Week 7 lecture_math_221_apr_2012

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Week 7 Lecture for Math 221

Week 7 Lecture for Math 221

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  • 1. B. Heard (This material can not be copied or postedwithout the author’s consent. Students may download one copy for personal use.)
  • 2.  Standard Normal Distribution  The “standard” normal distribution is a normal distribution with mean zero and where the standard deviation (and variance) equals one.  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).
  • 3.  Using Minitab for Normal Distribution calculations. Use Calc >> Probability Distributions >> Normal Examples Follow
  • 4.  Example Theaverage 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?
  • 5. Since this is the CumulativeDistribution Function, it “fills”from left to right.Therefore, you could say hiscatch was in the “Top 1 %”
  • 6.  Example Theaverage 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?
  • 7. Since this is the CumulativeDistribution Function, it “fills”from left to right.Therefore, you could say hiscatch was in the “Bottom 10 %”
  • 8.  Other types of questions If you have a normal distribution with a mu = 100 and sigma = 15, what number corresponds to a z = -2 -2 = (x – 100)/15 Multiply both sides by 15 to get -30 = x – 100 Add 100 to each side to get 70 = x So “70” is my answer, I just did a little Algebra.
  • 9.  Another type of question 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. (Answer next chart)
  • 10.  Answer 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. Variance = 25/81 or 0.309 Standard Deviation = Square Root(25/81) = 5/9 = 0.556
  • 11.  Findingz scores Example  The area to the left of the “z” is 0.6262. What z score corresponds to this area.  Use Calc >> Probability Distributions >> Normal  (Set Mean = 0 and Standard Deviation to 1 and use “INVERSE Cumulative Probability”
  • 12. Answer is 0.322rounded to threedecimals. Rememberthe distribution fillsfrom left to right.
  • 13.  Another type of question In a normal distribution with mu = 40 and sigma = 10 find P(32 < x < 44)  Easy, but this takes a couple of steps.  Using Calc >> Probability Distributions >> Normal find the probabililties that x < 32 and x < 44 using the Cumulative Probability option.
  • 14.  Continued Get results for Both 32 and then 44.
  • 15.  Answer Subtract 0.655422 – 0.211855 To get 0.443567 Or 0.444 rounded to three decimals P(32 < x < 44) = 0.444 based on the given mean and std deviation.
  • 16.  Confidence Intervals and Examples  Charts follow
  • 17.  Interpreting Confidence Intervals  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)  It’s really that simple.
  • 18.  Finding Confidence Intervals  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.
  • 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… Use Stat >> Basic Statistics >> 1 sample Z  Make sure to click Options and set to 95%
  • 20.  Click your OK buttons… 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.
  • 21.  FindConfidence Intervals of Proportions Example  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.
  • 22.  Example Solution  p hat = 120/200 = 0.60  q hat = 1- 0.60 = 0.40  n p hat = 200 * 0.60 = 120  n q hat = 200 * 0.40 = 80  Using E = Zc* Square Root ((p hat * q hat)/n) = 1.96 * Square Root ((0.60*0.40)/200) =0.0679 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) 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%.
  • 23.  Link to charts will be posted at  www.facebook.com/statcave PLEASENOTE THAT I WILL BE BACK HERE NEXT SUNDAY NIGHT FOR A BONUS LECTURE TO HELP YOU PREPARE FOR THE FINAL EXAM. ENTER IN THE WEEK 7 iConnect AREA JUST LIKE YOU DID TONIGHT.