AP Stat: Ch. 23 notes
Inference:
 Confidence Intervals and Tests of Significance
 Definition: Making conclusions about a population from statistics with a known degree of
confidence
Chapters 19 – 22: Inference about proportions
Chapters 23 – 25: Inference about means
Ch. 23:
 Estimating the population mean 

We always list _standard deviation__ with the mean 

 (x  )
2
i

n 1
So if we don’t know the mean, then we CAN’T know… the standard deviation

This chapter will we be estimating 2 things: mean and standard deviation

What statistics do we use to estimate mean and std. dev?
x
sx

What is the sampling distribution for a sample mean?

 

N  ,

n

But we don’t know __  ____ and __  ____ so we will use:
x

s 

N  x, x 
n

Our new model: Student’s t-distibution (model)
-used when you cannot use Z-scores
Student’s t-model:
 Family …
There is more than one distribution

Similar to normal model
In the normal model there is only one model center at a mean of 0 with a standard
deviation of 1

Model changes based on … sample size and degrees of freedom
(because s changes based on sample size)

Degrees of freedom (df) = n – 1

Generally … wider than the Normal model
(2 statistics yield more variation and S varies in each sample,  does not)

Larger the sample size…. (df), the closer the Student’s t-model approaches the Normal model
(page 549)
1-sample t-interval:
CONDITIONS:
1) Randomization (SRS)
2) 10 % condition
3)
Normal Population OR n  30
Checking #3:
* Look at a histogram of the sample values
- want a normal shape (symmetric and unimodal)
* Look at a normal probability plot (on calculator)
- want a straight line
* Sample size greater than 30
-CLT is applied and the sample distribution will be normal
Conditions met Student’s t-model 1 sample t-Test/Interval
Normal Probability Plot:
Finds the Z-score of each value compared to the sample mean and standard deviation. Plots
these values against themselves.

No curve (straight line) in the plot shows a unimodal and symmetric distribution
MECHANICS:
 Calculate (or given): x , sx , n

Find degrees of Freedom: df = n – 1

Find critical value t * (A- 76) or invT or Table B)
Note: You MUST write this somewhere in the problem!
Formula for Interval:
statistic  (critical value)(standard deviation of statistic)
 s 
x  (t * )  x   (a, b)
 n
T-Interval (on calculator)
INTERPRETATION: We are ____% confident that the true mean of ______ is between ___a____ and
__b____ units.
Example: A coffee vending machine dispenses coffee into a paper cup. You’re supposed to get 10
ounces of coffee, but the amount varies slightly from cup to cup. Below are the amounts measured in a
random sample of 20 cups. Is there evidence that the machine is shortchanging customers? Construct a
95% confidence interval.
9.9
9.6
9.7
9.8
x  9.845
9.7
9.8
10.1
10.0
n = 20
10.0
9.8
9.9
9.9
10.1
10.0
9.6
9.5
9.9
9.5
10.2
9.9
df = 19
Conditions:
1. SRS
1. Stated SRS
2. 10% condition
2. There are more than 200 cups of coffee that the vending machine can
dispense so we are sampling less than 10% of the population
3. Normal Population
OR
n  30
3. Normal population assumed normal prob plot is linear
Conditions met  t- distribution 1 sample t- Interval
 .199 
9.845  (2.093) 
  (9.752, 9.938)
 20 
We are 95% confident that the true average amount of coffee dispensed from the vending machine is
between 9.752 and 9.938 oz.
1-sample t-Test:
 Inference about the mean of a population 
HYPOTHESES:
#

 
  #

 
CONDITIONS:
Same as 1 sample t- interval
Don’t forget to write conditions mett-distribution1 sample t-test
MECHANICS:
Calculate (or given): x , sx , n
Find: degrees of freedom: n - 1
Formula:
t
statistic  parameter
x 

s
std . dev. of statistic ( SE )
n
P-Value: P(t <, > test statistic) = tcdf(LB, UB, df)
CONCLUSION:



Compare P-Value to Alpha
Same two sentences as always
State conclusion in context
-
We reject/ fail to reject ….
We have sufficient/insufficient evidence that the true mean is ….(use Ha)
Calculator: T-Test
Example: The EPA wants to show that “the mean carbon monoxide level of air pollution is higher than
4.9.” Does a random sample of 50 readings (with sample mean of 5.1 and sample std. deviation of 1.17)
present sufficient evidence at the .05 level of significance to support the EPA’s claim? Previous studies
have indicated that such readings have an approximately normal distribution.
s = 1.17
n = 50
x  5.1
Hypotheses:
  4.9
  4.9
Conditions:
Conditions met  t- distribution  1 sample t-test
Mechanics:
t
5.1  4.9
 1.209
1.17
50
P(t  1.209)  .116
Conclusions:
We fail to reject the Ho because our P-Value = .116 which is greater than   0.5 .
We have insufficient evidence that the true mean carbon monoxide level of air pollution is
greater than 4.9 units.
T- Test and T-Interval Practice Problems
1. A survey was conducted involving 250 families living in a city. The average amount of income tax
paid per family in the sample was $3540 with a standard deviation of $1150. Establish and interpret
a 99% confidence interval estimate for the taxes paid by families in this city.
2. The estimated U.S. intake of trans-fatty acids is 8 g per day. Consider a research project involving
150 individuals in which their daily intake of trans-fatty acids was measured. Suppose the average
fatty acid intake from this sample was 12.5 g, with a standard deviation of 7.7 g. Test the research
hypothesis that the average intake has increased at  = 0.05.
3. Suppose that in a sample of 36 bottles from a certain bottling machine, the machine filled the bottles
with an average of 16.1 ounces of cola. The sample had a standard deviation of 0.11 ounces. Give a
90% confidence interval for the mean number of ounces. Interpret this interval.
4. The average stay in days for nongovernmental not-for-profit hospitals is given to be 7.2 days. A
sample of 50 such hospitals was selected to test the hypothesis that the average stay is different
from the national average. The data collected is below. Is this sufficient evidence to reject the null
hypothesis? Use  = 0.01.
5
6
8
3
4
6 10 11
7 3 13
5 4 4
8 7 2
9 9 1
6
5
8
7
7
8 9
4 10
5 2
6 9
7 1
2
7
3
8
4
8
9
6
2
5
9
4
6
2
1