This module roughly corresponds to chapter 7 in the textbook.
This module is divided into two sections.
A.
The Sampling Distribution of
(Sample Mean)
1. Population size (N) is unknown
2.
Population size (N) is known
B.
The Sampling Distribution of (Sample Proportion)
1.
Sample proportion ( is known
2.
Sample proportion ( is unknown
Central Limit Theorem (CLT)
The
central limit theorem
states that the sampling distribution of any statistic will be normal or nearly normal, if
the
sample size is large enough.
For our purposes, what this means is that we can use the normal distribution (Z) if our sample size is sufficiently
large.
For this class, if the sample size is greater than or equal to 30, we can use the normal distribution.
(n
•

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
The CLT tells us that the standard deviation of the sample means is the standard deviation divided by the square root
of the sample size:
=
What does this mean for us?
It means that our Z-formula has changed from this: Z =
to this:
Z =
If the population standard deviation is not available substitute the sample standard deviation (s) for the population
standard deviation (
).
Thus Z =
becomes
Z =
•

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
1. Population size (N) unknown
Example 1
: Suppose that the mean expenditure per customer at a tire store is $85 with a standard deviation tire of $9.00.
If a random sample of 40 customers is taken, what is the probability that the sample mean expenditure per customer for
this sample will be $87.00 or more.
P(
$87 |
= $85 and
= $9)
From here the process is very similar to what we've done in Modules 8 and 9:
Steps 1 & 2
: Calculate the Z-Score using the new formula and use Excel to determine the Z-Probability:
Z
=
=
=
=
1.4055
[Z-Probability =
.9201
]
Step 3:
Interpretation: .9201 is the area under the curve to the left of 87. We
want to determine the area to the
right of 87 P(X ≥ $87). To determine this we subtract .9201 from 1.
1 - .9201 =
.0799
→ P(
$87 |
= $85 and
= $9) =
.0799
•
Z-Score
1.4055
Z-Probability
0.9201
=NORMSDIST(1.4055)

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
1. Population size (N) unknown
Example 1
:
P(
$87 |
= $85 and
= $9) =
.0799
The graph looks like this:
Again, you don't have to graph it.
But it may help you to understand what to do in step 3 on the previous slide better if you
can visualize what's going on.
•

Module 10: Introduction to Sampling
A.
The Sampling Distribution of
(Sample Mean)
1. Population size (N) unknown
Example 2
: Suppose that during any hour in a large department store, the average number of shoppers is 448 with a
standard deviation of 21 shoppers. What is the probability that a random sample of 49 different shopping hours will yield a
sample mean between 441 and 446 shoppers?

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- DebraACasto
- Standard Deviation