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1) Create a 99% for the difference of the means of the variable WAR (wins above replacement) for those players who did not make the playoffs compared to those players who made the playoffs. Based on this interval, could you make the conclusion that the players who made the playoffs had a higher mean WAR than those who did not make the playoffs? Explain.

2) Create a 90% confidence interval for the percent of players who were considered to be outfielders. For position players, 37.5% of position players on the field are outfielders. Based on the sample provided, explain why this is (is not) a good estimate for the percentage of players who are outfielders.

3) Using hypothesis testing, explain if there is enough evidence at the 5% level of significance to say that the mean number of homeruns (HR) is less than 25.

4) At the 2% level of significance, is there any evidence to support the claim an individual’s batting average in September is going to be less than his batting average in April. Explain your answer.

5) Create a new variable called “TeamPlayer” where the variable has the value “Yes” if the player has more than 80 runs batted in (RBI) and has the value of “No” if not. Is there a difference in the percent of players in each league who are considered to be a “TeamPlayer”? Use a 5% level of significance and explain your...

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...Stat 141 R1 - Lecture #35 Announcements: 1) Assignment #11 Question 5: The answer is wrong … should be “fail to reject” but MyStatLab wants “reject”... so give the wrong answer for full marks in this question 2) Exam: STAT 141 R1 3 hrs 1400 Wed Apr 17 MAIN GYM, ~45 Multiple Choice Questions Chapters 7, 8, 18-28 …. some pre MT skills will be required. Simple Linear Regression …. continued Last time: Ex) Predicting final exam marks (%) from midterm exam marks (%) in a class of 88 students: Student | #1 | #2 | … | #88 | Midterm mark | 67% | 72% | … | 88% | Final mark | 62% | 50% | … | 91% | Stat 141 R1 - Lecture #35 page 2 Given x = midterm percentage, y = final percentage, n=88, x = 67.812, y = 52.643, sx = 17.922, sy = 25.430, r = 0.718 ∑(yi–ŷi)2 = 27278.82 We had calculated: The slope and intercept of the sample line of best fit: * sample line of best fit: y= -16.443 + 1.019 x An estimate for σ (standard deviation about the population line): se Given SSE = ∑(yi–ŷi)2 = 27278.82 : se2=SSEn-2=27278.8288-2=317.196 →σ≈se=317.196=17.810 Stat 141 R1 - Lecture #35 page 3 Inference for the population slope β1 When the 4 basic assumptions of the SLR model are satisfied: o The relationship between x and y is sufficiently linear. Presuming linearity, this means, at any x, με = 0. o The std. dev. of ε is the same for any particular x (constant). o The...

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...expected to be positive. The fact that one of them has the wrong sign (βS) may reflect negative omitted variable bias. Multicollinearity obviously is a problem and may be responsible for the wrong sign of βS too. Variable S looks irrelevant but it should be relevant in theory. Its small t-stat may be due to omitted variable or multicollinearity. Q2. Using the intuition from simple case of omitted variable: (a) y = apple consumption, x1 = price of banana, x2 = price of orange yt = 0 + 1 x1,t + 2 x2,t + ut If we leave out x2, bias on β1 depends on (1) whether x1 and x2 are correlated in the sample which mostly likely is positive (e.g. inflation), and whether price of orange has positive or negative effect on the annual consumption of apple in the population (sign of 2) which should be positive too (law of demand). Bias is (+) (+) = (+) (b) y = student grades on the midterm, x1 = hours studies for the midterm x2 = hours slept before the midterm If we leave out x2, bias on β1 depends on (1) whether x1 and x2 are correlated in the sample which mostly likely is negative, and (2) whether hours slept before the midterm has positive or negative effect on the midterm grade in...

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...recent study of 460 drivers age 70 and over by the National Highway Traffic Safety Administration reported that 75% of those drivers had uncorrected vision problems. Answer: The report of the ‘traffic safety administration’ reported that 75% (345 people) was the reported statistic from the 460 drivers of 70 or older drivers. I feel the assumed parameter would be the population of drivers that had uncorrected vision that were 70 years old or older. 4. What type of sampling procedure was used to collect the data in the MM207 Student Data Set? Answer: The sampling procedure used, believe was the ‘Conveyance’ sampling because this was a population which was readily available. STATISTICS, MM207, ‘Midterm Project’ Andre Bergman, 4-20-2014 Page, 2 of 3 5. From the MM207 Student Data Set identify one...

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...ca Lectures: Monday, Wednesday, Friday: 1.30pm to 2.20pm. Room: ARTS 143 Coordinator: Dr. Lawrence Chang, 236 McLean Hall. E-mail: chang@math.usask.ca Office hours: Please e-mail your instructor or the coordinator to make an appointment at a mutually convenient time. You should e-mail him a day or two in advance and not at the last minute. Lab Schedules: Every student is required to register in either one of the following 4 labs. 1. L01: Thursday, 2.30pm to 3.50pm. Room: ARTS 133 2. L03: Thursday, 4.00pm to 5.20pm. Room: ARTS 133 3. L05: Thursday, 2.30pm to 3.50pm. Room: THORV 271 4. L07: Thursday, 4.00pm to 5.20pm. Room: ARTS 134 First lab: Thursday 13 Sept 2012. Last lab: Thursday 29 November 2012. The lab periods will be devoted to midterm tests and to discussing homework problems. The lab is an essential part of this course. Previous terms have shown that students who skip labs tend not to do well and have a high chance of failing this course. Therefore, you are strongly advised to attend labs. Textbook: Mathematical Analysis for Business and Economics by Schelin and Bange. Customised copies are available in the U. of S. Bookstore. This course will cover Chapters 1 to 6 and part of Chapter 7. Website: To access the course website, go to https://bblearn.usask.ca and log in using your NSID and password. In the blackboard MATH 121...

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...STAT 2000 Midterm Exam # 1 Review Sheet Descriptive Statistics 1. Data, categorical variable, quantitative variable, identiﬁer 2. Population, sample, census, parameter, statistic 3. Sampling designs: simple random, stratiﬁed and cluster samplings 4. Data presentation (a) Categorical variable: frequency tables, bar chart, pie chart (b) Quantitative variable: histogram, ﬁve-number summary, boxplots 5. Descriptive statistics: mean x, median, mode, range, interquartile range (IQR), variance s2 , ¯ standard deviation s. They are used to describe shape, center and spread of the distribution. 6. Percentiles: 25th = Q1 , 50th = median, 75th = Q3 etc. Correlation and Linear Regression 1. Scatterplot, y (dependent, response) variable, x (independent, explanatory) variable 2. Correlation (linear association between 2 variables) and correlation coeﬃcient r = interpretations, properties, conditions, lurking variables, eﬀect of outliers 3. Linear model: y = b0 + b1 x, where b1 = r(sy /sx ) and b0 = y − b1 x. ˆ ¯ ¯ 4. Interpretations of predicted value y , slope b1 and intercept b0 in problem context. ˆ 5. R2 = r2 : fraction of y’s variability accounted for by linear regression on x Probability 1. Trial, sample space, sample points, events 2. Three types of probability: theoretical, empirical and personal 3. Contingency table: joint probability and marginal probability 4. Notation: A, Ac , A ∩ B, A ∪ B, Venn diagram 5. Complement rule: P (Ac ) = 1 − P (A) 6. Addition rule: P (A ∪ B) = P (A)...

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...Question 1 # sold | P(x) | 0 | 0.1 | 1 | 0.11 | 2 | 0.2 | 3 | 0.23 | 4 | 0.3 | 5 | 0.06 | Question 2: The average rate of occurrence of accidents was 8.15 per week. A) during that time, what is the probability of getting less than 3 accidents a week? Use poisson for this question. Since it is less that 3 accidents, x = 0,1,2 (Px=o) = [(8.15^0) (e^8.15)] / 0! = 0.00288735. (Px=1) = [(8.15^1) (e^8.15)] / 1! = 0.002353193. (Px=2) = [(8.15^2) (e^8.15)] / 2! = 0.009589362. Add all these results together, and p(x<3) is 0.01223119. B) The probability if an accident on the road is 0.295. Out of 10 reportable accidents, what is the probability that 2 occurred on the road? Use Binomial for this. Since it is exactly 2, we need to find P(x=2). Use the binomial formula, n = 10, x = 2, p = 0.295, and q is 0.705. = 10!/2!(10-2)! * 0.205 ^2 * q^ 10-2. Answer is 0.23898436. Question 3: 15% of drivers have a probationary license and 85% have a regular license. In a given year, 10% of drivers with probationary licenses are involved in an accident. 2.5% of regular drivers are involved in an accident. For this question it is better to draw a tree diagram. Question 4: Mean values for income for men are 100K, for woman, 80K. SD deviations of 20K and 15K respectively. Variables are normally distributed and independent. Calculate the 95th percentile, for men income. 95th p = 0.95. In the zscore table, we find a value of 1.64. Us e the formula for z-score (population) : ......

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... The calculation we made to compute the debt is as follows: = Total current liabilities + Total long-term debt and other non-current liabilities - Cash and cash equivalents - Trades payables = 2647961 + 6410751 - 1330441 - 1202398 D = 11.5915 billions The proportion of debt is then: D/D+E = 0.3328 B. Estimation of the cost of debt rD According to the financial statement, the average interest rate for Debt and Financing Arrangements, which corresponds actually to long-term notes, was 4.28% for the fiscal year 2012. After that we wanted to estimate the cost of bonds. To do so we found the Heinz’s credit rating, which is BBB . We then looked at the future bonds that Heinz is going to issue, and decided to take the bond with a midterm maturity as representative debt of the company. P=400 C=3.125% Price=101.4 Since the company is rated BBB, the default rate is d=0.4%. We have found a y = 3.38% using the formula seen in the course. To get the final rD we...

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