This is just a follow up to the Week 8 Lecture with additional explanations on some problems.

1. B Heard Lecture for the Final ExamStatistics For Decision Making Follow-Up With Selected Excel Explanations Note these are not all of the charts originally presented Not to be used, posted, etc. without my expressed permission. B Heard

2. Be able to understand the normal distribution and how it relates to the mean, standard deviation, variance, etc. 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. How many standard deviations is 10 from the mean? 10 – 7 = 3, 3/1.5 = 2 (your answer) How many standard deviations is 6.25 from the mean? 6.25 – 7 = - .75, - .75/1.5 = -0.5 (your answer) Final Exam Review Use Calculator

3. Be able to use the Standard Normal Distribution Tables or Excel to find probability values and z scores. Examples: Find the following probability involving the Standard Normal Distribution. What is P(z<1.55)? .9394 (from the table or use Excel “=NORMDIST(1.55,0,1,TRUE)”) Find the following probability involving the Standard Normal Distribution. What is P(z > -.60)? 1 – 0.2743 = 0.7257 (using table or Excel 1 – “=NORMDIST(-0.6,0,1,TRUE)”) Final Exam Review See Following Charts

4. Explanation Go to Template Site http://highered.mcgraw-hill.com/sites/0070620164/student_view0/excel_templates.html Save “Normal Distribution File” to your computer Open the file and make sure you “Unprotect” as I described in previous presentations Take out all the values in the green cells

5. Explanation I put 0 in for the mean and 1 for the standard deviation. I worked the first one on the top left and the second on the top middle. (Only input into green cells)

6. A researcher is performing a hypothesis test on a claim about a population proportion. Using an alpha = .04 and n = 80, what is the rejection region if the alternate hypothesis is Ha: p > 0.70? Alternate hypothesis test shows that this is a Right Tailed test (since it’s p > 0.70) with a right tail area of .04 (since alpha = .04). Therefore we are going to reject Ho if z > 1.75 (looked in standard normal table to find z score for a probability of 0.96, z of 1.75 was the closest) Final Exam Review Covered in Week 7 Lecture

7. A researcher is performing a hypothesis test on a claim about a population proportion. Using an alpha = .03 and n = 95, what two critical values determine the rejection region if the null hypothesis is: Ho: p = 0.44? Since Ho: p = 0.44, this is a two tailed test. Each tail has an area = .03/2 = .015. The z-values that correspond to this area in the tail is +/- 2.17. (You can see this by finding the z score for either .015 or .985 realizing it’s two tailed) Final Exam Review Covered in Week 7 Lecture

8. A manufacturer claims that the mean lifetime of its computer component is 1100 hours. A buyer’s researcher selects 49 of these components and finds the mean lifetime to be 1105 hours with a standard deviation of 30 hours. Test the manufacturer's claim. Use alpha = .02. Answer on following chart Final Exam Review

9. Ho: mu = 1100 hours (claim); Ha: mu does not = 1100 hours; two tailed test, therefore, .01 is in left tail and .01 is in right tail; thus critical values are ± 2.33; test statistic is z = (xbar - mu) ÷ [sigma ÷ sqrt(n)] = (1105 - 1100) ÷ [30 ÷ sqrt(49)] = 5 ÷ [30 ÷ 7] = 5 ÷ 4.29 = 1.17 which is in the do not reject area because p value corresponding to z= +1.17 is 0.879 Fail to reject Ho. (Because 1.17 is in the bounds of the critical values ± 2.33)There is not enough evidence to reject the manufacturer's claim that the mean lifetime is 1100 hours. Final Exam Review Covered in Week 7 Lecture

10. 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.) Answer following chart Final Exam Review

11. Ho: mu >= 30 min. Ha: mu < 30 min. (claim). Therefore, it is a left-tailed test. n=49; x-bar=28.6; s=4.7; alpha=0.10 Since alpha = 0.10, then the critical z value will be zc = -1.28 since n>30 then s can be used in place of sigma. Standardized test statistic z = (x-bar - mu)/(s/sqrt(n)) z = (28.6-30)/(4.7/sqrt(49)) z = -2.085 since -2.085 < -1.28, we REJECT Ho. That is, at alpha = 0.10, There is enough evidence to support the Pizza Delivery Service’s claim. (p-value method could have also been used) Final Exam Review Covered in Week 7 Lecture

12. 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. n = (Zc*sigma/E)^2 = [(1.645 * 8.4)/ 5]^2 = (2.7636)^2= 7.64= 8 (always round up sample sizes) Final Exam Review Covered in Week 7 Lecture

13. 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? mean = 169.3 median =190 mode = 200 variance = 4553.6 range = 185 Doubtful it came from a normal, compare mean, median, mode, etc. Final Exam Review Use "average", "median", "mode", "var" etc. functions in Excel (Don't forget to visually inspect the mode for multiples)

14. The random variable X represents the annual salaries in dollars for a group of entry level accountants. Find the expected value E(X). X = {$30,000; $38,000; $42,000}. P(30,000) = .2; P(38,000) = .7; P(42,000) = .1 E(X) = 30000*.2 + 38000*.7 + 42000*.1 = $36,800 Final Exam Review Use Calculator

15. 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? Answers follow Final Exam Review

16. Using Tables or EXCEL (I used an Excel Template like one I showed you):a) P(550 < x < 650) = 0.5953 b) P(x < 700) =.9522 c) P(x > 630) = .3085 Final Exam Review See Following Charts, Use Normal Template

17. Explanation I put 600 in for the mean and 60 for the standard deviation. I worked the part a on the top right, part b on the top left and part c on the top middle. (Only input into green cells AND don’t forget to hit the “Enter” key)

18. The earnings per share (in dollars) for The Very Pretty Products Company are given by the equation y-hat = 0.863 + 0.029a - 0.011b where "a" represents total revenue (in billions of dollars) and "b" represents total net worth (in billions of dollars). Predict the earnings per share when total revenue is $6 billion and net worth is $1 billion. y-hat = 0.863 + 0.029a - 0.011b y-hat = 0.863 + 0.029*6 - 0.011*1 y-hat = 1.026 Final Exam Review Use Calculator

19. The time required to produce a product was normally distributed with a mean 10.5 days and a standard deviation of 1.5 days (i.e., 36 hours). What is the probability that it will take more that 11 days to produce the product? We want P(X > 11)z = (x - mu)/sigma= (11 – 10.5)/1.5 = .5/1.5 = .33P(z > .33) = 1 - P(z < .33)= 1 – 0.6293= 0.3707 Final Exam Review See Following Charts, Use Normal Template

20. Explanation I put 0 in for the mean and 1 for the standard deviation. Remember I found the z in question to be “0.33” Result is shown on middle top. In thinking about this one, I would have gotten better results by using “0.3333” (not rounding so much). I could have also used the mean of 10.5 and the standard deviation of 1.5 and found the probability of it being greater than 11. This would have given me 0.3694 rather than the 0.3707 noted. In other words, be careful not to round too much.

21. A shipment of 30 Widgit’s contains 7 defective Widgits. How many ways can a Widget buying company buy 3 of these units and receive no defective units? There are 23 Widgets which are not defective.Thus there are 23C3 ways to get 3 sets of Widgets with none defective.23C3 = 1771 (using “combin” function in Excel) Final Exam Review Use Excel

22. For the following statement, write the null hypothesis and the alternative hypothesis. Then, label the one that is the claim being made.A car manufacturer claims that the mean life time of a car is more than 7 years. Ho: mu <= 7 yearsHa: mu > 7 years (claim) Remember that Ho always contains equality and the claim can be either Ho or Ha Final Exam Review Self-explanatory

23. 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? Poisson with average of 3. want P(2) P(2) = .2240 (Use Excel or Excel Template) Final Exam Review Use Poisson Template as described in previous lecture

24. 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? a. Is this a binomial experiment? 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. Answer Follows Final Exam Review

25. 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. b) n = 50; p = .40 = P(success) = house has a dog We want P(22) --> P(22) = .0959 (I used a Binomial Template) Final Exam Review Use Binomial Template as described in previous lecture

26. The stem and leaf plot for the following data is displayed below: {11,11,12,16,17,24,25,26,26,34,36,37,44,46,51,53,62} Stem and Leaf Plot:1|112672|45663|4674|465|136|2 Discuss the shape of the data distribution. It is right skewed. (other choices would be left skewed, symmetric, etc). This one is right skewed because if you turned it 90 degrees counter clockwise the “tail” would be on the right. Final Exam Review By hand

27. 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? 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) Final Exam Review Use Normal Template Explanation follows

28. Explanation I put 60 in for the mean and 12 for the standard deviation. Put in 0.15 for the bottom middle to get 72.44 (remember to round up to 73).

29. 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. Answer Follows Final Exam Review

30. (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 E = z * s / sqrt(n) = 1.96 * 8.50/sqrt(100) = 1.666 xbar + E = 50.00 + 1.67 = 51.67xbar - E = 50.00 - 1.67 = 48.33 Thus, 95% confidence interval = ($48.33,$51.67) (b) We are 95% confident that the population mean amount contributed is between $48.33 and $51.67 Final Exam Review Use confidence template from lab (See next chart)

31. Explanation From part 3 of the Week 6 lab (Excel provided there – tab labeled “Conf Intervals”) Put in 95% (or your value may already be there) and see the z-score. Then use formula to complete.

32. Private Johnson earned a 78 on his history test and 82 on his math test. In the history class the mean score was 79 with standard deviation of 6. In the math class the mean score was 84 with standard deviation of 5. Convert each score to a standard score (Z score) Which score was the higher with respect to the rest of the class? Show your work and explain your answer. history: mu = 79; sigma = 6; z = -0.167 (78-79/6) math: mu = 84; sigma = 5; z = -0.400 (82-84/5) thus an 78 in history is better than an 82 in math. (because -0.167 is greater than -0.400) Final Exam Review By hand

33. Last Chart Please forgive me for any typo’s I wish you the best!!! Please stay with me in the “Stat Cave” so you can be an inspiration to future students and let me know how you are doing. “How do I stop this recording?” (Please note these are only charts and there is no audio)