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Week 7 Lecture for help on the Week 7 Quiz.

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- 1. Week 7 LectureStatistics for Decision Making<br />B. Heard<br />(This material can not be copied or posted without the author’s consent. Students may download one copy for personal use.)<br />
- 2. Standard Normal Distribution<br />The “standard” normal distribution is a normal distribution with mean zero and where the standard deviation (and variance) equals one.<br />The Total Area under the curve is one (1) or 100% (This is true for all normal distributions regardless of the mean and standard deviation).<br />Week 7 Lecture<br />
- 3. Using Minitab for Normal Distribution calculations.<br />Use Calc >> Probability Distributions >> Normal<br />Examples Follow<br />Week 7 Lecture<br />
- 4. Example<br />The average fish in Happy Lake weighs 2 pounds with a standard deviation of 0.5 pounds. If Bob catches a fish that weighs 3.2 pounds. What could you say about the catch?<br />Week 7 Lecture<br />
- 5. Week 7 Lecture<br />Since this is the Cumulative Distribution Function, it “fills” from left to right. Therefore, you could say his catch was in the “Top 1 %”<br />
- 6. Example<br />The average fish in Happy Lake weighs 2 pounds with a standard deviation of 0.5 pounds. If Bob catches a fish that weighs 1.35 pounds. What could you say about the catch?<br />Week 7 Lecture<br />
- 7. Week 7 Lecture<br />Since this is the Cumulative Distribution Function, it “fills” from left to right. Therefore, you could say his catch was in the “Bottom 10 %”<br />
- 8. Other types of questions<br />If you have a normal distribution with a mu = 100 and sigma = 15, what number corresponds to a z = -2<br />Week 7 Lecture<br />-2 = (x – 100)/15<br />Multiply both sides by 15 to get<br />-30 = x – 100<br />Add 100 to each side to get<br />70 = x<br />So “70” is my answer, I just did a little Algebra.<br />
- 9. Another type of question<br />Say we take 120 samples of size 81 each from a distribution we know is normal. Calculate the standard deviation of the sample means if we know the population variance is 25.<br />(Answer next chart)<br />Week 7 Lecture<br />
- 10. Answer<br />The Central Limit Theorem tells us the variance is the Population variance divided by the Sample Size. We can just take the square root to get the standard deviation.<br />Week 7 Lecture<br />Variance = 25/81 or 0.309<br />Standard Deviation = Square Root(25/81) = 5/9 = 0.556<br />
- 11. Finding z scores<br />Example<br />The area to the left of the “z” is 0.6262. What z score corresponds to this area.<br />Use Calc >> Probability Distributions >> Normal<br />(Set Mean = 0 and Standard Deviation to 1 and use “INVERSE Cumulative Probability”<br />Week 7 Lecture<br />
- 12. Week 7 Lecture<br />Answer is 0.322 rounded to three decimals. Remember the distribution fills from left to right.<br />
- 13. Another type of question<br />In a normal distribution with mu = 40 and sigma = 10 find P(32 < x < 44)<br />Easy, but this takes a couple of steps.<br />Using Calc >> Probability Distributions >> Normal find the probabililties that x < 32 and x < 44 using the Cumulative Probability option.<br />Week 7 Lecture<br />
- 14. Continued<br />Week 7 Lecture<br />Get results for <br />Both 32 and then 44.<br />
- 15. Answer<br />Week 7 Lecture<br />Subtract <br />0.655422 – 0.211855<br />To get<br />0.443567<br />Or 0.444 rounded to<br />three decimals<br />P(32 < x < 44) = 0.444 based on<br />the given mean and std deviation.<br />
- 16. Confidence Intervals and Examples<br />Charts follow<br />Week 7 Lecture<br />
- 17. Interpreting Confidence Intervals<br />If you have a 90% confidence interval of (15.5, 23.7) for a population mean, it simply means “There is a 90% chance that the population mean is contained in the interval (15.5, 23.7) <br />It’s really that simple.<br />Week 7 Lecture<br />
- 18. Finding Confidence Intervals<br />A luxury car company wants to estimate the true mean cost of its competitor’s automobiles. It randomly samples 180 of its competitors sticker prices. The mean cost is $65,000 with a standard deviation of $3200. Find a 95% confidence interval for the true mean cost of the competitor’s automobiles. Write a statement about the interval.<br />Week 7 Lecture<br />
- 19. It randomly samples 180 of its competitors sticker prices. The mean cost is $65,000 with a standard deviation of $3200. Find a 95% confidence interval…<br />Use Stat >> Basic Statistics >> 1 sample Z<br />Make sure to click Options and set to 95%<br />Week 7 Lecture<br />
- 20. Week 7 Lecture<br />
- 21. Click your OK buttons…<br />Week 7 Lecture<br />Confidence Interval is (64533, 65467), which means we can be 95% confident the true mean cost of the competitor’s vehicles are between those two values.<br />
- 22. Find Confidence Intervals of Proportions<br />Example<br />An student wants to estimate what proportion of the student body eats on campus. The student randomly samples 200 students and finds 120 eat on campus. Using a 95% confidence interval, estimate the true proportion of students who eat on campus. Write a statement about the confidence level and interval.<br />Week 7 Lecture<br />
- 23. Example Solution<br />p hat = 120/200 = 0.60<br />q hat = 1- 0.60 = 0.40<br />n p hat = 200 * 0.60 = 120<br /> n q hat = 200 * 0.40 = 80<br />Using E = Zc* Square Root ((p hat * q hat)/n)<br /> = 1.96 * Square Root ((0.60*0.40)/200)<br />=0.0679<br />Now we subtract this from the mean for the left side of the interval and add it to the mean for the right side. (0.60 – 0.0679, 0.60 + 0.0679) = (0.5321, 0.6679) <br />So with 95% confidence, we can say the population proportion of students who eat lunch on campus is (0.5321, 0.6679) or between 53.21% and 66.79%.<br />Week 7 Lecture<br />
- 24. Link to charts will be posted at<br />www.facebook.com/statcave<br />PLEASE NOTE THAT I WILL BE BACK HERE NEXT SUNDAY NIGHT FOR A BONUS LECTURE TO HELP YOU PREPARE FOR THE FINAL EXAM.<br />ENTER IN THE WEEK 7 iConnect AREA JUST LIKE YOU DID TONIGHT.<br />Week 7 Lecture<br />

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