Download Sampling Distribution Models AP Stats Chapter 18 Notes part

Chapter 18 Notes: Sampling Distribution Models
A.
Sampling Distribution Model:
1.
Shows the behavior of a distribution (statistic) over all the possible samples of a
population (should look like a Normal model).
2.
Should follow a Bernoulli Trials:
a.
There are only two possible outcomes: The probability of success (p) and the
probability of failure (q).
b.
The probability of success is the same for every trial.
c.
The trials are independent. The outcome of one event does not change the
outcome of successive events.
3.
Sampling Variability: The slight changes (variability) from one sample to the next
a.
Should follow a Normal model.
b.
Standard deviation is important
c.
Can find the changes in variability (differences) by using z-score models.
B.
Assumption and Conditions: In order to verify that a Normal model is appropriate
for a Sampling Distribution Model, then specific assumptions must be true:
1.
The sample values must be independent of each other.
2.
The sample size, n, must be large enough (but not too large, see 10% Condition)
Since assumption are often impossible to check, there are specific conditions we must
check in order to back-up the assumptions:
3.
Randomizing Condition: The sample should be a simple random sample of the
population,
Or, the distribution represents a random sample of the population.
4.
10% Condition: the sample size must not be any larger than 10% of the entire
population the sample can be chosen from.
C.
The Central Limit Theorem (The Fundamental Theorem of Statistics):
1.
The mean of all sample distributions should have a shape that can be
approximated by a Normal model.
2.
The larger the sample, the better the Normal approximation will be.
D.
Sampling Distribution Model for a Proportion:
1.
The Model: (p, SD(p))
2.
Mean: The probability of success (p)
3.
SD(p): σ(p) = √(pq/n)
4.
The sampling distribution model should follow a Normal model with the center
(mean) at the proportion of success (p).
5.
Use z-scores to find the actual probability of success comparing the mean
probability to the actual probability of success.
6. Success/Failure Condition: The sample size has to be big enough so that np and nq
are at least 10. We need to expect at least 10 successes and 10 failures to have
enough data for sound conclusions.
E. Example: Suppose that about 13% of the population is left-handed. A 200-seat
auditorium has been built with 15 “lefty seats,” seats that have the built-in desk on the
left rather than the right arm of the chair. In a class of 90 students, what’s the probability
that there will not be enough seats for the left-handed students?
a.
Check the appropriate conditions for the model:
The 90 students represent a random sample of all students.
The 90 students represent less than 10% of the total population of all students.
Success/Failure: np = 0.13(90) = 11.7 ≥ 10,
nq = 0.87(90) = 78.3 ≥ 10
Since the conditions are satisfied, the sampling distribution model can be represented
using a Normal model, where N(p, σ):
p = 0.13 and SD(p) = √((0.13 ∙ 0.87)/90) = 0.035
b. Clearly label the model:
N(0.13, 0.035)
c.
Plot the Normal model using the 68-95-99.7 rule.
d.
Find the z-score of the proportion for ^p = 15/90 = 0.167
z = (^p – p)/SD(p) = (0.167 – 0.13)/0.035 = 1.06
2nd, DISTR, normalcdf(1.06, 6) = 0.1452 = ~14.5%
There is about a 14.5 % chance that there will not be enough seats for the left-handed
students in the class.
e.
Is this an unusually large proportion of lefties without left-handed seats?
No, since the result of 14.5% is only a little more than 1 standard deviation above the
mean proportion of 13%.
F.
Sampling Distribution for a Mean:
1.
The shape of the sampling distribution is approximately Normal.
2.
The sample size must be large enough (Large Enough Sample Condition).
3.
There still must be independence.
4.
For a sampling distribution model, both the mean and the standard deviation are
given: N(y, σ(y))
5.
When using a sampling distribution model for an actual population sample, you
must calculate the standard deviation of the sample: σ(y) = SD
√n
G.
The Conditions for the Sampling Distribution for a Mean (based on the Central
Limit Theorem):
1.
Randomizing Condition: The data values must be sampled randomly, or
represent a random sample of the population.
2.
Independence Assumption: The sampled values must be independent of each
other..
3.
10% condition: The sample size is no more than 10% of the total population.
4.
Large Enough Sample Condition (Chapter 24): Think about your sample size in
the context of what you know about the population, and then tell whether you believe
the sample is large enough (Usually, a sample size of 30 would be considered large
enough).
H.
Example: Suppose that mean adult weight is 175 pounds with a standard
deviation of 25 pounds. An elevator in a building has a weight limit of 10 persons or
2000 pounds. What’s the probability that the 10 people who get on the elevator overload
the weight limit?
1.
Check the conditions:
a.
The 10 people getting on the elevator are a random sample of the population.
b.
The weights of the 10 people are independent.
c.
The 10 people represent less than 10% of the total possible elevator riders.
d.
The distribution of weights is approximately Normal, so the sample may be large
enough.
2.
State the parameters of the sampling model, and perform the mechanics of the
test: Mean Weight = 175 pounds, SD(y) = 25 pounds
Sample Mean = 2000/10 = 200 pounds, Sample SD = 25/√(10) = 7.91 pounds
N(175, 7.91)
z = (200 – 175)/7.91 = 3.162
2nd, DISTR, normalcdf(3.162, 10) = 0.000783
3.
State your conclusion: The chance that a random sample of 10 people will
exceed the elevator’s weight limit is only 0.000783. Since this random sample is more
than 3 standard deviations above the mean, this would be an unusual circumstance.