1. Warm Up The average amount of meat that an American consumes per year is 218.4 lbs. Assume that the standard deviation is 25 and that the distribution of meat eaten by Americans is approximately normal. If an individual is selected at random, find the probability that he or she consumes less than 210 lbs of meat per year.

2. Ch. 18 – Sampling Distribution Models (Day 2 –Sample Means) Part V – From the Data at Hand to the World at Large

3. Sample Means Yesterday, we examined the sampling distribution of sample proportions Today, we will look at a similar idea – the distribution of sample means We are still imagining the results of repeated samples, but instead of recording how often a categorical outcome occurs (proportion), we are recording the average value (mean) of a quantitative variable

4. Individual Values v. Sample Means In the warm-up, we calculated the probability that a randomly selected individual American eats less than 210 lbs of meat per year Let’s look at the picture we drew again: What if we had selected 40 Americans at random, and found the average amount of meat eaten per year by this group? Do you think this value has a higher or lower chance of being below 210 lbs? 210218.4.3669

5. Individual Values v. Sample Means The distribution of sample means has less variability than the distribution of individual values

6. What do you think would happen if we looked at a sample of 100 people? 1000? As the sample size gets bigger, the distribution becomes less variable However, the average value stays the same Mean and standard deviation for a sampling distribution of sample means: Bigger Sample Sizes

7. If a random sample of 40 Americans is selected, what is the probability that the mean amount of meat eaten by those sampled is less than 210 lbs per year? Back to our problem… 218.4210

8. One more example… The scores of students on the ACT college entrance examination in a recent year had the normal distribution with a mean score of 18.6 and a standard deviation of 5.9. a)What is the probability that a single randomly chosen student has a score 21 or higher? 18.621

9. One more example (continued)… The scores of students on the ACT college entrance examination in a recent year had the normal distribution with a mean score of 18.6 and a standard deviation of 5.9. b)What is the probability that the mean score for a random sample of 50 students is 21 or higher? 18.621

10. Conditions for Sample Means For today, some of your problems will involve sample means for normally distributed variables We are going to adjust these a little tomorrow, but for now, you should check: 1) Randomization: The sample must be selected randomly 2) 10% Condition: The sample size must be less than 10% of the population size 3) Normal Sampling Distribution: The sampling distribution must be approximately normal. This happens automatically when the variable being studied is stated to be normally distributed. We will revisit this 3 rd condition tomorrow.

11. Assignment 18-2 Pg. 432 # 37, 38