Please turn off cell phones, pagers, etc. The lecture will begin shortly.

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. • Review Central Limit Theorems (Sections 19.2-3) • Empirical Rules for proportions and • means (Section 19.2-3)

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

4. CLT for sample proportion No Yes many samples one sample sqrt of p×(1-p) / n p No Yes Population: True proportion of “Yes” in the population is p (fixed) Sample proportions: Sample data: The proportion of “Yes” in one sample of size n will be close to p (random) 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

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

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

8. 0.95 mean + 2 SD mean – 2 SD • In a normal distribution, 95% of the observations lie • within two standard deviations of the mean. • 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 = 0.47 1 – 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 = 100 σ = 16 σ / sqrt(n) = 16 / sqrt(25) = 3.2 100 – (3 × 3.2) = 90.4 100 + (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 = 65.5 σ = 2.5 σ / 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.