Lesson Distribution of the Sample Mean

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

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

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

Central Limit Theorem 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 or x-bar Distribution Population Distribution Random Samples Drawn from Population xxxxxxxxxxxxxxxx

Shape, Center and Spread of Population Distribution of the Sample Mean ShapeCenterSpread Normal with mean, μ and standard deviation, σ Regardless of sample size, n, distribution of x-bar is normal μ x-bar = μ σ σ x-bar =  n Population is not normal with mean, μ and standard deviation, σ As sample size, n, increases, the distribution of x-bar becomes approximately normal μ x-bar = μ σ σ x-bar =  n Summary of Distribution of x

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

Example 1 The height of all 3-year-old females is approximately normally distributed with μ = 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. μ = σ = 3.17 n = 10 σ x = 3.17 /  10 = x - μ Z = σ x 40 – = = = normalcdf(1.277,E99) = normalcdf(40,E99,38.72,1.002) = a P(x-bar > 40)

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? μ = 2400 σ = 300 n = 50 σ x = 300 /  50 = x - μ Z = σ x 2600 – 2400 = = = normalcdf(4.714,E99) = normalcdf(2600,E99,2400, ) = a P(x-bar > 2600)

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

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