1. Ch. 8 Central Limit Theorem 8.1: Sample means 8.2: Sample proportions
Ma260notes_Sull_ch8_CLT.pptx

2. Central Limit Theorem- Intro Experiments to visualize the CLT
Experiment #1: Pick 5 numbers out of the bag. Average them. Report your results, and return the numbers.
Experiment #2: Plot the probability distribution for tossing 1 dice. Now do it for the average number you toss when you toss 2 dice.
Experiment #3: I’ll present a skewed distribution on home prices
Experiment #4: TV watching in families

3. Expt #1: Numbers experiment = 80, = 21.9

4. CLT Expt #1-numbers Original distribution (at left) For n = 5,  = 80
 = 80
 = /n = 21.9/5 = 9.8
For n = 25,
 = 80
 = /n =
= 21.9/25 = 4.4

5. Expt #2: CLT on dice
When tossing one die,
each probability is 1/6– uniform distribution
Average on the dice when tossing 2 dice– graph starts to approach normal

6. Experiment #3– taking sample means of size n from a skewed distribution
Home price
freq
500,000 5
300,000
200,000 10
100,000
80,000 15
70,000
60,000
50,000
40,000
30,000
10,000
original distribution:
Mean = 109,500
St. dev.  114,800

7. CLT on skewed Ex #3 Plot the Original distribution  = 109,500
 = 109,500
 =114,821
Distribution of sample means when n=10
 = 109,500
 = /n = 114,821/ 10 = 36,309
Sample means when n = 30
 = /n = 114,821/ 30 = 20,963

8. Experiment #4
Collect data on the number of hours everyone in your family or household watches TV
Calculate the average for your household
Plot the entire group
Plot the means
Note each distribution

9. The Standard Error
The standard error is just another name for the standard deviation of the sampling distribution

10. Central Limit Theorem
Special Case: If x is a random variable with a normal distribution, with mean = µ, and standard deviation = σ, then the following holds for any sample size n:
In General, If a random variable has any distribution with mean = µ and standard deviation = σ, then as sample size n gets large, the sampling distribution of will:
approach a normal distribution
with mean = µ and standard deviation = /n

11. Sample Size Considerations for the CLT
For the Central Limit Theorem (CLT) to be applicable:
If the x distribution is symmetric or reasonably symmetric, n ≥ 30 should suffice.
If the x distribution is highly skewed or unusual, even larger sample sizes will be required.
If possible, make a graph to visualize how the sampling distribution is behaving.

12. Normal/ CLT problems- Example 1
Ex. #1: Assume that the cholesterol level of a group of patients is normally distributed with = 180 and = 20.
A. (normal ex) Find the probability that an individual selected at random would have a cholesterol level of:
Less than 170
B. (CLT ex) If a sample of 35 patients is selected, find the probability that this sample will have a mean of

13. Ex #2
Assume the average amount of meat consumed among a group of people is normally distributed with = 60 pounds/year and = 10.
A. (normal) Find the probability that an individual selected at random would consume more than 65 pounds a year.
B. (CLT ex) If a sample of 35 people is selected, find the probability that this sample will have a mean of more than 65 pounds a year.

14. Ex #3
Assume that the number of hours medical students work is normally distributed with = 80 and = 20.
A. (normal) Find the probability that an individual selected at random would work :
Less than 83 hours a week
More than 83 hours a week
B. (CLT ex) If a sample of 32 residents is selected, find the probability that this sample will work an average of

15. Ex #4- skewed home prices
Recall the experiment (#3) we did in class, where home prices were skewed right and had a mean of = 109,500 and = 114,821.
Consider a random sample of 40 homes.
Describe the distribution of the sample means:
Shape
Mean =
Standard error =/n=
Find the probability that the sample mean is less than $150,000

16. Ex 4c-- = 109,500 and = 114,821
With n=40, find that probability the sample mean is
Is Greater than $150,000
Less than $90,000
Greater than $90,000
Between $90,000 and $150,000

17. 8.2 Sample Proportions- Notation
size
Mean
Standard deviation
Proportion
Sample n s
read “p-hat”
Population N  p

18. Calculating a sample proportion
Assume n=100 people are surveyed. If x=81 say they like their jobs, the sample proportion is = x/n =

19. Distribution of the sample proportion
Ex: Let p=.80 and n=100
Mean= p
Standard error= [(p*(1-p))/n]=

20. Central Limit Theorem -- proportions
For these conditions:
np(1-p)>10
n<.05N
The distribution of is approximately normal with
Mean= p
Standard error= (pq/n)=
Note: q=1-p
Ex: Try these formulas for p=.80 and n=100 and N=1,000,000

21. Example #1
A national study claims that 80% of people like their jobs. In a random sample of 100 people, what is the probability that is greater than 82%? (Note: first find z)

22. Example #2
A national study claims that 15% of college students stay for summer school. In a random sample of 80 students, what is the probability that more than 13 stay for summer school?
Note: x=13, so first find , then z, then prob