Assignment Code: 2012 GM03 B2 Last Date of Submission: 15th November 2012
Maximum Marks: 100

Attempt all the questions. All questions are compulsory and carry equal marks.

SECTION – A

1.a) Explain the need for different forecasting techniques. How can we evaluate as to how good is our forecast.
b)Collect data on the amount of expenditure you do each day for the next 25 days and based on the same forecast using 7day moving average the forecast for the 26th day. Evaluate the accuracy of your forecast. c) Explain one application where you can apply Queuing Theory in daily life.

2. Customers from the Higher Income Group were sampled to test the quality of foods at one of the four major restaurant run by the Taj Group of Hotels. The four types of foods for which the quality were to be checked were Chicken Platter, Honey Chicken, Chicken Spinach, and Tandoori Chicken. Each of the respondent were asked to rate in a scale of 1- 10. The following were the data. Chicken PlatterHoney ChickenChicken SpinachTandoori Chicken 6876

7866
8967
5866
9758
8977
7876
Is there sufficient evidence to indicate that the quality of food being served is different. Use 5% level of significance.

3. Maxwell’s Hot Chocolate is concerned about the effect of the recent year long coffee advertising campaign on hot chocolate sales. The average weekly hot chocolate sales two years ago was 984.7 pounds and the standard deviation was 72.6 pounds. Maxwell’s has randomly selected 30 weeks from the past year and found average sales of 912.1 pounds. 1.State appropriate hypotheses for testing weather hot chocolate sales have decreased. 2.At the 2 percent significance level, test the hypotheses.

4. A manufacturer was making sport shirts for men. The manufacturer was mainly catering to the needs of Baseball, Basketball, and Football...

...STAT 600 Statistics and Quantitative Analysis
PROJECT: Stock return estimation
The project must be done by 6-15 a.m. October, 16th. You should submit your projects before the class begins. This is a group project. Read the course outline for general guidelines. Good luck!
The project is closely related to Lectures 1-5 of the class.
Today is September 15, 2013 and you have just started your new job with a financial planning firm. In addition to studying for all your license exams, you have been asked to review a portion of a client’s stock portfolio to determine the risk/return profiles of 12 stocks in the portfolio. Unfortunately, your small firm cannot afford the expensive databases that would provide all this information with a few simple keystrokes, but that’s why they hired you. Specifically, you have been asked to determine the monthly average returns and standard deviations for the 12 stocks for the past five years.
The stocks (with their symbols in parentheses) are:
Apple Computer (AAPL) Hershey (HSY)
Archer Daniels Midland (ADM) Motorola (MOT)
Boeing (BA) Procter and Gamble (PG)
Citigroup (C) Sirius XM radio (SIRI)
Caterpilar (CAT) Wal-Mart (WMT)
Deere&Co. (DE)...

...typically have? You take a random sample of 51 reduced-fat cookies and test them in a lab, finding a mean fat content of 4.2 grams. You calculate a 95% confidence interval and find that the margin of error is ±0.8 grams. A) You are 95% confident that the mean fat in reduced fat cookies is between 3.4 and 5 grams of fat. B) We are 95% confident that the mean fat in all cookies is between 3.4 and 5 grams. C) We are 95% sure that the average amount of fat in the cookies in this study was between 3.4 and 5 grams. D) 95% of reduced fat cookies have between 3.4 and 5 grams of fat. E) 95% of the cookies in the sample had between 3.4 and 5 grams of fat. Determine the margin of error in estimating the population parameter. 12) How tall is your average statistics classmate? To determine this, you measure the height of a random sample of 15 of your 100 fellow students, finding a 95% confidence interval for the mean height of 67.25 to 69.75 inches. A) 1.5 inches B) 0.25 inches C) 1.06 inches D) 1.25 inches E) Not enough information is given. 12) 11) 10)
3
Construct the indicated confidence interval for the difference between the two population means. Assume that the assumptions and conditions for inference have been met. 13) The table below gives information concerning the gasoline mileage for random samples of trucks of two different types. Find a 95% confidence interval for the difference in the means m X - m Y. Brand X Brand Y 50 50 20.1 24.3 2.3 1.8 13)...

...1. A radio station that plays classical music has a “By Request” program each Saturday night. The percentage of requests for composers on a particular night are listed below:
Composers Percentage of Requests
Bach 5
Beethoven 26
Brahms 9
Dvorak 2
Mendelssohn 3
Mozart 21
Schubert 12
Schumann 7
Tchaikovsky 14
Wagner 1
a. Does the data listed above comprise a valid probability distribution? Explain.
The individual probabilities are all between 0 & 1 and the sum = 100%
b. What is the probability that a randomly selected request is for one of the three B’s?
P(one of the B’s) = P(Bach) + P(Beethoven) + P(Brahms) = 5 + 26 + 9 = 40%
c. What is the probability that a randomly selected request is for a Mozart piece?
P(Mozart) = 21%
d. What is the probability that a randomly selected request is not for one of the two S’s?
P(not Schubert or Schumann) = 1 – P(Schubert or Schumann)
= 1 – (12 + 7)
= 81%
e. Neither Bach nor Wagner wrote any symphonies. What is the probability that a randomly selected request is for a composer who wrote at least one symphony?
P(Symphony) = 1 – P(Bach or Wagner)
= 1 – (5 + 1)
= 94%
f. What is the probability that a randomly selected request is for a composer other...

...1. Introduction
This report is about the case study of PAR, INC. From the following book: Statistics for Business an Economics, 8th edition by D.R. Anderson, D.J. Sweeney and Th.A. Williams, publisher: Dave Shaut. The case is described at page 416, chapter 10.
2. Problem statement
Par, Inc. has produced a new type of golf ball. The company wants to know if this new type of golf ball is comparable to the old ones. Therefore they did a test, which consists out of 40 trials with the current and 40 trials with the new golf balls. The testing was performed with a mechanical fitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the design. The outcomes are given in the table of appendix 1.
3. Hypothesis testing
The first thing to do is to formulate and present the rationale for a hypothesis test that Par, Inc. could use to compare the driving distance of the current and new golf balls. By formulation of these hypothesis there is assumed that the new and current golf balls show no significant difference to each other. The hypothesis and alternative hypothesis are formulated as follow:
Question 1
H0 : µ1 - µ2 = 0 (they are the same)
Ha : µ1 - µ2 ≠ 0 (the are not the same)
4. P-value
Secondly; analyze the data to provide the hypothesis testing conclusion. The p-value for the test is:
Question 2
Note: the statistical data is provide in § 5.
-one machine
-two...

...
Final Project: Nyke Shoe Company
Barbara Greczyn
STA 201 - Principles of Statistics
Instructor Alok Dihtal
April 26, 2015
Introduction
Nyke Shoe Company has been in business for over 50 years. Over the last five years, the company has been undergoing some financial hardship due to an erratic market and an inability to understand what the consumer actually needs. In a last ditch effort to avoid bankruptcy, they have adopted a new business model which entails the development of only one shoe size. In order to achieve this goal, statistical data must be utilized and applied to make the best choice. The data used will be explained to the fullest and a conclusion will be then obtained.
Methodology
A sample group of 35 participants was gathered, 18 females and 17 males. Their heights and shoe sizes were gathered and their data was processed in three categories: shoe size, height, gender. Descriptive statistics was applied to three separate data sets, one with all participants included, one sets with just female participants, and one with just male participants. Then a two sample t-test was conducted with the assumption that there were unequal variances amongst both male and female data sets.
Results
There is a normal distribution of the data with ranges in size from size 5 to size 14 amongst the participants. With these ranges, the mean is 9.142, with a standard deviation of 2.583 and a variance of 6.670....

...that we are accepting the alternative hypothesis and this statement works vice versa. In this case that means that the null hypothesis can be rejected or disproving. For the data set that was given the null hypothesis also known as H-nought was µ1=µ2, while the alternative hypothesis is µ1<µ2. Null hypothesis states that the amount of rural nurse homes was equal to the average amount of beds used. Alternative hypothesis states that rural area nursing homes uses fewer amounts of beds. The claim indicated to what kind of test was going to be used and since I claimed that the rural area were going to have a lower average number of beds it states that the shaded area on the critical value test will be less than zero.
Table 1. Descriptive statistics for the given null and alternative hypothesis that includes the sample, mean, median, standard deviation, maximum values, and minimum values.
Sample Size
Mean
Median
Standard Deviation
Maximum Value
Minimum Value
Rural Area
34
0.6538
1.0000
0.4803845
1
0
Bed
4850
93.27
88.00
40.85273
244.00
25.00
Figure 1. This figure illustrates the critical value test for the left-tailed test. The critical value that was needed for the test was -1.692 according to the t-table since our sample size was 34. Used the degree of freedom formula to find the critical value.
Figure 2. This figure reflects to the p-value. When we figured out the p-value we used pt(t,33). Since pt(t,33) equaled 0.0137855 that indicated...

...MM207 Statistics
Unit IV Mid Term Project
1. In the following situation identify the implied population.
A recent report on the weekly news presented the findings of a study on the effectiveness of Onglyza, along with diet and exercise, for treating diabetes.
According to Bennett (2009), a population is defined as “the complete set of people or things being studied” in a statistical study. Given that the information is in relation to finding the success of a drug used to care for diabetes, the population would be individuals who experience diabetes. Therefore, the implied population is the entire individual’s who diabetics are. Individuals who are diabetics are those who were used to test the effectiveness of Onglyza, diet and exercise.
2. In the following scenario identify the type of statistical study that was conducted.
A Gallop poll surveyed 1,018 adults by telephone, and 22% of them reported that they smoked cigarettes within the past week.
I would say that this would be an observational study which is a study when specific characteristics of the subject are observed, but the characteristics are in no way customized by the researcher. The reason I say that this would be an observational study is because the sample population that was studied was not influenced by the researcher themselves. In addition based on the fact that this study was a poll, in which people were asked to answer the questions but no responses were...

...Charismatic Condition
Mean 4.204081633
Standard Error 0.097501055
Median 4.2
Mode 4.8
Standard Deviation 0.682507382
Sample Variance 0.465816327
Kurtosis 5.335286065
Skewness -1.916441174
Range 3.5
Minimum 1.5
Maximum 5
Sum 206
Count 49
Confidence Level(95.0%) 0.196039006
In both the Charismatic and the punitive condition data sets there were 49 people surveyed. We know this because we were able to use descriptive statistics to show the count and that shows the number of people surveyed. The average or the mean of the charismatic condition is 4.20. The standard error is saying that 0.0975 % is the error that will normally occur if two different people are comparing results. The middle value or the median of the data set is 4.2 and the most frequently occurring value or mode is 4.8. The charismatic condition is skewed to the left because we are getting a negative number for the skewness data. The skewness is at -1.916441174. The difference between the largest and the smallest value which is also called the range is 3.5. The minimum score is 1.5 and the maximum score was a 5 given by the students for the charismatic condition.
The Frequency for the charismatic condition is telling us the summary of the data that is presented is the form of class intervals and frequency. This bar graph will show you the values of the scores starting at 1.5 and going up to 5. The frequency shows us the number of students who picked the...