Lesson 8 - 1

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-barDistribution

Population Distribution

Random Samples Drawn from Population

x x x x x x x x x x x x x x x x

Shape, Center and Spread of Population

Distribution of the Sample Mean

Shape Center Spread

Normal with mean, μ and standard deviation, σ

Regardless of sample size, n, distribution of

x-bar is normalμx-bar = μ

σσx-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

μx-bar = μ σσx-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 = 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.

μ = 38.72 σ = 3.17 n = 10 σx = 3.17 / 10 = 1.00244

x - μZ = ------------- σx

40 – 38.72= ----------------- 1.00244

1.28= ----------------- 1.00244

= 1.277

normalcdf(1.277,E99) = 0.1008

normalcdf(40,E99,38.72,1.002) = 0.1007 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 σx = 300 / 50 = 42.4264

x - μZ = ------------- σx

2600 – 2400= ----------------- 42.4264

200= ----------------- 42.4264

= 4.714

normalcdf(4.714,E99) = 0.000015

normalcdf(2600,E99,2400,42.4264) = 0.0000001 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 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)