1. Sample Means: Target Goal: I can find the mean and standard deviation of the sampling distribution. 7.3a h.w: pg 441:43 – 46; pg 454: 49, 51, 53, 55

2. Because sample means are just averages of observations, they are among the most common statistic. Bull Market or Bear Market Rates of return for NYSE in 1987. What do you notice compared to the next “portfolio”?

3. Rates of return for NYSE in 1987 Data is wide spread and variability is high. Rates of return for “5 stock portfolios” Data is less spread with less variation. Sample means are: Averages are more normal than individual observations. Averages are less variable than individual observations.

4. The Mean and Standard Deviation of the Sample Mean Suppose that is the mean of an SRS of size n drawn from a large population with mean μ and standard deviation σ.

5. Note: We can only use when the population is at least 10 times the sample. (This is almost always the case.)

6. The Shape of the distribution of depends on the shape of the distribution. If drawn from a population that is normal then: Sample mean has:

7. Note: the sample mean will not be as exact as the population mean because of sampling variability.

8. Example : Young Woman’s Heights The heights of young women vary according to N(64.5, 2.5). This is the distribution of one individual chosen at random. Take a SRS of 10 women.

9. Find the Mean and Standard Deviation = 64.5 The average height of individual women vary widely (2.5inches) about the population mean. But, the average height of a sample of 10 women is less variable (.79 inches).

10. What if n = 100? = 0.25 The average height of a sample of 100 women is has even less variable.

11. What is the probability that a randomly selected woman is taller than 66.5 inches? Let x = the height of a randomly selected women. This follows a N(64.5,2.5) We want: P(X > 66.5)

12. Standardize: P(X > 66.5) = 0.80 = 1 – normcdf(-E99,0.80) = 1 -.7881 = 0.2119

13. What is the probability that a SRS of 10 women has a mean height greater than 66.5? We want : P( > 66.5) for n = 10 We know that

14. Standardize Sample Mean: z = 2.53 P(x > 66.5) = P(Z>2.53) = 1 – P(Z≤2.53) = 1 – normcdf(-E99,2.53) = 0.0057 Or normcdf(2.53,E99)

15. The probability is much less likely (less than 1% chance) compared to the individual probability which we calculated to be approx, 21% to have a height > 66.5. 1% 21%

16. Conclusion: The averages of several observations are less variable than individual observations. So, the sample mean is less variable than a single measurement.

17. Receive CLT hand out and discuss tomorrow. Work on carnival calculations or hw the rest of class.