Chapter 23: Inferences About
Means
AP Statistics
VERY Important Idea for Sample
Means
Problem, however, is that we don’t know the population standard
deviation σ, and we cannot determine it from the sample mean.
So we end up estimating σ to be the sample standard deviation,
s. Therefore SE y  s

n
Student’s (Gosset’s) t
Using an estimated standard deviation (standard error)
for the sampling distribution of sample means,
however, creates problems, especially when our
sample size it small.
When the sample size was big, Gossert found that a
normal model could be used for the sampling distribution
of sample means
However, when that sample size was small, he noticed the
normal model was inappropriate (Guinness).
Therefore, a NEW model was adopted to take care of this
problem: Student’s t-model
t2
As df increases, the t-model
approximates the Normal
Model.
The t-models are a whole family of related
distributions that depend on a parameter known
as degrees of freedom (df). Degrees of
freedom = n-1
t-distribution
When we have sample means and do not
know the population standard deviation,
we use the t-distribution.
We now find t-scores and find the area
under the t-model.
Basically, acts like the Normal Model
Sampling Distribution Model for
Sample Means
The standardized sample mean:
y 
t
SE y

Follows a t-model with n-1 degrees of freedom.
We estimate the standard error with:

s
SE y 
n
One-Sample t-Interval for the Mean
Assumptions/Conditions
Independence Assumption:
Randomization Condition
10% Condition
Normal Population Assumption:
Cannot assume—most times is NOT true
We can check:
Nearly Normal Condition
Nearly Normal Condition
To Check this condition, ALWAYS DRAW A
PICTURE, EITHER A HISTOGRAM OR A
NORMAL PROBABILITY PLOT.
Normality of t-model depends of sample size
(think Central Limit Theorem).
Nearly Normal Condition
• For very small sample size (n<15), the data
should follow Normal model pretty closely. If
you find outliers or strong skewness don’t use tmodel.
• For moderate sample sizes (15<n<40 or so), tmodel will work well if as long as data are
unimodal and reasonably symmetric. Make
Histogram.
• For large sample size (n > about 40 or 50) tmodel is good no matter what the shape—be
careful, however, of outliers—analyze with and
without them
Example
Make and interpret a 95% confidence interval for the mean number of
chips in an 18 oz bag of Chips Ahoys.
Check Conditions
In order to create a 1-Proportion t-Interval, I need
to assume Independence and a Normal
population. To justify the Independence
Assumption I need to satisfy both the
Randomization Condition and the 10%
Condition:
Check Conditions
To justify the Normal Population Assumption I need
to satisfy the Nearly Normal Condition:
Mechanics (Calculations)
t15=±2.13
invT(percentile, df)
InvT(.025,15)
Conclusion
Conclusion
What do you think about Chips Ahoys claim of an
average of 1000 chips per bag?
Why?
One-Sample t-test for the Mean
One-Sample t-test for the Mean
• A one-sample t-test is performed just like a
one-proportion z-test.
• Use the 4 steps used in a test for
proportions—the only thing that changes is
the model. Instead of a Normal Model and
z-scores, you use a t-Model and t-scores.
Example
In order to conduct a one-sample t-test, I
need to assume Independence and a
Normal population. To justify the
Independence Assumption I need to
satisfy both the Randomization Condition
and the 10% Condition:
Randomization Condition: It states that the
sample was chosen randomly.
10% Condition: The 25 students represent less
than 10% of all the students in the school.
To justify the Normal Population Assumption
I need to satisfy the Nearly Normal
Condition:
Nearly Normal Condition: The histogram
of the number of hours of TV watched is
unimodal and reasonably symmetric.
Since the conditions are satisfied, we can
use a t-model with 24 degrees of freedom
to do a one-sample t-test for the mean.

s
7.96
SE x 

 1.5923
n
25
x   0 14.32  13
t 24 

 .8289
1.5923
SE x

P  px  14.32
 pt 24  .8289
 0.2077
Given a P-value of 0.2077, I will fail to reject
the null hypothesis at α=0.05. This Pvalue is not small enough for me to reject
the hypothesis that the true mean number
of hours that the students in this high
school watch TV is 13 hours. Therefore,
the difference between the observed mean
of 14.32 hours and 13 hours is probably
due to random sampling error.
Sample Size Computation
Just like with Normal Model
ME  t
*
df
s
n

s
Remember : SE y 
n
Computer Printout
Computer Printout