Lesson 8 - 1

1. Lesson 8 - 1 Distribution of the Sample Mean

2. Objectives • Understand the concept of a sampling distribution • Describe the distribution of the sample mean for sample obtained from normal populations • Describe the distribution of the sample mean from samples obtained from a population that is not normal

3. Vocabulary • Statistical inference – using information from a sample to draw conclusions about a population • Standard error of the mean – standard deviation of the sampling distribution of x-bar

4. Conclusions regarding the sampling distribution of X-bar • Shape: normally distributed • Center: mean equal to the mean of the population • Spread: standard deviation less than the standard deviation of the population • Law of Large numbers tells us that as n increases the difference between the sample mean, x-bar and the population mean μ approaches zero

5. Central Limit Theorem X or x-barDistribution Regardless of the shape of the population, the sampling distribution of x-bar becomes approximately normal as the sample size n increases. Caution: only applies to shape and not to the mean or standard deviation x x x x x x x x x x x x x x x x Random Samples Drawn from Population Population Distribution

6. Summary of Distribution of x

7. What Happens to Sample Spread If the random variable X has a normal distribution with a mean of 20 and a standard deviation of 12 • If we choose samples of size n = 4, then the sample mean will have a normal distribution with a mean of 20 and a standard deviation of 6 • If we choose samples of size n = 9, then the sample mean will have a normal distribution with a mean of 20 and a standard deviation of 4

8. a Example 1 The height of all 3-year-old females is approximately normally distributed with μ = 38.72 inches and σ = 3.17 inches. Compute the probability that a simple random sample of size n = 10 results in a sample mean greater than 40 inches. P(x-bar > 40) μ = 38.72 σ = 3.17 n = 10 σx = 3.17 / 10 = 1.00244 x - μ Z = ------------- σx 1.28 = ----------------- 1.00244 40 – 38.72 = ----------------- 1.00244 = 1.277 normalcdf(1.277,E99) = 0.1008 normalcdf(40,E99,38.72,1.002) = 0.1007

9. a Example 2 We’ve been told that the average weight of giraffes is 2400 pounds with a standard deviation of 300 pounds. We’ve measured 50 giraffes and found that the sample mean was 2600 pounds. Is our data consistent with what we’ve been told? P(x-bar > 2600) μ = 2400 σ = 300 n = 50 σx = 300 / 50 = 42.4264 x - μ Z = ------------- σx 200 = ----------------- 42.4264 2600 – 2400 = ----------------- 42.4264 = 4.714 normalcdf(4.714,E99) = 0.000015 normalcdf(2600,E99,2400,42.4264) = 0.0000001

10. Summary and Homework • Summary: • The sample mean is a random variable with a distribution called the sampling distribution • If the sample size n is sufficiently large (30 or more is a good rule of thumb), then this distribution is approximately normal • The mean of the sampling distribution is equal to the mean of the population • The standard deviation of the sampling distribution is equal to σ / n • Homework • pg 431 – 433; 3, 4, 6, 7, 12, 13, 22, 29

11. Homework • 3) standard error of the mean • 4) zero • 6) population is normal • 7) four • 12) μ=64, σ= 18, n=36 μx-bar =64, σx-bar= 18/36 = 18/6 = 3 • 13) μ=64, σ= 18, n=36 μx-bar =64, σx-bar= 18/36 = 18/6 = 3 • 22) μ=81.7, σ= 6.9 • a) P(x < 75) = 0.1658 normalcdf(-E99,75,81.7,6.9) • b) n=5 P(x<75) =0.01496 normalcdf(-E99,75,81.7,6.9/5) • c) n=8 P(x<75) =0.00301 normalcdf(-E99,75,81.7,6.9/8) • d) very unlikely (or unusual) • 29) P(x < 45) = 0.0078 normalcdf(-E99,45,50,16/60)