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# Week8finalexamlivelecture2011 mt

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Week 8 Lecture, Review for Final Exam.

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### Week8finalexamlivelecture2011 mt

1. 1. Lecture for the Final ExamStatistics For Decision Making<br />B Heard<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
2. 2. Week 3 Quiz<br />Some Things to Remember<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
3. 3. This data shows the Lab Report scores of 8 selected students and the number of hours they spent preparing their Statistics Lab Report. 40 was the highest score the student could make.<br />(hours, scores), <br />(3,34), (2,30), (4,38), (4,40), (2,32), (3,33), (4,37), (5,39) <br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
4. 4. Understand the equation of the regression line for the given data.<br />What does the correlation coefficient “r” for the data mean?<br />Predict a Lab Report Score for someone who spent one hour on it.<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
5. 5. Predicted score for someone who spent one hour would be:<br />y = 3.158(1) +24.71<br />y = 27.9 or I would say 28 since all scores are in whole numbers<br />“r” of 0.9248 means strong positive correlation. <br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
6. 6. Stronger Positive Correlation<br />Stronger Negative Correlation<br />0<br />+1<br />-1<br />“r”<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
7. 7. Know the difference between Binomial and Poisson<br />Remember we talked about these in a previous lecture!<br />
8. 8. A State Trooper notes that at a certain intersection, an average of three cars run the red-light per hour.  What is the probability that the next time he is there exactly two cars run the red-light?<br /> <br />Poisson with average of 3. want P(2) P(2) = .2240 (Use Minitab) and also be able to find probability values for less than, less than or equal to, etc. <br />Final Exam Review<br />
9. 9. The probability that a house in a neighborhood has a dog is 40%.  If 50 houses in the neighborhood are randomly selected what is the probability that one (or a certain number) of the houses will have a dog? <br /> <br />a. Is this a binomial experiment? <br /> <br />b. Use the correct formula to find the probability that, out of 50 houses, exactly 22 of the houses will have dogs. Show your calculations or explain how you found the probability.<br />Answer Follows<br />Final Exam Review<br />
10. 10. a) Fixed number of independent trials, only two possible outcomes in each trial {S,F} (dog or not), probability of success is the same for each trial, and random variable x counts the number of successful trials. So YES it is.<br /> <br />b) n = 50; p = .40 = P(success) = house has a dog    We want P(22) --> P(22) = .0959 (Use Minitab) (Also be able to find “at least” , “at most”, “or” etc.<br />Final Exam Review<br />
11. 11. Know basic terms like mean, median, mode, standard deviation, variance, range etc.<br />Mean is the “average”<br />Median is the “center”<br />Mode is the “most frequently occurring”<br />Know the variance is the standard deviation squared<br />Know the standard deviation is the square root of the variance<br />
12. 12. Be able to understand the normal distribution and how it relates to the mean, standard deviation, variance, etc.<br />For example, I did an analysis and found the mean number of failures was 7 and the standard deviation was 1.5. Answer the two questions below.<br />How many standard deviations is 10 from the mean? 10 – 7 = 3, 3/1.5 = 2 (your answer)<br />How many standard deviations is 6.25 from the mean? 6.25 – 7 = - .75, - .75/1.5 = -0.5 (your answer)<br />Final Exam Review<br />
13. 13. Be able to use the Standard Normal Distribution Tables or <br />Minitab to find probability values and z scores.<br />Examples:<br />Find the following probability involving the Standard Normal Distribution.  What is P(z<1.55)?<br />.9394 (Use Minitab)<br />Find the following probability involving the Standard Normal Distribution.  What is P(z > -.60)?<br />1 – 0.2743 = 0.7257 (Use Minitab)<br />Final Exam Review<br />
14. 14. The mean number of teachers in a Virginia public school is said to be 42.7. A hypothesis test is performed at a level of significance of 0.05 and a P-value of .06. How would you interpret this?<br />Final Exam Review<br />
15. 15. The mean number of teachers in a Virginia public school is said to be 42.7. A hypothesis test is performed at a level of significance of 0.05 and a P-value of .06. How would you interpret this?<br />Fail to reject the null hypothesis, because there is not enough evidence to reject the claim that there are 42.7 teachers per school.<br />Final Exam Review<br />
16. 16. I am buying parts for a new project. I have two vendors to choose from. Vendor X has a customer satisfaction rating of 8.7 with a standard deviation of 1.9. Vendor Y has a customer satisfaction rating of 8.6 with a standard deviation of 0.2 Which should I choose?<br />Final Exam Review<br />
17. 17. I am buying parts for a new project. I have two vendors to choose from. Vendor X has a customer satisfaction rating of 8.7 with a standard deviation of 1.9. Vendor Y has a customer satisfaction rating of 8.6 with a standard deviation of 0.2 Which should I choose?<br />I think I would go with Vendor Y who seems to be more consistent (smaller standard deviation)<br />Final Exam Review<br />
18. 18. I am playing a game that has four different outcomes in terms of how much money I could win. Determine my expected gain if I played this game 5 times.<br />Outcomes/Probability \$10 (10%), \$6 (20%), \$2 (30%), \$1 (40%)<br />Final Exam Review<br />
19. 19. I am playing a game that has four different outcomes in terms of how much money I could win. Determine my expected gain if I played this game 5 times.<br />Outcomes/Probability \$10 (10%), \$6 (20%), \$2 (30%), \$1 (40%)<br />10(0.10)+6(0.20) + 2(0.30) + 1 (0.40) = \$3.20 (this would be the expected gain for playing the game once)<br />For five times it would be 5(\$3.20) = \$16.00<br />Final Exam Review<br />
20. 20. How would you describe the following stem and leaf plot?<br />2| 5<br />3| 2<br />4| 1899<br />5| 13668<br />6| 0227<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
21. 21. How would you describe the following stem and leaf plot?<br />2| 5<br />3| 2<br />4| 1899<br />5| 13668<br />6| 0227<br />Skewed to the Left<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
22. 22. Determine the minimum required sample size if you want to be 90% confident that the sample mean is within 5 units of the population mean given sigma = 8.4.  Assume the population is normally distributed.<br />Final Exam Review<br />
23. 23. Determine the minimum required sample size if you want to be 90% confident that the sample mean is within 5 units of the population mean given sigma = 8.4.  Assume the population is normally distributed.<br />n = (Zc*sigma/E)^2 = [(1.645 * 8.4)/ 5]^2 = (2.7636)^2= 7.64= 8 (always round up sample sizes)<br />Final Exam Review<br />
24. 24. Scores on an exam for entering a military training program are normally distributed, with a mean of 60 and a standard deviation of 12. To be eligible to enter, a person must score in the top 15%. What is the lowest score you can earn and still be eligible to enter?<br />mu = 60; sigma = 12we want top 15% or an area greater than 1 - .15 or .85z = 1.04 ---> x = (1.04)(12) + 60 = 72.48 orneed a score of 73 (Round it up)<br />Final Exam Review<br />I showed this on Minitab in a previous lecture.<br />
25. 25. An airplane has 50 passengers. There are 4 celebrities on the plane. How many ways can a reporter choose 3 of these passengers at random and not pick a celebrity?<br />Final Exam Review<br />
26. 26. An airplane has 50 passengers. There are 4 celebrities on the plane. How many ways can a reporter choose 3 of these passengers at random and not pick a celebrity?<br />This is a Combination 46C3 which is 15180<br />Final Exam Review<br />
27. 27. The average (mean) monthly grocery cost for a family of 4 is \$600.  The distribution is known to be “normal” with a standard deviation = 60.  A family is chosen at random.  a) Find the probability that the family’s monthly grocery cost purchases will be between \$550 and \$650.b)Find the probability that the family’s monthly grocery cost purchases will be less than \$700.c) What is the probability that the family’s monthly grocery cost purchases will be more than \$630?<br />Answers follow<br />Final Exam Review<br />
28. 28. Using Tables or Minitab (I used Minitab to show you some previously):a) P(550 < x < 650) = 0.5953 b) P(x < 700) =.9522  c) P(x > 630) = .3085  <br />Final Exam Review<br />Use Minitab, we have done these<br />
29. 29. As an instructor, I have been collecting data to see if I can model a student’s performance on a standardized entrance exam. I determined that the multiple regression equation y = -250+ 16a + 30b, where a is a student’s grade on a quiz, b is the student’s rank on a class list, gives y, the score on a standardized entrance exam. Based on this equation, what would the standardized entrance exam score for a student who makes a 7 on the quiz and had a ranking of 10 be?<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
30. 30. y = -250+ 16a + 30b<br />Substitute 7 for “a” and 10 for “b”y = -250+ 16*7 + 30*10y = -250 + 112 + 300y = -250 + 412y = 162<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
31. 31. Be able to write the null and alternative hypothesis and know which is the claim.<br />Not to be used, posted, etc. without my expressed permission. B Heard<br />
32. 32. A Pizza Delivery Service claims that it will get its pizzas delivered in less than 30 minutes. A random selection of 49 service times was collected, and their mean was calculated to be 28.6 minutes. The standard deviation was 4.7 minutes. Is there enough evidence to support the claim at alpha = .10. Perform an appropriate hypothesis test, showing each important step.  (Note:  1st Step:  Write Ho and Ha; 2nd Step:  Determine Rejection Region; etc.)<br />Answer following chart<br />Final Exam Review<br />
33. 33. Ho: mu >= 30 min.<br />Ha: mu < 30 min. (claim). Therefore, it is a left-tailed test. <br />n=49; x-bar=28.6; s=4.7; alpha=0.10<br />Since alpha = 0.10, then the critical z value will be zc = -1.28<br />since n>30 then s can be used in place of sigma.<br />Standardized test statistic z = (x-bar - mu)/(s/sqrt(n))                                       z = (28.6-30)/(4.7/sqrt(49))                                      z = -2.085<br />since -2.085 < -1.28, we REJECT Ho.  <br />That is, at alpha = 0.10, There is enough evidence to support the<br />Pizza Delivery Service’s claim.<br />(p-value method could have also been used)<br />Final Exam Review<br />
34. 34. A polling company wants to estimate the average amount of contributions to their candidate. For a sample of 100 randomly selected contributors, the mean contribution was \$50 and the standard deviation was \$8.50.(a)  Find a 95% confidence interval for the mean amount given to the candidate (b)  Interpret this confidence interval and write a sentence that explains it.<br />Answer Follows<br />Final Exam Review<br />
35. 35. (a).  Since sample size = n = 100> 30, we can use a z-value.  For a 95% confidence level, z-value = 1.96.  Also, sample mean = xbar = 50; population standard deviation is estimated by sample standard deviation (since n > 30) = s = 8.50 <br />E = z * s / sqrt(n) = 1.96 * 8.50/sqrt(100) = 1.666 <br />xbar + E = 50.00 + 1.67 = 51.67xbar - E = 50.00 - 1.67 = 48.33<br />Thus, 95% confidence interval = (\$48.33,\$51.67)<br />(b)  We are 95% confident that the population mean amount contributed is between \$48.33 and \$51.67<br />Final Exam Review<br />
36. 36. The failure times of a component are listed in hours.  {100, 95, 120, 190, 200, 200,280}.Find the mean, median, mode, variance, and range.Do you think this sample might have come from a normal population?  Why or why not?<br />mean = 169.3<br />median =190<br />mode = 200<br />variance = 4553.6<br />range = 185<br />Doubtful it came from a normal, compare mean, median,<br />mode, etc.<br />Final Exam Review<br />
37. 37. Drop by and see me on Facebook<br />http://www.facebook.com/statcave<br />You can find a link to these charts there!<br />