Chapter 23: accuracy of averages

Transcription

1 Chapter 23: accuracy of averages Context: previous chapters Context: previous chapters Context: Ch Expected value and SE for average 5 Example ,000 repetitions on computer Strategy Example Formulas for average Normal approximation What happens if we increase the number of draws? Compare sample size 25 and 100, for 10,000 repetitions Not knowing the box 14 Example Solution part Solution part Solution part Interpretation Interpretation Which formula to use? 21 Which SE? Warnings 23 Warnings

2 Context: previous chapters Chapter 20: we looked at chance variability in sampling, and learned formulas for the expected value and SE for a percentage. Chapter 21: we dropped the assumption that we knew the contents of the box (= population). We looked at estimating the population percentage: The problem is about classifying, so we change the box into a 0-1 box Estimate for population percentage: sample percentage Estimate is likely to be off by SE for percentage: Estimate SE for count, using bootstrap method Convert to SE for percentage 2 / 24 Context: previous chapters Confidence intervals for the population percentage: 68% confidence interval: sample percentage ± 1 SE for percentage 95% confidence interval: sample percentage ± 2 SEs for percentage 99.7% confidence interval: sample percentage ± 3 SEs for percentage What is the correct interpretation of a 95% confidence interval? We are 95% confident that the interval captures the sample percentage Or: We are 95% confident that the interval captures the population percentage 3 / 24 Context: Ch 23 We will learn formulas for the expected value and the SE for the sample average, if we know the box. We then drop the assumption that we know the contents of the box. From the sample, we want to learn about the box (= population). We look at estimating the population average: The problem is not about classifying and counting, so we do not change the box in a 0-1 box. Estimate for population average: sample average Estimate is likely to be off by SE for average: Estimate SE for sum, using bootstrap method Convert to SE for average Confidence intervals 4 / 24 2

3 Expected value and SE for average 5 / 24 Example 25 draws are made with replacement from a box with the following tickets: 1, 2, 3, 4, 5, 6, 7 The sum of the draws will be around..., give or take... or so. Fill in the blank. The average of the draws will be around..., give or take... or so. Fill in the blank. See computer simulation 6 / 24 10,000 repetitions on computer... sum of the draws, sample size = 25 average of the draws, sample size = sum of the draws average of the draws 7 / 24 3

4 Strategy We first look at the sum of the draws. Average of the box: = 4 SD of the box: Note that we cannot use the shortcut formula! So compute deviations from the average: -3, -2, -1, 0, 1, 2, 3 SD is r.m.s. size of the deviations: ( 3) 2 + ( 2) 2 + ( 1) = 2 Expected value for sum of the draws: (nr of draws) (average of the box) = 25 4 = 100 SE for sum of the draws: square root formula: nr of draws (SDof box) = 25 2 = 10 8 / 24 Example The sum of the draws is around 100, give or take 10 or so. Convert to averages: Expected value for average of the draws: = 4 This is just the average of the box! SE for average of the draws: = 0.4 So the average of the draws is around 4, give or take 0.4 or so. Note that this matches the histograms on slide 7. 9 / 24 Formulas for average Expected valuefor average = averageof thebox SEfor sum SEfor average = nr of draws These formulas are exact if we draw with replacement If the sample is small compared to the population (say, less then 1/10th), then the formulas are good approximations for drawing without replacement. Why? 10 / 24 4

5 Normal approximation Central limit theorem works for sums and averages of draws That s why the histograms on slide 7 follow the normal curve Hence, we can use the normal curve to approximate probabilities Example: Estimate the chance that the average of the draws is more than 4.4. Method: New average: Expected value for average of the draws = 4 New SD: SE for the average of the draws = 0.4 Then use normal approximation as usual, see overhead. Answer is 16%. 11 / 24 What happens if we increase the number of draws? SEfor average = = = SEfor sum nr of draws nr of draws (SDof box) nr of draws SDof box nr of draws Suppose we multiply the number of draws by a factor 4, so that we now make 100 draws. Then the SE for the average of the draws is divided by a factor 4 = 2. With a larger sample size we can estimate the population average more accurately. See computer simulations on next slide. 12 / 24 5

6 Compare sample size 25 and 100, for 10,000 repetitions... sum of the draws, sample size = 25 average of the draws, sample size = sum of the draws average of the draws sum of the draws, sample size = 100 average of the draws, sample size = sum of the draws average of the draws 13 / 24 Not knowing the box 14 / 24 Example Example 3 on page 417 We take a simple random sample of 400 persons age 25 and over is taken in a certain town. Their average educational level is 11.6 years. The SD of the educational level in the sample is 4.1 years. If possible, find a 68% confidence interval for the average educational level of all persons age 25 and over in this town. 15 / 24 6

7 Solution part 1 Identify population, parameter, sample, statistics: Population: all persons age 25 and over in the town Parameter: average educational level in years Sample: simple random sample of 400 persons age 25 and over in the town Statistic: average educational level of people in the sample: 11.6 years Make box model, see overhead The average educational level in the sample is like the average of 400 draws without replacement from the box. 16 / 24 Solution part 2 Fill in the blanks: The average educational level is estimated to be... This estimate is likely to be off by... or so. First blank: estimate for population average = sample average = 11.6 years Second blank: SE for average = (SE for sum)/(nr of draws) Since the sample is small relative to the population, we can use the formulas for drawing with replacement So the SE for sum = nr of draws (SD of the box) But we don t know the SD of the box! Since the sample is reasonably large, we can use the bootstrap method: SD of box SD of sample = 4.1 years. SE for sum of draws = = 82 SE for average = 82/400 = / 24 7

8 Solution part 3 The average educational level is estimated to be This estimate is likely to be off by 0.2 years or so. Construct a confidence interval: If the normal approximation works, a 68% confidence interval for the population average is: sample average ± 1 SE for average = 11.6 ± 0.2. Education in town does not have a normal distribution, see overhead. So can we use the normal approximation? Yes! The distribution of the educational level in the town does not follow the normal curve. But because of the central limit theorem, the distribution of sample averages does follow the normal curve quite closely. 18 / 24 Interpretation 68% confidence interval for population average: (11.4, 11.8) Which of the following is/are true: About 68% of the people in the town have an educational level between 11.4 and 11.8 years. We are about 68% confident that the interval (11.4, 11.8) contains the population average. 19 / 24 Interpretation What does it mean: we are about 95% (or 68% or..%) confident that the interval contains the population parameter? Remember that the population parameter is a fixed number Each time we take a different sample, we get a different sample average, and also a different estimate for the SE So each time we take a different sample, the confidence interval turns out a bit different See computer simulation If we repeat this many many times, then 95% of the confidence intervals will contain the true population percentage, and 5% won t. Difficulty: after computing a confidence interval, we don t know if it is one that contains the true parameter or not. 20 / 24 8

9 Which formula to use? 21 / 24 Which SE? See page 422 of the book Four operations to think about: sum of the draws: SE for sum = nr of draws (SD of box) average of the draws: SE for average = SE forsum nrof draws classifying and counting: SE for count = SE for sum, from a 0-1 box taking percentages: SE for percentage= SE forsum, from0 1box nrof draws 100% In each case, the SE shows the likely amount off. The book calls this a give or take number. 22 / 24 Warnings 23 / 24 Warnings The methods in this chapter only work for simple random samples For more complicated sampling methods like cluster sampling, we need more complicated formulas For non-probability sampling methods, we basically have no formulas The sample size should be small relative to the population (say < 1/10th), so that we can ignore that we draw without replacement For the bootstrap method to work, the sample size should be reasonably large For the normal approximation to work, the sample size should be large. The more asymmetric the box is, the larger the sample size we need. 24 / 24 9