Topics 19 - 20
Unit 4 – Inference from Data: Principles
TOPIC 19
CONFIDENCE INTERVALS: MEANS
Topic 19 - Confidence Interval: Mean, σ is unknown
The purpose of confidence intervals is to use the sample statistic to construct an
interval of values that you can be reasonably confident contains the actual, though
unknown, parameter.
The estimated standard deviation of the sample statistic X-bar is called the standard error
S
n
Confidence Interval for a population proportion :
est imat e margin of error  X  t *
where n >= 30
t * is calculated based on level of confidence
When running for example 95% Confidence Interval:
95% is called Confidence Level and
we are allowing possible 5% for error, we call this alpha
(α )= 5% where α is the significant level
S
n
Topic 19 - Confidence Interval: Mean, σ is unknown
Use if the sample data is given, use the Stat,
Edit and enter data in the calculator before
running the Confidence Interval
L1 is where data is entered by you
C-Level: is the level you are running the
Confidence Interval
Use if the information about sample data is
given.
X-Bar mean of sample data
Sx is Standard deviation of the sample
n is sample size
C-Level: is the level you are running the
Confidence Interval
Activity 19-3: M&M Consumption
Travel time to work.
• A study of commuting times reports the travel times to work of a
random sample of 20 employed adults in New York State. The mean
is = 31.25 minutes and the standard deviation is s = 21.88
minutes. What is the standard error of the mean?
• s/√n = 21.88/√20 = 4.8925 minutes.
Ancient air.
The composition of the earth’s atmosphere may have changed
overtime. To try to discover the nature of the atmosphere long ago,
we can examine the gas in bubbles inside ancient amber. Amber is
tree resin that has hardened and been trapped in rocks. The gas in
bubbles within amber should be a sample of the atmosphere at the
time the amber was formed. Measurements on specimens of amber
from the late Cretaceous era (75 to 95 million years ago) give these
percents of nitrogen:
63.4
65.0
64.4
63.3
54.8
64.5
60.8
49.1
51.0
Assume (this is not yet agreed on by experts) that these observations are an
SRS from the late Cretaceous atmosphere. Use a 90% confidence interval
to estimate the mean percent of nitrogen in ancient air.
Ancient air.
Enter data for L1.
95% confidence Interval:
Using TI83, under Stat, TEST,
Choose option 8:TInterval
Mean of the sample = 59.6
Standard deviation = 6.26
Degree of freedom = df= 8
Confidence interval for mean percent of
nitrogen is between 54.8 and 64.4.
Exercise 19-15 Page 414
Exercise 19-23 Page 417
Exercise 19-24 Page 417
TOPIC 20
TEST OF SIGNIFICANCE: MEANS
Topic 20 – Test of Significant: Mean
The purpose of Test of Significant is when we do know the population Parameter but
we do not necessary agree with it or we have question about it. To do the test we
need to run a sample and we use the statistic to test its validity.
Step 1: Identify and define the parameter.
Step 2: we initiate hypothesis regarding the
question – we can not run test of significant
without establishing the hypothesis
H 0 :
H :
 a








 0
 0
 0
 0
Step 3: Decide what test we have to run, in case of proportion, we use t-test
x  0
t 
S
n
or
or
Topic 20 – Test of Significant: Mean
Step 4: Run the test from calculator
Step 5: From the calculator write
down the p-value T-test
Step 6: Compare your p-value with α – alpha – Significant Level
If p-value is smaller than α
we “reject” the null hypothesis, then it is statistically significant based on data.
If p-value is greater than the α
we “Fail to reject” the null hypothesis, then it is not statistically significant based on data.
Last step: we write conclusion based on step 6 at significant level α
•
•
•
•
•
p- value > 0.1: little or no evidence against H0
0.05 < p- value <= 0.10: some evidence against H0
0.01 < p- value <= 0.05: moderate evidence against H0
0.001 < p- value <= 0.01: strong evidence against H0
p- value <= 0.001: very strong evidence against H0
Few Possible cases to look at:
A teacher suspects that the mean for older students is higher than 115
Higher than means (> 115)
The opposite of higher than is less than or equal to 115 ( 115)
Comparing the two, null hypothesis is the comparison that includes equality (=)
Ho: µ = 115
One-sided alternative
Ha: µ > 115
A teacher suspects that the mean for older students is same or more than 115
Same or more than means (> 115)
The opposite of same or more than is less than 115 (< 115)
Ho: µ = 115
Ha: µ < 115
One-sided alternative
A teacher suspects that the mean for older students is also 115
Same means (= 115)
The opposite of same is not equal to 115 ( 115)
Ho: µ = 115
Ha: µ  115
Two-sided alternative
Fuel economy.
According to the Environmental Protection Agency (EPA),
the Honda Civic hybrid car gets 51 miles per gallon (mpg)
on the highway. The EPA ratings often overstate true fuel
economy. Larry keeps careful records of the gas mileage of
his new Civic hybrid for 3000 miles of highway driving.
His result is x-bar= 47.2 mpg. Larry wonders whether the
data show that his true long-term average highway mileage
is less than 51 mpg. What are his null and alternative
hypotheses?
Answer
Larry wonders whether the data show that his true long-term
average highway mileage is less than 51 mpg.
H0: µ = 51 mpg;
Ha: µ < 51 mpg.
Problem
If a researcher is interested in testing
whether the mean is different from some
claimed value, 55, then the null and
alternative are
test the hypotheses
H0: μ = 55,
Ha: μ ≠ 55
Stating hypotheses.
In planning a study of the birth weights of babies
whose mothers did not see a doctor before
delivery, a researcher states the hypotheses as
H0 : x-bar = 1000 grams
Ha : x-bar < 1000 grams
What’s wrong with this?
Hypotheses should be stated in terms of µ, not x-bar .
Topic 20 – Test of Significant: Mean, σ is unknown
Use if the sample data is given, use the Stat,
Edit and enter data in the calculator before
running the T-test
µ0 is mean–value in question
List: L1 where the raw data is entered by you
µ: is the alternative hypothesis
Use if the information about sample data is
given.
µ0 is mean–value in question
X-bar is sample mean
Sx is Sample Standard deviation
n is sample size
µ: is the alternative hypothesis
Improving your SAT score.
We suspect that on the average students will score higher on their
second attempt at the SAT mathematics exam than on their first
attempt. Suppose we know that the changes in score (second try
minus first try) follow a Normal distribution. Here are the results for
46 randomly chosen high school students:
Do these data give good evidence that the mean change in the
population is greater than zero?
−30
24
47
70
−62
55
−41
−32
128
−11
−43
122
−10
56
32
−30
−28
−19
1
17
57
−14
−58
77
27
−33
51
17
−67
29
94
−11
2
12
−53
−49
49
8
−24
96
120
2
−33
−2
−39
99
Activity 20- 2: Sleeping Times
The null hypothesis is that the mean sleep time of the population is 7
hours. In symbols, the null hypothesis is
H0 : µ = 7.0 hours.
The alternative hypothesis is that the mean sleep time of the
population is not 7 hours. In symbols, the alternative hypothesis is
Ha : µ ≠ 7.0 hours.
Sample
Number
Sample
Size
Sample
Mean
Sample
SD
1
10
6.6
0.825
2
10
6.6
1.597
3
30
6.6
0.825
4
30
6.6
1.597
Test
Statistic
p- value
Exercise 20-8: UFO Sighters’ Personality – Page 432
Exercise 20-10: Credit Card Usage - Page 433
Exercise 20-21: Pet Ownership - Page 436
Exercise 20-14: Age Guesses – Page 434
EXTRA PROBLEMS
Problem
Assume that you are conducting a test of
significance using a significance level of α =
0.10. If your test yields a P-value of 0.08,
what is the appropriate conclusion?
P-value = 0.08 < 0.10 Reject Null, It is statistically
significant
Problem
The nicotine content in cigarettes of a certain brand is normally
distributed with mean (in milligrams) μ and standard deviation σ
= 0.1. The brand advertises that the mean nicotine content of
their cigarettes is 1.5, but measurements on a random sample of
400 cigarettes of this brand gave a mean of x = 1.52. Is this
evidence that the mean nicotine content is actually higher than
advertised?
at significance level α = 0.01. You conclude
Is this evidence that the mean nicotine content is actually higher
than advertised? State the hypothesis
test the hypotheses
H0: μ = 1.5,
Ha: μ > 1.5
Problem
A researcher wants to know if the average time in jail for robbery
has increased from what it was several years ago when the
average sentence was 7 years. He obtains data on 400 more
recent robberies and finds an average time served of 7.5 years. If
we assume the standard deviation of sample is 3 years, what is
the p-value of the test? at significance level α = 0.05. You conclude