Mathematics
MIS301 San Diego State Statistical Analysis for Business Case Questions

MIS301

San Diego State University

Question Description

I don’t know how to handle this Statistics question and need guidance.

My last name starts with a B. My Red ID is 822591309. My section # 22421. I sent you the link with the topics you choose. I will provide a link of everything and even samples of what my professor is looking for. These are the directions:

Please choose 3 topics under the letter of your last name and create 3 multiple choice questions.I will grade your best 2.Write a multiple choice question that pertains to the topic you chose, and list the correct answer first, and then 3 or 4 possible wrong answers.Show your solution for the correct answer, and then also explain how you went about choosing the wrong answer.Have good reasoning behind it, just don’t make up out of thin air.

Please include your name, section number and Red ID at the top of the page.For each multiple choice question you do, clearly label which topic it relates to and follow my examples.All work needs to be typed, no hand written cases.Any pen or pencil marks will be ignored when I grade it.It is due at the beginning of class.If it is turned it late during class you will lose 10%, and if you turn in at the end of class of the close of the day Monday it will be 20% off.Each day after will be an additional 10% deduction.Make sure you do a good job, if you like your grade on this project, then you can count this grade for your second case which will be assigned later in the month.


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MIS 301 Statistical Analysis for Business – Fall 2019, Section 1 CASE Assignment 1 Due November 20, 2019 Overview The objective of this case is to help you better understanding the last 7 chapters by constructing multiple choice questions and then solving them. Last name begins with this letter: A-F G-L M-R S-Z CH 3 Percentile Empirical Rule Coefficient of Var Weighted Avg CH 4 Permutation Combination Factorial Distinct Permutation CH 5 Expected Value Binomial Poisson Poisson CH 6 Uniform Normal Distribution Top & Bottom % Normal Distribution Between Normal Distribution Above & Below CH 7 Sample Mean Sample Proportion Sample Mean Sample Proportion CH 8 Sample Size Margin of Error C I for sigma Known C I for proportion C I for sigma unknown CH 9 Hypothesis Test Proportion Type 1 & Type 2 Errors Hypothesis Test Sigma Known Hypothesis Test Sigma Unknown Please choose 3 topics under the letter of your last name and create 3 multiple choice questions. I will grade your best 2. Write a multiple choice question that pertains to the topic you chose, and list the correct answer first, and then 3 or 4 possible wrong answers. Show your solution for the correct answer, and then also explain how you went about choosing the wrong answer. Have good reasoning behind it, just don’t make up out of thin air. Please include your name, section number and Red ID at the top of the page. For each multiple choice question you do, clearly label which topic it relates to and follow my examples. All work needs to be typed, no hand written cases. Any pen or pencil marks will be ignored when I grade it. It is due at the beginning of class. If it is turned it late during class you will lose 10%, and if you turn in at the end of class of the close of the day Monday it will be 20% off. Each day after will be an additional 10% deduction. Make sure you do a good job, if you like your grade on this project, then you can count this grade for your second case which will be assigned later in the month. I have given you 2 examples to show you what I am looking for. Examples: Chebyshev’s: The average amount of money students at SDSU spend at the bookstore during the semester is $346 with a standard deviation of $58. What is the minimum proportion of students that spend between $172 and $520? A. B. C. D. E. 88.89%  Correct Answer 66.67% 33.33% 99.7% 50.28% Explanation of Answers: A. 172  346  520 174 346  174 k= 174 =3 58 1− 1 = .8889 = 88.89% 32 174 1 = .6667 = 66.67% - Forgetting to square k 3 B. Using 1 − C. 58 = .3333 = 33.33% 174 D.   3  99.7% Normal Distribution E. 174 = .5028 = 50.28% 346 Hypothesis Testing: It is stated by SDSU Bookstore that at least 78% of students use/shop in the bookstore during the semester. You think this proportion is high and would like to test SDSU’s Bookstore claim at the level of significance of .05. So you randomly select 120 students and find that 89 used the bookstore during the semester. State the Null Hypothesis Po  .78  Correct Answer A. B. C. D. E. Po Po Po Po  .78  .78  .78 = .78 State the Alternative Hypothesis A. Po  .78  Correct Answer B. C. D. E. Po Po Po Po  .78  .78  .7417  .7417 Test statistics equals A. -1.01  Correct Answer B. -.96 – using pbar in the denominator C. -1.645 – critical value D. 1.01 E. .7417 Probability of the test statistic for this specific problem? A. .1562  Correct Answer B. .3124 - if it was a 2-tail test this would be the correct answer C. .8438 - compliment of the 1-tail probability D. .6876 - compliment of the 2-tail probability Using the Critical Value approach at what critical value would you start rejecting the null hypothesis? A. -1.645  Correct Answer B. -1.96 - if alpha was .025 and lower tail test C.  1.96 - if it was a 2-tail test D. 1.645 - if it was an upper tail test E. 1.96 - if alpha was .025 and an upper tail test Conclusion of this hypothesis test A. There is not enough evidence to prove that less than 78% of students use bookstore  Correct B. Reject the null and accept the alternative C. Accept the null and reject the alternative D. Reject the alternative You only need to do 1 of these questions. If you do more than 1 it will still only count as 1. But you will need to solve the whole hypothesis test as I did regardless of which question you did. Null Hypothesis: Po  .78 At least 78% Alternative Hypothesis: Po  .78 Less than 78% Test statistic: z= .7417 − .78 .78 * .22 120 = −1.01 bad calculation for z -→ z = .7417 − .78 .7417 * .2583 120 = −.96 P( p  .7417) = P( Z  −1.01) = .1562 Compare: P > Alpha .1562 > .05 1-tail, so need to double the probability Conclusion: Accept the null hypothesis (Fail to reject the null hypothesis). There is no statistically significant evidence to show that less than 78% of students use the book store. Example: Poisson Distribution The help desk at SDSU gets 4.5 calls per hour. Calls are independent of one another and the amount of calls or lack of calls in one hour does not influence or affect the amount of calls in any other hour. Calls are equally likely at any point in any and all the one hour periods (poisson distribution). What is the probability of getting less than 3 calls in an hour? A. B. C. D. E. .173578  Correct Answer .342296 .168718 .657704 .826422 Explanation of the Answers: A. P(x<3) = P(0) + P(1) + P(2) P(0) = 4.5 0 e −4.5 4.51 e −4.5 = .011109 P(1) = = .04999 0! 1! B. Include P(3) => P ( x  3) C. Just P(3) D. P(x>3) = 1- [ P(0) + P(1) + P(2) + P(3) ] = 1 - .342296 = .657704 E. 3 and more - P ( x  3) = 1- [ P(0) + P(1) + P(2) ] = 1 - .173578 = .826422 P(3) = P(2) = 4.5 2 e −4.5 = .112479 2! 4.53 e −4.5 = .168718 3! Example: Poisson Distribution (Adjusting mu) The help desk at SDSU gets 4.5 calls per hour. Calls are independent of one another and the amount of calls or lack of calls in one hour does not influence or affect the amount of calls in any other hour. Calls are equally likely at any point in any and all the one hour periods (poisson distribution). What is the probability of getting at most 1 call in a 20 minute? A. .557825  Correct Answer B. .061099 C. .183297 D. .020366 E. .334695 Explanation of the Answers: Need to adjust the mean for the new interval  = 4.5 per hour or  = 4.5 per 60 minutes Divide both by 3  = 4.5  3 per 60 minutes  3  = 1.5 per 20 minutes How do you get from 60 minutes to 20 minutes? Divide by 3 A. P( x  1) = P(0) + P(1) P(0) = 1.5 0 e −1.5 = ..22313 0! B. P(0) + P(1) using  = 4.5 and not adjusting C. Using answer B. (did not adjust the mean due to the interval change) and multiplying by 3 D. Using answer B. (did not adjust the mean due to the interval change) and dividing by 3 E. Adjust the mean properly for the new time interval but only use P(1) P(1) = 1.51 e −1.5 = .334695 1! Above are just some examples and how to answer and what materials, formulas and information that should be included to consider the problem complete. I have also listed below some things to avoid or not to do for this case. 1. 2. 3. 4. 5. 6. Do your own work, do not work or collaborate with any of your classmates. Make sure it is typed and detailed (identify the topic and such) Show computations and work wherever possible or feasible. If you have questions, make sure they are specific, I only grade your case once, so don’t ask me to look it over and see how it is. Specific Questions! Error to the side of doing more than less. Your questions need to be original, can not use any of my questions or work. ...
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Final Answer

Please find answer.Let me know for any clarifications.Pleasure working with you.Good Bye.

Running head: STATISTICS

1

Statistics
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University Name
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psumanrec (4252)
UIUC

Anonymous
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