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AP Statistics Monday, 24 March 2014

AP Statistics Monday, 24 March 2014. OBJECTIVE TSW investigate properties and characteristics of Sampling Distributions of Proportions. Read pp. 461-466. Everyone needs a calculator. E A A C B C A. B A A B A A D C. TEST: Inference with Means and Errors. E E A B E D B.

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AP Statistics Monday, 24 March 2014

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  1. AP StatisticsMonday, 24 March 2014 • OBJECTIVETSW investigate properties and characteristics of Sampling Distributions of Proportions. • Read pp. 461-466. Everyone needs a calculator.

  2. E A A C B C A B A A B A A D C TEST: Inference with Means and Errors • E • E • A • B • E • D • B • C • C • A • E • D • B • C • A • C • A • C • E • A

  3. Sampling Distributions of Proportions

  4. The dotplot is a partial graph of the sampling distribution of all sample proportions of sample size 20. If I found all the possible sample proportions – this would be approximately normal! • Toss a penny 20 times and record the number of heads. • Calculate the proportion of heads & mark it on the dot plot on the board. What shape do you think the dot plot will have?

  5. Sampling Distribution • Is the distribution of possible values of a statistic from all possible samples of the same size from the same population • In the case of the pennies, it’s the distribution of all possible sample proportions (p) We will use: p for the population proportion and p-hat for the sample proportion

  6. Suppose we have a population of six people: Alice, Ben, Charles, Denise, Edward, & Frank What is the proportion of females? What is the parameter of interest in this population? Draw samples of two from this population. How many different samples are possible? 1/3 gender 6C2 =15

  7. Alice & Ben 0.5 Alice & Charles 0.5 Alice & Denise 1 Alice & Edward 0.5 Alice & Frank 0.5 Ben & Charles 0 Ben & Denise 0.5 Ben & Edward 0 Ben & Frank 0 Charles & Denise 0.5 Charles & Edward 0 Charles & Frank 0 Denise & Edward 0.5 Denise & Frank 0.5 Edward & Frank 0 Find the 15 different samples that are possible & find the sample proportion of the number of females in each sample. How does the mean of the sampling distribution (mp-hat) compare to the population parameter (p)? mp-hat = p Find the mean & standard deviation of all p-hats.

  8. AP StatisticsTuesday, 25 March 2014 • OBJECTIVETSW investigate properties and characteristics of Sampling Distributions of Proportions. • Read pp. 461-466. Everyone needs a calculator.

  9. These are found on the (yellow) formula chart! Formulas:

  10. Correction factor – multiply by Does the standard deviation of the sampling distribution equal the equation? NO - So – in order to calculate the standard deviation of the sampling distribution, we MUST be sure that our sample size is less than 10% of the population! WHY? We are sampling more than 10% of our population! If we use the correction factor, we will see that we are correct.

  11. Assumptions (Rules of Thumb) • Sample size must be less than 10% of the population (independence) • Sample size must be large enough to insure that a normal approximation can be used. np > 10 & n (1 – p) > 10

  12. Why does the second assumption insure an approximate normal distribution? Remember back to binomial distributions Suppose n = 10 & p = 0.1(probability of a success), a histogram of this distribution is strongly skewed right! np > 10 & n(1-p) > 10 insures that the sample size is large enough to have a normal approximation! Now use n = 100 & p = 0.1 (Now np > 10!) While the histogram is still strongly skewed right – look what happens to the tail!

  13. Suppose one student tossed a coin 200 times and found only 42% heads. Do you believe that this is likely to happen? np = 200(.42) = 84 & n(1-p) = 200(.58) = 116 Since both > 10, I can use a normal curve! Find m & s using the formulas. No – since there is approximately a 1% chance of this happening, I do not believe the student did this.

  14. Based on past experience, a bank believes that 7% of the people who receive loans will not make payments on time. The bank recently approved 200 loans. What are the mean and standard deviation of the proportion of clients in this group who may not make payments on time? Are assumptions met? What is the probability that over 10% of these clients will not make payments on time? Yes – np = 200(.07) = 14 n(1 - p) = 200(.93) = 186 Ncdf(0.10, 1E99, 0.07, 0.01804) = 0.0482

  15. Assume that 30% of the students at JVHS wear contacts. In a sample of 100 students, what is the probability that more than 35% of them wear contacts? Check assumptions! mp-hat = 0.3 & sp-hat = 0.045826 np = 100(0.3) = 30 & n(1-p) =100(0.7) = 70 Ncdf(0.35, 1E99, 0.3, 0.045826) = 0.1376

  16. Assignment • Read pp. 461-466. • WS Sampling Distributions of Proportions • Due on Thursday, 27 March 2014.

  17. Inference on Proportions

  18. Assumptions: • SRS • Normal distribution np> 10 & n(1 – p) > 10 • Population is at least 10n

  19. Normal curve Formula for Confidence interval: Note: For confidence intervals, we DO NOT know p – so we MUST substitute p-hat for pin both the SD & when checking assumptions.

  20. A May 2000 Gallup Poll found that 38% of a random sample of 1012 adults said that they believe in ghosts. Find a 95% confidence interval for the true proportion of adults who believe in ghosts.

  21. Assumptions: • Have an SRS of adults • np =1012(0.38) = 384.56 andn(1 – p) = 1012(0.62) = 627.44 Since both are greater than 10, the distribution can be approximated by a normal curve • Population of adults is at least 10,120. Step 1: check assumptions! Step 2: make calculations We are 95% confident that the true proportion of adults who believe in ghost is between 35% and 41%. Step 3: conclusion in context

  22. AP StatisticsThursday, 27 March 2014 • OBJECTIVETSW create confidence intervals with proportions. • ASSIGNMENTS DUE BEFORE YOU LEAVE TODAY • WS Sampling Distributions of Proportions wire basket • WS Confidence Intervals for Proportions black tray • BOOKWORK is due on Tuesday, 01 April 2014. • QUIZ: One-Sample Inference Test will be given at the end of today’s lesson. • 6th PERIOD • POSSIBLE RELOCATION (Due to STAAR Testing) • Tuesday, 01 April 2014: Room 2264

  23. To find sample size: However, since we have not yet taken a sample, we do not know a p-hat (or p) to use! Another Gallop Poll is taken in order to measure the proportion of adults who approve of attempts to clone humans. What sample size is necessary to be within + 0.04 of the true proportion of adults who approve of attempts to clone humans with a 95% Confidence Interval?

  24. 0.1(0.9) = 0.09 0.2(0.8) = 0.16 0.3(0.7) = 0.21 0.4(0.6) = 0.24 0.5(0.5) = 0.25 By using 0.5 for p-hat, we are using the worst-case scenario and using the largest SD in our calculations. What p-hat (p) do you use when trying to find the sample size for a given margin of error?

  25. Another Gallop Poll is taken in order to measure the proportion of adults who approve of attempts to clone humans. What sample size is necessary to be within + 0.04 of the true proportion of adults who approve of attempts to clone humans with a 95% Confidence Interval? Use p-hat = 0.5 Divide by 1.96 Square both sides Round up on sample size

  26. Hypotheses for proportions: • Sometimes π will be used instead of p for the true proportion. H0: p = value Ha: p > value where p is the true proportion of context Use >, <, or ≠

  27. Formula for hypothesis test:

  28. A company is willing to renew its advertising contract with a local radio station only if the station can prove that more than 20% of the residents of the city have heard the ad and recognize the company’s product. The radio station conducts a random sample of 400 people and finds that 90 have heard the ad and recognize the product. Is this sufficient evidence for the company to renew its contract?

  29. Assumptions: • Have an SRS of people • np = 400(0.2) = 80 & n(1 – p) = 400(0.8) = 320 - Since both are greater than 10, this distribution is approximately normal. • Population of people is at least 4000. Use the parameter in the null hypothesis to check assumptions! H0: p = 0.2 where p is the true proportion of people who Ha: p > 0.2 heard the ad Use the parameter in the null hypothesis to calculate standard deviation! Since the p-value > a, I fail to reject the null hypothesis. There is not sufficient evidence to suggest that the true proportion of people who heard the ad is greater than 0.2.

  30. WS Confidence Intervals for Proportions 2) Assumptions: • Have SRS of adult drivers • Since np =361.44 & n(1-p) = 642.56 are both > 10, the distribution is approximately normal • Population of adult drivers is at least 10,040. We are 98% confident that the true proportion of adult drivers who have often or sometimes talked on a cell phone when driving is between 32.476% & 39.524%.

  31. WS Confidence Intervals for Proportions 4) p-hat = 68/300 = 0.22667 (REMEMBER: In theory we don’t know p so we MUST use p-hat.) np = 300(0.22667) = 68 ≥ 10 n(1 – p) = 300(1 – 0.22667) = 232 ≥ 10  approximately normal There are at least 3000 households in Jersey Village 95% CI = (17.929%, 27.404%) No, since the interval contains the national proportion of 25%, there is no evidence to indicate that the proportion of households who experience some sort of crime is less in Jersey Village.

  32. Assignments Make sure you turn inWS Sampling Distributions of Proportions and WS Confidence Intervals for Proportionsbefore you leave! • WS One-Sample Proportions • Due on Friday, 28 March 2014. • WS AP Review – Mixed • Due on Monday, 31 March 2014. CHECKERBOARD ! ! ! • QUIZ: One Sample Inference with Proportions • You may use notes, but no worksheets.

  33. AP StatisticsFriday, 28 March 2014 • OBJECTIVETSW demonstrate comprehension by completing worksheets about One-Sample Inference. • ASSIGNMENT DUE • WS One-Sample Proportions wire basket • Bookwork is due on Tuesday, 01 April 2014. • RELOCATION NEXT TUESDAY, 01 APRIL 2014 • 4th Period: Room 2266 • Possibly 6th Period: Room 2264 • TODAY’S ASSIGNMENT • WS One-Sample Inference Mixed Review due before you leave today.  black tray Water for sale ! ! !

  34. WS Sampling Distributions of Proportions • a) b)np = 80(0.7) = 56 ≥ 10 n(1 – p) = 80(1 – 0.7) = 24 ≥ 10 Yes, both np and n(1 – p) are ≥ 10 c) d) e) 95th percentile: z = 1.645 2) 0.0809430 3) 0.0878139 4) 0.150490

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