by your graph.

3. Which method (experimental or nonexperimental) would you use to test this hypothesis?

Why?

4. Which is the independent or “cause” variable and which is the dependent or “effect”

variable?

5. State an operational definition for each variable. If experimental, describe how the

independent variable might be manipulated and the dependent variable measured. If

nonexperimental, describe how both variables might be measured.

Hypothesis 4: Frequency of child abuse is related to the parents’ ages when they married.

1. What type of relationship would you predict (positive linear, negative linear, and so on)?

2. Graph and label the relationship you predict. State a prediction based on the relationship

shown by your graph.

3. Which method (experimental or nonexperimental) would you use to test this hypothesis?

Why?

4. Which is the independent or “cause” variable and which is the dependent or “effect”

variable?

5. State an operational definition for each variable. If experimental, describe how the

independent variable might be manipulated and the dependent variable measured. If

nonexperimental, describe how both variables might be measured.

Hypothesis 5: Political orientation is related to attitude toward gun control.

1. What type of relationship would you predict (positive linear, negative linear, and so on)?

2. Graph and label the relationship you predict. State a prediction based on the relationship

shown by your graph.

3. Which method (experimental or nonexperimental) would you use to test this hypothesis?

Why?

4. Which is the independent or “cause” variable and which is the dependent or “effect”

variable?

5. State an operational definition for each variable. If experimental, describe how the

independent variable might be manipulated and the dependent variable measured. If

nonexperimental, describe how both variables might be measured.

A student is given 20 multiple choice questions. Each multiple choice question has 4 options & only one correct answer. The student answers all the questions in a random manner.

a) What is the probability the student scores 80% or above?

b) What is the probability the student scores 50% or above?

Read the article below and complete the task at hand.

LISTERIOSIS OUTBREAK IN SOUTH AFRICA In March 2018, Tiger Brands suspended operations at its factories in Polokwane, Germiston and Pretoria, which the department of health had linked to the listeriosis outbreak and had claimed over 200 lives in South Africa, and infected over 1,000 others, mostly pregnant women, children, the elderly and people with chronic health conditions fell ill because of the infection. Tiger Brands was thrown into the centre of the listeriosis storm after South Africa’s National Institute for Communicable Diseases announced that its investigation had traced the origins of the disease to one of the company’s biggest meat processing plants. The culprit was identified as polony from the Enterprise Foods facility that produced a range of cold meats. Tiger Brands, a $2.5-billion Johannesburg Stock Exchange-listed business, owns Enterprise Foods among another continent-wide popular food brands. South Africa has been struggling with the listeriosis outbreak for 14 months. Unable to find the source of the affected products, the outbreak developed into the worst case of listeriosis in the world. By the end of February 2018, health authorities had confirmed 948 cases with 180 fatalities. The repercussion was always going to be unforgiving. But Tiger Brands did not help the situation. It overlooked a number of the accepted protocols of handling a crisis of this nature. As a result, the company’s brand equity took a serious strain. The SA government decisively responded to stop the outbreak of Listeriosis. All manufacturers were instructed to recall processed, ready-to-eat meat products; exports of such products were suspended; the public was advised to avoid all processed, ready-to-eat meat products; the products were isolated, returned to the stores for a refund and thorough cleaning was carried out both at the stores and at home where they were kept. This intervention was welcomed by all, but more decisive measures were required. Tiger Brands recalled all products across the country. At the time, it projected that the recall would cost it as much as R377 million. The costs excluded ongoing trading losses and were calculated on the assumption that the facilities would be reopened by 30 September 2018. All frozen stock tested proved negative for Listeria monocytogenes. The remaining carrying value of frozen stock exposed to potential future write-offs is R80 million. All other stock of product, raw materials, packaging and ingredients, which were expected to be utilised following a recommencement of operations amounted to R103 million at 31 March 2018. In March, the group warned that the monthly impact of a complete cessation in production equals an ‘adverse movement’ of approximately R50 million at earnings, before interest and tax level relative to the existing earnings base. Tiger Brands published its interim financial results for the six months ended March 2018, showing what effect the devastating listeriosis outbreak had on its bottom line. The group recorded a 4% decline in revenue for the period to R15.7 billion (from R16.4 billion before) while operating income (before impairments and ‘abnormal items’) declined 8% to R2 billion. The recall associated with its value-added meat processing (VAMP) amounted to R415 million but balanced at R363 million, net of the R50 million insurance claim, and R2 million in profit from the disposal of related property. HEPS was down 16% to 868 cents, while an interim dividend was declared at 378cps. “With the exception of VAMP, the group’s core domestic food businesses delivered a steady performance in the six-month period ended 31 March 2018, notwithstanding intense competition and ongoing pressure on pricing as consumers continually search for value,” the group said. However, the R363 million blow was certainly felt, with a further R183 million hanging in the balance as tests for Listeria continue on remaining stock. Tiger Brands faced serious brand erosion as a result of the way it had handled the unfolding crisis. It could have responded better, by being more rapid in its responses. The company only held a media briefing a day after the minister’s announcement of listeriosis.However, the Tiger Brands noted with concern the media statement released by the Department of Health and the National Institute of Communicable Diseases (NICD) on 4 March 2018, which essentially concluded that the present outbreak was traced to a food production facility in Polokwane whilst raising further concerns about a facility in Germiston. They claimed media released was devoid of detail and this lack of details resulted in misinformation which was not only detrimental to the consumer, but also the South African Red Meat Industry. In this regard, the average consumer was being led into a Listeria hysteria which had unfortunate consequences for families who relied on processed meat as their source of protein. At the outset, Tiger Brands pointed out that food safety remained at the heart of their business and assured the consumer that everything possible is being done with the utmost urgency to ensure that consumer’s personal health and well-being were protected not only as a matter of routine but with increased vigilance. The outbreak emphasized the responsibility of the Tiger Brands together with other food industries to provide for proper and improved hygiene during the production, processing, packing and preparation of red meat and red meat products. However, according to Tiger Brands, during the processing of livestock to meat at the abattoir, particular attention is given to slaughter procedures, personal hygiene and sterilization of equipment to minimize bacterial contamination during this process. Furthermore, meat inspection of each animal and carcass ensure the health of the animal and removal of any possible contamination that might have occurred. Microbiological testing of water, product, contact surfaces and hands is a prerequisite at a registered abattoir in terms of the Meat Safety Act, 2000 and supporting regulations. Whilst Tiger Brands was responsible for producing food that was safe for human consumption, it was also the responsibility of consumers not to content themselves, that the only contamination can come from the facilities implicated, but to adopt basic hygienic practices when buying, transporting food home, preparing and storing food to protect their health and to ensure that cross-contamination does not occur between cooked and raw products or from human hands and equipment. Listeria presents a particular concern with respect to food handling because it can multiply at refrigerator temperatures. It is therefore imperative that consumers ensure that the meat they purchase is sourced only from registered abattoirs that have an Independent Meat Inspection Service and that once purchased the cold chain is maintained at all times; as well as to avoid eating raw or undercooked meat products. The SACP called on the government to hold Tiger Brands accountable. They protested that heavy penalties must be imposed for the negligence that resulted in the failure to prevent the bacteria at Tiger Brands Enterprise Foods in Polokwane, in Gauteng and elsewhere at the monopoly’s facilities, and at other companies that failed to uphold food health and safety, Rainbow Chicken included. The union further requested that all workers at Tiger Brands’ Enterprise Foods must receive urgent priority for screening, paid for by the monopoly. The workers’ conditions of employment, including hours of work, must be looked into and be addressed.

TASK: “Unfortunately, the aforesaid media release is devoid of detail and this lack of detail has resulted in misinformation which is not detrimental to the consumer. In this regard, the average consumer is being a Listeria hysteria which is having unfortunate consequences for families who rely on processed meat as their source of protein.” Having been appointed as Marketing Manager of Tiger Brands, you are required to advise the board on the most appropriate brand crisis management and provide a revised marketing strategy going forward to restore confidence in the brand and prevent any further negative impact on the brand. You are required to research this incident and the company in detail. Your report should focus on the following issues:

1. Situation analysis: assess the current situation that Tiger Brands is facing

2. The impact the incident had on the brand

3. Tiger Brands’ positioning strategy before and after the incident and assess how this situation has affected the positioning of the brand in the mind of the consumers.

4. Public Relations steps and approach to be taken to diffuse the public outcry.

5. Revised Marketing strategy GOING FORWARD

Determine the mean, median, and mode for each of the variables. What is the variance for each set of data for each of the variables? What is the standard deviation for each of the variables? What is the probability that each event occurs in each of the two variables.

Data: Time Stamp with questions with 3 multiple choice options

Timestamp 1. If given the opportunity would you work from home? 2. Do you consider working from home more of a employee convenience or employer benefit?    3. Who benefits more from the "Work from Home" opportunity?
8/11/2019 11:24:03   Maybe   Employee Convenience   Employee
8/11/2019 11:26:37   Maybe   Employee Convenience   Employee
8/11/2019 12:14:33   Maybe   Employee Convenience   Employee
8/11/2019 14:07:45   Maybe   Employee Convenience   Employer
8/11/2019 20:01:33   Maybe   Employee Convenience   Employee
8/12/2019 23:30:00   Maybe   Employee Convenience   Employee
8/14/2019 15:21:15   Maybe   Employee Convenience   Employee
8/14/2019 21:13:15   Maybe   Employee Convenience   Employee
8/14/2019 13:50:32   No   Employer Benefit   Employer
8/11/2019 11:21:21   Yes   Employee Convenience   Employee
8/11/2019 11:22:28   Yes   Employer Benefit   Employer
8/11/2019 11:29:04   Yes   Employee Convenience   Employee
8/11/2019 11:36:23   Yes   Employer Benefit   Employee
8/11/2019 11:36:55   Yes   Employee Convenience   Employee
8/11/2019 11:43:05   Yes   Employee Convenience   Employee
8/11/2019 11:59:13   Yes   Employer Benefit   Employee
8/11/2019 12:22:02   Yes   Employer Benefit   Employer
8/11/2019 12:39:02   Yes   Employee Convenience   Employee
8/11/2019 12:47:51   Yes   Employee Convenience   Employee
8/11/2019 13:12:20   Yes   Employee Convenience   Employer
8/11/2019 13:49:33   Yes   Employer Benefit   Employer
8/11/2019 15:16:18   Yes   Employer Benefit   Employer
8/11/2019 18:55:11   Yes   Employer Benefit   Employer
8/11/2019 19:07:52   Yes   Employer Benefit   Employer
8/11/2019 20:03:24   Yes   Employer Benefit   Employer
8/11/2019 21:38:25   Yes   Employer Benefit   Employer
8/12/2019 5:52:13   Yes   Employer Benefit   Employer
8/12/2019 6:56:04   Yes   Employee Convenience   Employee
8/12/2019 12:16:08   Yes   Employer Benefit   Employer

A sample of n = 9 individuals is selected from a population with µ = 50. After a treatment is administered to the individuals, the sample mean is found to be M = 54. The sums of squares is SS =72. The researchers want to address whether the obtained sample mean is different from the population mean at α = .05, two-tailed.

a. Following the steps of a hypothesis test, determine whether the obtained sample mean is different from the population mean.

b. Calculate Cohen’s d and r2.

c. Write a sentence demonstrating how the results of the hypothesis test and the measure of effect size (use either d or r2) would appear in a research report.

1. SIMPLE LINEAR REGRESSION. For this and the next 3 parts. The journal, Fisheries Science (Feb 1995) reported on a study of the variables that affect endogenous nitrogen excretion (ENE) in carp raised in Japan. Carp were divided into groups of 2 to 15 fish each according to body weight and each group placed in a separate tank. The carp were then fed a protein-free diet three times daily for a period of 20 days. One day after terminating the feeding experiment, the amount of ENE in each tank was measured. The table below gives the mean body weight (in grams) and ENE amount (in milligrams per 100 grams of body weight per day) for each carp group [Source: Watanabe, T. and Ohta, M. "Endogenous Nitrogen Excretion and Non-Fecal Energy Loss in Carp and Rainbow Trout," Fisheries Science, Vol. 61, No. 1, Feb 1995, p. 56]. You should be able to determine which variable is the response variable and which is the explanatory variable. Which of the following is true? [I] Plot of the residuals shows an upward curvature [II] Plot of the residuals shows a downward curvature [III] The regression is significant at the 1% level [IV] Correlation coefficient of the two variables = -0.68 [V] Standard error of the estimate is 0.0297.

   Tank	Body Weight	ENE
   1	11.7	15.3
   2	25.3	9.3
   3	90.2	6.5
   4	213.0	6.0
   5	10.2	15.7
   6	17.6	10.0
   7	32.6	8.6
   8	81.3	6.4
   9	141.5	5.6
   10	285.7	6.0

   		I and V only
   		II, IV, V
   		I and IV only
   		II and III only
   		None of the above

Part B.

1. SIMPLE LINEAR REGRESSION (above data). Give a 99% prediction interval for the expected (mean) value of the dependent variable with X = 0 (note alpha = 0.01).

   		8.3724; 14.4355
   		6.9928; 15.8151
   		-0.0508; -0.0034
   		-0.0615; 0.0073
   		None of the above

Part C.

1. SIMPLE LINEAR REGRESSION (above data). Give a 99% confidence interval for the slope (note alpha = 0.01).

   		8.3724; 14.4355
   		6.9928; 15.8151
   		-0.0508; -0.0034
   		-0.0615; 0.0073
   		None of the above

Part D.

1. MULTIPLE REGRESSION (refer to above data). Conduct a multiple regression by introducing a quadratic term to the model. Which of the following is true? [I] About 74% of the variation in Y explained the regression [II] At the 1% level, the regression is statistically significant [III] At the 5% level, the regression is statistically significant [IV] Both coefficients are statistically significant at the 5% level [V] At least one of the independent variables is not significant

   		I, III, V
   		III, IV, V
   		I, II, IV
   		None of the above

Rosenberg Land Development (RLD) is a developer of condominium properties in the Southwest United States. RLD has recently acquired a 40.625 acre site outside Phoenix, Arizona. Zoning restrictions allow at most 8 units per acre. Three types of condominiums are planned: one-, two-, and three-bedroom units. The average construction costs for each type of unit are $450,000, $600,000 and $750,000, respectively. These units will generate a net profit of 10%. The company has equity and loans totaling $180 million dollars for this project. From prior development projects, senior managers have determined that there must be a minimum of 15% one-bedroom units, 25% two-bedroom units, and 25% three-bedroom units.

a. Develop a linear optimization model to determine how many of each type of unit the developer should build.

b. Find an optimal solution by Solver in Excel.

Let X represent the standard normal random variable. The P{X > 2.07} is equal to:

Suppose that 19 measurements of the failure stress of carbon fiber specimens results in a sample mean of 562.68 MPa and sample standard deviation of 180.874 MPa and that the data is normally distributed. Calculate two-sided confidence intervals for the failure stress for the following cases:

a) 95% confidence interval for a single measurement (n = 1) of the failure stress. Assume that the population standard deviation is known to be 181 MPa.

b) 95% confidence interval for the sample mean with n = 19. Repeat for the 99% confidence interval. Assume that the population standard deviation is known to be 181 MPa

c) 95% confidence interval for the sample mean with n = 19. Repeat for the 99% confidence interval. Assume that the population standard deviation is UNKNOWN.

d) Suppose that an alternative process of producing carbon fibers results in a sample of failure stress measurements with n = 12, sample mean = 650.32 MPa and sample standard deviation of 130.470 MPa. What is the 95% confidence interval on the difference in failure stress between the two populations?

( i need Unique answer, don't copy and paste, please) (dont' use handwriting, please).

(i need references URL Link)

General Question

** How to perform logistic regression in SPSS?

Define G_N to be the game which pays the longest run of heads or tails observed in N flips of a fair coin. Let E_N be the expected value of the game G_N.

What is the difference between E₅ and E₄, to four decimal places?

Suppose that your hypotheses are H0: π = 0.25 and Ha: π < 0.25. In the context of these hypotheses, which of the following standardized statistics would provide the strongest evidence against the null hypothesis and for the alternative hypothesis? Why? A. z = –1 B. z = 0 C. z = 3 D. z = –1.80 E. z=25

4. One hundred students were interviewed. Forty–two are Monthly Active Users (MAU) of Facebook (F), and sixty–five are MAU of Snapchat(S). Thirty–four are MAU of both Facebook and Snapchat. One of the100 students is randomly selected, all 100 students having the same probability of selection (1/100).

(a) What is the probability that the student is an MAU of Facebook?

(b) What is the probability that the student is an MAU of Facebook given that the student is an MAU of Snapchat?

(c) Find Pr(S|F).

d) Express the probability in Part (b) in symbols, that is in a similar fashion to Part (c).

5. The students in Question 4 were also asked if they were MAU of Twitter. Twenty five were MAU of Twitter, including 15 who were MAU of Facebook and Twitter and 16 who were MAU of Twitter and Snapchat. Twenty four students were not MAU of any of Facebook, Snapchat or Twitter.

(a) What is the probability that a randomly selected student is an MAU of Facebook, Snapchat, and Twitter?

(b) What is the probability that a randomly selected student is an MAU of Twitter, conditional on that student being an MAU of both Facebook and Snapchat?

The data set  Roslyn in the accompanying workbook gives appraised values (In $1000’s) and size (in square feet)  for thirty houses in the Roslyn neighborhood.

a)     Use Excel’s Data Analysis ToolPak to produce output for the simple linear regression, with Valueas the response (y) variable and Size as the predictor (x) variable.

b)     Write out the equation of the regression of Value (y) on Size (x)

c)      State the numerical value of the slope of the regression line. What does it tell you in this context?

d)     State the numerical value of the standard error of the estimate? What does it tell you in this context?

e)     State the numerical value of the coefficient of determination? What does it tell you in this context?

f)      Give the 95% confidence interval for the intercept of the regression equation.   




Are there any negative numbers in this interval?   What practicalconclusion can you draw from your answer?

g)     Here is a muddled, inaccurate “explanation” of what it means to be “95% confident” in the interval for the slope.   “For 95% of samples the population slope, b,will be in the interval [0.1399, 0.2336]”
Give an accurateexplanation of “95% confidence” in this context.

h)     What practical information does the regression equation give a realtor about a 3000 square foot house ?

i)       What practical information does the regression equation give a realtor about a 30000 square foot house ?

j)       Conduct an appropriate statistical test for the significance of the regression of value on size, including a clear statement of the hypotheses.  (Note that for a Simple Linear Regression you can use either the Test for Individual Significance or the Test of Joint Significance.   You will reach identical conclusions)

k)              What practical conclusion can a realtor draw from the hypothesis test in j)?

Data:

Address	Appraised Value	House Size (square feet)
182 Village Road	681.4	2194
108 Burnham Avenue	606.0	3032
143 Powerhouse Road	457.9	1970
55 Hummingbird Drive	912.7	3356
40 Maple Street	416.7	2070
47 Magnolia Lane	726.6	2826
35 Harding Avenue	393.1	1606
100 Crescent Lane	612.4	2063
222 Garden Street	355.4	1392
6 Church Street	299.0	1120
12 Ridge Drive	471.0	1817
24 Madison Place	510.7	2496
18 Rockhill Road	517.7	1615
65 Elm Drive	873.3	4067
30 Wren Drive	854.7	3130
54 Lambert Street	374.8	1423
38 Magnolia Lane	543.0	1799
75 Burnham Avenue	554.0	2936
19 Oxford Street	365.2	1439
215 Elm Drive	811.8	4065
34 The Oaks	711.8	2191
2 Circle Lane	598.7	2008
70 Rugby Road	651.3	2070
150 Warner Avenue	511.1	2710
31 West Court	379.3	1416
7 The Locusts	786.0	3244
65 Starling Court	768.7	2493
106 Barberry Lane	679.9	2473
17 South Drive	615.8	1968
8 Woodland Road	766.4	3136