1. Please turn off cell phones, pagers, etc. The lecture will begin shortly. There will be a quiz at the end of today’s lecture. Friday’s lecture has been canceled.

2. Lecture 31 Today’s lecture will finish material related to Chapter 19. 1.Review Central Limit Theorems (Sections 19.2-3) 2.Empirical Rules for proportions and means (Section 19.2-3)

3. 1. Review of Central Limit Theorems A Central Limit Theorem describes the sampling variability of an estimate over hypothetical repeated samples from the same population. What does a Central Limit Theorem do? Last time, we learned about Central Limit Theorems for a sample proportion a sample mean

4. CLT for sample proportion Population: True proportion of “Yes” in the population is p (fixed) Sample data: p sqrt of p×(1-p) / n The proportion of “Yes” in one sample of size n will be close to p (random) Yes No Yes No one sample many samples The proportions of “Yes” in many samples of size n will be approximately normally distributed with mean = p and SD = square root of p×(1 – p) / n Sample proportions:

5. CLT for sample mean Population: True mean in the population is μ and true standard deviation is σ (fixed) Sample data: μ σ / sqrt(n) The mean and SD in one sample of size n will be close to μ and σ (random) one sample many samples The sample means of many samples of size n will be approximately normal with mean = μ and SD = σ / sqrt(n) μ σ close to μ close to σ Sample means:

6. Points to remember about CLT’s The true value in the population (the proportion or the mean) is fixed The estimated value from a sample (the proportion or the mean) is random The estimated values over many samples are approximately normally distributed This normal distribution is centered at the true population value The SD of this normal distribution depends on n (the size of each sample)

7. 2. Empirical Rules Recall that we had several empirical rules associated with the normal distribution. mean + SDmean - SD 0.68 In a normal distribution, 68% of the observations lie within one standard deviation of the mean.

8. In a normal distribution, 95% of the observations lie within two standard deviations of the mean. mean + 2 SD mean – 2 SD 0.95 In a normal distribution, 99.7% of the observations lie within three standard deviations of the mean.

9. Empirical rules for a proportion Suppose that we take many random samples of size = n from a population with true proportion = p In 68% of the samples, the sample proportion will lie between p – sqrt( p×(1 – p) / n ) and p + sqrt( p×(1 – p) / n ) In 95% of the samples, the sample proportion will lie between p – 2 × sqrt( p×(1 – p) / n ) and p + 2 × sqrt( p×(1 – p) / n ) In virtually all (99.7%) of the samples, the sample proportion will lie between p – 3 × sqrt( p×(1 – p) / n ) and p + 3 × sqrt( p×(1 – p) / n )

10. Example Suppose that the true proportion of support for a candidate among likely voters is.47 or 47%. Suppose we take a sample of 900 likely voters. There’s a 95% chance that the percentage of support in our sample will lie between _______ and _______ (fill in the blanks).

11. Solution First, find the SD of the sample proportion. The SD of the sample proportion is sqrt( p× (1 – p) / n ). p = 0.471 – p = 0.53 p × (1 – p) = 0.47 × 0.53 = 0.2491 p × (1 – p) / n = 0.2491 / 900 = 0.0002768 sqrt( p × (1 – p) / n ) = sqrt( 0.0002768 ) = 0.0166

12. Solution Next, identify the empirical rule. In a normal distribution, 95% lies within two SD’s of the mean. Finally, compute the bounds. two SD’s = 2 × 0.0166 = 0.0332 two SD’s below the mean = 0.47 – 0.0332 = 0.4368 two SD’s above the mean = 0.47 + 0.0332 = 0.5032 There’s a 95% chance that the percentage of support in our sample will lie between _______ and _______. 43.7% 50.3%

13. Another Example Suppose that you flip a fair coin 100 times. It’s virtually certain that the percentage of heads that you see will lie between _______ and _______ (fill in the blanks). Solution Because the coin is fair, the true proportion is p = 0.5. The SD of the sample proportion is sqrt( p × (1 – p) / n ) = sqrt(.5 ×.5 / 100 ) = sqrt(.0025 ) =.05

14. In a normal distribution, virtually all of the observations will lie within three SD’s of the mean. three SD’s = 3 ×.05 =.15 three SD’s below the mean =.5 –.15 =.35 three SD’s above the mean =.5 +.15 =.65 It’s virtually certain that the percentage of heads that you see will lie between _______ and _______. 35%65%

15. Empirical rules for a mean We have been discussing empirical rules for a proportion. But the same principle applies to a mean. In 68% of the samples, the sample mean will lie between μ – σ / sqrt( n ) and μ + σ / sqrt( n ) In 95% of the samples, the sample proportion will lie between μ – 2 × σ / sqrt( n ) and μ + 2 × σ / sqrt( n ) In virtually all (99.7%) of the samples, the sample proportion will lie between μ – 3 × σ / sqrt( n ) and μ + 3 × σ / sqrt( n )

16. Example Scores on the Stanford-Binet IQ test applied to the general population have a mean of 100 and a standard deviation of 16. Suppose we give the test to a sample of 25 individuals. It’s virtually certain that the average test score among the 25 subjects will lie between _______ and _______. Solution μ = 100σ = 16σ / sqrt(n) = 16 / sqrt(25) = 3.2 100 – (3 × 3.2) = 90.4100 + (3 × 3.2) = 109.6 Answer: Between 90.4 and 109.6

17. Another example The average height of American women age 18-24 is 65.5 inches, and the SD is 2.4 inches. Suppose you take a sample of 9 women from this age group. Would it be unusual to get an average height of 70 inches in your sample? Solution μ = 65.5σ = 2.5σ / sqrt(n) = 2.4 / sqrt(9) = 0.8 Three SD’s = 3 ×.8 = 2.4 It’s virtually certain that the sample average will lie between 65.5 – 2.4 = 63.1 and 65.5 + 2.4 = 67.9 inches. Yes, an average height of 70 inches would be very unusual.