Date post: | 02-Jun-2018 |
Category: |
Documents |
Upload: | parigreenlime |
View: | 237 times |
Download: | 0 times |
of 31
8/10/2019 triola stats chapter 12
1/31
Section 7.3-1Copyright 2014, 2012, 2010 Pearson Education, Inc.
Lecture Slides
Elementary StatisticsTwelfth Edition
and the Triola Statistics Series
by Mario F. Triola
8/10/2019 triola stats chapter 12
2/31
Section 7.3-2Copyright 2014, 2012, 2010 Pearson Education, Inc.
Chapter 7Estimates and Sample Sizes
7-1 Review and Preview
7-2 Estimating a Population Proportion
7-3 Estimating a Population Mean
7-4 Estimating a Population Standard Deviation orVariance
8/10/2019 triola stats chapter 12
3/31
Section 7.3-3Copyright 2014, 2012, 2010 Pearson Education, Inc.
Key Concept
This section presents methods for using thesample mean to make an inference about thevalue of the corresponding population mean.
8/10/2019 triola stats chapter 12
4/31
Section 7.3-4Copyright 2014, 2012, 2010 Pearson Education, Inc.
Key Concept
1. We should know that the sample mean isthe best point estimateof the populationmean .
2. We should learn how to use sample data toconstruct a confidence intervalfor estimatingthe value of a population mean, and weshould know how to interpret such confidenceintervals.
3. We should develop the ability to determinethe sample size necessary to estimate apopulation mean.
x
8/10/2019 triola stats chapter 12
5/31
Section 7.3-5Copyright 2014, 2012, 2010 Pearson Education, Inc.
Requirements
The procedure we use has a requirement that thepopulation is normally distributed or the sample size isgreater than 30.
8/10/2019 triola stats chapter 12
6/31
Section 7.3-6Copyright 2014, 2012, 2010 Pearson Education, Inc.
Margin of Error Efor Estimate of (WithNot Known)
where has n1 degrees of freedom.
Table A-3 lists values for .
8/10/2019 triola stats chapter 12
7/31Section 7.3-7Copyright 2014, 2012, 2010 Pearson Education, Inc.
If the distribution of a population is essentiallynormal, then the distribution of
is a Student tDistribution for all samples of size n.It is often referred to as a tdistributionand is usedto find critical values denoted by .
Student tDistribution
8/10/2019 triola stats chapter 12
8/31Section 7.3-8Copyright 2014, 2012, 2010 Pearson Education, Inc.
degrees of freedom= n1
for the methods in this section
Definition
The number of degrees of freedomfor a collectionof sample data is the number of sample valuesthat can vary after certain restrictions have beenimposed on all data values.
The degree of freedom is often abbreviated df.
8/10/2019 triola stats chapter 12
9/31Section 7.3-9Copyright 2014, 2012, 2010 Pearson Education, Inc.
= population mean
= sample mean
s = sample standard deviation
n = number of sample values
E = margin of error
t/2 = critical tvalue separating an area of /2 in the
right tail of the tdistribution
Notation
x
8/10/2019 triola stats chapter 12
10/31Section 7.3-10Copyright 2014, 2012, 2010 Pearson Education, Inc.
where
found in Table A-3.
Confidence Interval for theEstimate of (With Not Known)
df = n1
8/10/2019 triola stats chapter 12
11/31Section 7.3-11Copyright 2014, 2012, 2010 Pearson Education, Inc.
2. Using n1 degrees of freedom, refer to Table A-3 or usetechnology to find the critical value that corresponds to thedesired confidence level.
Procedure for Constructing a ConfidenceInterval for (With Not Known)
1. Verify that the requirements are satisfied.
3. Evaluate the margin of error .
4. Find the values of Substitute those values in
the general format for the confidence interval:
5. Round the resulting confidence interval limits.
x Eandx E.
x Ex E
8/10/2019 triola stats chapter 12
12/31Section 7.3-12Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example
A common claim is that garlic lowers cholesterol levels. In
a test of the effectiveness of garlic, 49 subjects weretreated with doses of raw garlic, and their cholesterollevels were measured before and after the treatment.
The changes in their levels of LDL cholesterol (in mg/dL)
have a mean of 0.4 and a standard deviation of 21.0.
Use the sample statistics of n= 49, = 0.4, and s= 21.0 toconstruct a 95% confidence interval estimate of the meannet change in LDL cholesterol after the garlic treatment.
What does the confidence interval suggest about theeffectiveness of garlic in reducing LDL cholesterol?
x
8/10/2019 triola stats chapter 12
13/31Section 7.3-13Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example - Continued
Requirements are satisfied: simple random
sample and n= 49 (i.e., n> 30).
2
21 02 009 6 02749
.. .E tn
95% implies = 0.05.With n= 49, the df = 491 = 48Closest df is 50, two tails, so = 2.009
Using = 2.009, s= 21.0 and n= 49 the marginof error is:
8/10/2019 triola stats chapter 12
14/31
Section 7.3-14Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example - Continued
Construct the confidence interval: x 0.4,E 6.027
We are 95%
confident that the limits of5.6 and 6.4 actually do containthe value of , themean of the changes in LDL cholesterol for thepopulation.
Because the confidence interval limits contain the value of 0, it is verypossible
that the mean of the changes in LDL cholesterol is equal to 0,
suggesting that the
garlic treatment did not affect the LDL cholesterollevels.
It does not appear that
the garlic treatment is effective in lowering LDLcholesterol.
x Ex E0.46.027 0.46.027
5.6 6.4
8/10/2019 triola stats chapter 12
15/31
Section 7.3-15Copyright 2014, 2012, 2010 Pearson Education, Inc.
Important Properties of theStudent tDistribution
1. The Student tdistribution is different for different sample sizes.(See the following slide for the cases n= 3 and n= 12.)
2. The Student tdistribution has the same general symmetric bellshape as the standard normal distribution but it reflects the greater
variability (with wider distributions) that is expected with smallsamples.
3. The Student t distribution has a mean of t= 0 (just as the standardnormal distribution has a mean of z= 0).
4. The standard deviation of the Student tdistribution varies with thesample size and is greater than 1 (unlike the standard normaldistribution, which has = 1).
5. As the sample size ngets larger, the Studenttdistribution getscloser to the normal distribution.
8/10/2019 triola stats chapter 12
16/31
Section 7.3-16Copyright 2014, 2012, 2010 Pearson Education, Inc.
Student tDistributions forn= 3 and n= 12
8/10/2019 triola stats chapter 12
17/31
Section 7.3-17Copyright 2014, 2012, 2010 Pearson Education, Inc.
Point estimate of :
=(upper confidence limit) + (lower confidence limit)
2
Margin of Error:
= (upper confidence limit) (lower confidence limit)
2
Finding the Point Estimateand Efrom a Confidence Interval
8/10/2019 triola stats chapter 12
18/31
Section 7.3-18Copyright 2014, 2012, 2010 Pearson Education, Inc.
Finding a Sample Size for Estimating aPopulation Mean
= population mean
= population standard deviation
= sample mean
= desired margin of error
= zscore separating an area of in the right tail ofthe standard normal distribution
x
8/10/2019 triola stats chapter 12
19/31
Section 7.3-19Copyright 2014, 2012, 2010 Pearson Education, Inc.
Round-Off Rule for Sample Size n
If the computed sample size nis not a whole
number, round the value of nup to the next
largerwhole number.
8/10/2019 triola stats chapter 12
20/31
Section 7.3-20Copyright 2014, 2012, 2010 Pearson Education, Inc.
Finding the Sample Size nWhen is Unknown
1. Use the range rule of thumb (see Section 3-3) toestimate the standard deviation as follows:
2. Start the sample collection process without knowing and, using the first several values, calculate the samplestandard deviation sand use it in place of . Theestimated value ofcan then be improved as moresample data are obtained, and the sample size can be
refined accordingly.
3. Estimate the value of by using the results of someother earlier study.
.
8/10/2019 triola stats chapter 12
21/31
Section 7.3-21Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example
Assume that we want to estimate the mean IQ score for thepopulation of statistics students. How many statistics studentsmust be randomly selected for IQ tests if we want 95%confidence that the sample mean is within 3 IQ points of thepopulation mean?
= 0.05
/2 = 0.025z
/2 = 1.96E = 3
= 15With a simple random sample of only 97statistics students, we will be 95%
confident that the sample mean is within3 IQ points of the true population mean .
8/10/2019 triola stats chapter 12
22/31
Section 7.3-22Copyright 2014, 2012, 2010 Pearson Education, Inc.
Part 2: Key Concept
This section presents methods forestimating a population mean. In additionto knowing the values of the sample dataor statistics, we must also know the valueof the population standard deviation, .
8/10/2019 triola stats chapter 12
23/31
Section 7.3-23Copyright 2014, 2012, 2010 Pearson Education, Inc.
= population mean
= sample mean
= population standard deviation
n = number of sample values
E = margin of error
z /2 = critical zvalue separating an area of /2 in the
right tail of the standard normal distribution
Confidence Interval for Estimating aPopulation Mean (with Known)
x
8/10/2019 triola stats chapter 12
24/31
Section 7.3-24Copyright 2014, 2012, 2010 Pearson Education, Inc.
Confidence Interval for Estimatinga Population Mean (with Known)
1. The sample is a simple random sample. (All samplesof the same size have an equal chance of beingselected.)
2. The value of the population standard deviation isknown.
3. Either or both of these conditions is satisfied: The
population is normally distributed or n> 30.
8/10/2019 triola stats chapter 12
25/31
Section 7.3-25Copyright 2014, 2012, 2010 Pearson Education, Inc.
Confidence Interval for Estimatinga Population Mean (with Known)
x Ex E where E z 2
n
or x E
orx
E,x
E
8/10/2019 triola stats chapter 12
26/31
Section 7.3-26Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example
People have died in boat and aircraft accidents because anobsolete estimate of the mean weight of men was used.
In recent decades, the mean weight of men has increasedconsiderably, so we need to update our estimate of thatmean so that boats, aircraft, elevators, and other suchdevices do not become dangerously overloaded.
Using the weights of men from a random sample, we obtainthese sample statistics for the simple random sample:
n= 40 and = 172.55 lb.
Research from several other sources suggests that thepopulation of weights of men has a standard deviation givenby = 26 lb.
x
8/10/2019 triola stats chapter 12
27/31
Section 7.3-27Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example - Continued
a. Find the best point estimate of the mean weight of thepopulation of all men.
b. Construct a 95% confidence interval estimate of themean weight of all men.
c. What do the results suggest about the mean weight of
166.3 lb that was used to determine the safe passengercapacity of water vessels in 1960 (as given in theNational Transportation and Safety Board safetyrecommendation M-04-04)?
8/10/2019 triola stats chapter 12
28/31
Section 7.3-28Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example - Continued
a. The sample mean of 172.55 lb is the best point estimate ofthe mean weight of the population of all men.
226
1.96 8.057483540E z n
b. A 95% confidence interval implies that = 0.05, so z/2= 1.96.
Calculate the margin of error.
Construct the confidence interval.
x Ex E172.55 8.0574835 172.55 8.0574835
164.49 180.61
8/10/2019 triola stats chapter 12
29/31
Section 7.3-29Copyright 2014, 2012, 2010 Pearson Education, Inc.
Example - Continued
c. Based on the confidence interval, it is possible that themean weight of 166.3 lb used in 1960 could be the
mean weight of men today.
However, the best point estimate of 172.55 lb suggeststhat the mean weight of men is now considerably greater
than 166.3 lb.
Considering that an underestimate of the mean weightof men could result in lives lost through overloaded
boats and aircraft, these results strongly suggest thatadditional data should be collected.
8/10/2019 triola stats chapter 12
30/31
Section 7.3-30Copyright 2014, 2012, 2010 Pearson Education, Inc.
Choosing the Appropriate Distribution
8/10/2019 triola stats chapter 12
31/31
C i ht 2014 2012 2010 P Ed ti I
Choosing the Appropriate Distribution
Use the normal (z)
distributionknown and
normally distributedpopulation or n> 30
Use tdistribution not known and
normally distributedpopulation or n> 30
Use a nonparametricmethod or bootstrapping
Population is not normallydistributed andn 30