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Topics 19 - 20 Unit 4 Inference from Data: Principles TOPIC 19 CONFIDENCE INTERVALS: MEANS Topic 19 - Confidence Interval: Mean, is unknown The estimated ... – PowerPoint PPT presentation

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Title: Topics 19 - 20

1
Topics 19 - 20
• Unit 4 Inference from Data Principles

2
Topic 19 Confidence Intervals Means
3
Topic 19 - Confidence Interval Mean, s 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
Confidence Interval for a population proportion
where n gt 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 (a ) 5 where a
is the significant level
4
Topic 19 - Confidence Interval Mean, s 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
5
Activity 19-3 MM Consumption
6
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/vn 21.88/v20 4.8925 minutes.

7
Ancient air.
The composition of the earths 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.
8
Ancient air.
Enter data for L1.
95 confidence Interval Using TI83, under Stat,
TEST, Choose option 8TInterval 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.
9
Exercise 19-15 Page 414 Exercise 19-23 Page
417 Exercise 19-24 Page 417
10
Topic 20 Test of Significance Means
11
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
Step 3 Decide what test we have to run, in case
of proportion, we use t-test
12
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 a alpha
Significant Level
If p-value is smaller than a we reject
the null hypothesis, then it is statistically
significant based on data. If p-value is
greater than the a 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 a
• p- value gt 0.1 little or no evidence
against H0
• 0.05 lt p- value lt 0.10 some evidence
against H0
• 0.01 lt p- value lt 0.05 moderate evidence
against H0
• 0.001 lt p- value lt 0.01 strong evidence
against H0
• p- value lt 0.001 very strong evidence
against H0

13
Few Possible cases to look at
A teacher suspects that the mean for older
students is higher than 115
Higher than means (gt 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 Ha µ gt 115
One-sided alternative
A teacher suspects that the mean for older
students is same or more than 115
Same or more than means (gt 115) The opposite of
same or more than is less than 115 (lt 115)
Ho µ 115 Ha µ lt 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
14
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 lt 51 mpg.
15
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
16
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 lt 1000
grams Whats wrong with this?
Hypotheses should be stated in terms of µ, not
x-bar  .
17
Topic 20 Test of Significant Mean, s 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 meanvalue 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 meanvalue in question X-bar is
sample mean Sx is Sample Standard deviation n is
sample size µ is the alternative hypothesis
18
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
19
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 Test Statistic p- value
1 10 6.6 0.825
2 10 6.6 1.597
3 30 6.6 0.825
4 30 6.6 1.597
20
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
21
EXTRA Problems
22
Problem
Assume that you are conducting a test of
significance using a significance level of a
0.10. If your test yields a P-value of 0.08, what
is the appropriate conclusion?
P-value 0.08 lt 0.10 Reject Null, It is
statistically significant
23
Problem
The nicotine content in cigarettes of a certain
brand is normally distributed with mean (in
milligrams) µ and standard deviation s 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 a 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 gt 1.5
24
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 a 0.05. You conclude