Post on 04-Jan-2016
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Chapter 6Estimates and Sample Sizes
6-1 Estimating a Population Mean: Large Samples / σ Known
6-2 Estimating a Population Mean: Small Samples / σ Unknown
6-3 Estimating a Population Proportion
6-4 Estimating a Population Variance: Will cover with chapter 8
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Overview
Methods for estimating population means and proportions
Methods for determining sample sizes
This chapter presents:
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6-1
Estimating a Population Mean:Large Samples / σ Known
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Assumptions
Large Sample is defined as samples with n > 30 and σ known.
Data collected carelessly can be absolutely worthless, even if the sample is quite large.
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Estimatora formula or process for using sample data to
estimate a population parameter
Estimatea specific value or range of values used to
approximate some population parameter
Point Estimatea single value (or point) used to approximate a
population parameter
The sample mean x is the best point estimate of the population mean µ.
Definitions
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DefinitionConfidence Interval
(or Interval Estimate)
a range (or an interval) of values used to estimate the true value of the population
parameter
Lower # < population parameter < Upper #
As an exampleLower # < < Upper #
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DefinitionWhy Confidence Intervals
A couple of points
1. Even though x is the best estimate for and s is the best estimate for they do not give us an indication of how good they are.
2. A confidence interval gives us a range of values based on
a) variation of the sample data
b) How accurate we want to be
3. The width of the range of values gives us an indication of how good the estimate is.
4. The width is called the Margin of Error (E). We will discuss how to calculate this later.
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Proportion of times that the confidence interval actual contains the population parameter
Degree of Confidence = 1 - often expressed as a percentage value
usually 90%, 95%, or 99% So ( = 10%), ( = 5%), ( = 1%)
DefinitionDegree of Confidence
(level of confidence or confidence coefficient)
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Interpreting a Confidence Interval
Let: 1 - = .95
Correct: we are 95% confident that the interval from 98.08 to 98.32 actually does contain the true value of .
This means that if we were to select many different samples of sufficient size and construct the confidence intervals, 95% of them would actually contain the value of the population mean .
Wrong: There is a 95% chance that the true value of will fall between 98.08 and 98.32. (there is no way to calculate the probability for a population parameter only a sample statistic)
98.08o < µ < 98.32o
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Confidence Intervals from 20 Different SamplesSimulations
http://www.ruf.rice.edu/~lane/stat_sim/conf_interval/index.html
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The number on the borderline separating sample statistics that are likely to occur from those that are
unlikely to occur. The number z/2 is a critical value that
is a z score with the property that it separates an area /2
in the right tail of the standard normal distribution.
ENGLISH PLEASE!!!!
Definition
Critical Value
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The Critical Value
z=0Found from calculator
z2
z2-z2
2 2
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-z2z2
95%
.95
.025.025
2 = 2.5% = .025 = 5%
Critical Values
Finding z2 for 95% Degree of Confidence
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Finding z2 for 95% Degree of Confidence
.025.025
- 1.96 1.96
z2 = 1.96
.025
Use calculator to find a z score of 1.96
= 0.025 = 0.05
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Finding z2 for other Degrees of
Confidence
1. 1 -
2. 1 -
3. 1 -
4. 1 -
5. 1 - (will use on test for ease of calculation)
Find critical value and sketch
Examples:
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Margin of Error is the maximum likely difference observed between sample mean x and true population
mean μ.
denoted by E
μ
Definition
lower limit upper limit
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x - E < µ < x + E
Where
Confidence Interval (or Interval Estimate)
for Population Mean µ(Based on Large Samples: n >30)
E = z/2 •n
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When can we use zα/2?
If n > 30 and we know
If n 30, the population must have a normal distribution and we must know .
Knowing is largely unrealistic.
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1. When using the original set of data, round the confidence interval limits to one more decimal place than used in original set of data.
2. When the original set of data is unknown and only the
summary statistics (n, x, s) are used, round the confidence interval limits to the same number of decimal places used for the sample mean.
Round-Off Rule for Confidence
Intervals Used to Estimate µ
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n = 100
x = 43704
σ = 9879
= 0.95
= 0.05/2 = 0.025
z / 2 = 1.96
E = z / 2 • = 1.96 • 9879 = 1936.3n 100
x - E < < x + E
$41,768 < < $45,640
Example: A study found the starting salaries of 100 college graduates who have taken a statistics course. The sample mean was $43,704 and the sample standard deviation was $9,879. Find the margin of error E and the 95% confidence interval.
43704 - 1936.3 < < 43704 + 1936.3
Based on the sample provided, we are 95% confident the population (true) mean of starting salaries is between 41,768 & 45,640.
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TI-83 Calculator
Finding Confidence intervals using z
1. Press STAT
2. Cursor to TESTS
3. Choose ZInterval
4. Choose Input: STATS*
5. Enter σ and x and confidence level6. Cursor to calculate
*If your input is raw data, then input your raw data in L1 then use DATA
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Width of Confidence Intervals
Test QuestionWhat happens to the width of confidence
intervals with changing confidence levels?
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Finding the Point Estimate and E from a Confidence Interval
Point estimate of x:
x = (upper confidence interval limit) + (lower confidence interval limit)
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Margin of Error:
E = (upper confidence interval limit) - x
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Find x and E
26 < µ < 40
x = (40 + 26) / 2 = 33
E = 40 - 33 = 7
Example
Use for #4 on hw
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z/ 2 •E =
n
(solve for n by algebra)
z/ 2 E
n =
2
z/2 = critical z score based on the desired degree of confidence E = desired margin of error
= population standard deviation
Sample Size for Estimating Mean
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Example: If we want to estimate the mean weight of plastic discarded by households in one week, how many households must be randomly selected to be 99% confident that the sample mean is within 0.25 lb of the true population mean? (A previous study indicates the standard deviation is 1.065 lb.)
= 0.01
z = 2.575
E = 0.25
σ = 1.065
n = z = (2.575)(1.065) E 0.25
= 120.3 = 121 households
2 2
If n is not a whole number, round it up to the next higher whole number.
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Example: If we want to estimate the mean weight of plastic discarded by households in one week, how many households must be randomly selected to be 99% confident that the sample mean is within 0.25 lb of the true population mean? (A previous study indicates the standard deviation is 1.065 lb.)
= 0.01
z = 2.575
E = 0.25
σ = 1.065
n = z = (2.575)(1.065) E 0.25
= 120.3 = 121 households
2 2
We would need to randomly select 121 households to be 99% confident that this mean is within 1/4 lb of the
population mean.
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Example: How large will the sample have to be if we want to decrease the margin of error from 0.25 to 0.2? Would you expect it to be larger or smaller?
= 0.01
z = 2.575
E = 0.20
σ = 1.065
n = z = (2.575)(1.065) E 0.2
= 188.01 = 189 households
2 2
We would need to randomly select a larger sample because we require a smaller margin of error.
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What happens when E is doubled ?
Sample size n is decreased to 1/4 of its original value if E is doubled.
Larger errors allow smaller samples.
Smaller errors require larger samples.
/ 2 z
1
n = =2
1/ 2
(z ) 2
/ 2 z
2
n = =
2
4/ 2
(z ) 2
E = 1 :
E = 2 :
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Class Assignment
1. Use OLDFAITHFUL Data in Datasets File 2. Construct a 95% and 90% confidence interval for the mean
eruption duration. Write a conclusion for the 95% interval. Assume σ to be 58 seconds
3. Compare the 2 confidence intervals. What can you conclude?4. How large a sample must you choose to be 99% confident the
sample mean eruption duration is within 10 seconds of the true mean
Guidelines:1. Choose a partner2. Suggest having one person working the calculator and one writing3. Due at the end of class (5 HW points)4. Each person must turn in a paper
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6-2
Estimating a Population Mean:Small Samples / σ Unknown
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1. n 302. The sample is a random sample.3. The sample is from a normally
distributed population.
Case 1 ( is known): Largely unrealistic;
Case 2 (is unknown): Use Student t distribution if normal ; if n is very large use z
Small SamplesAssumptions
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Case 1 ( is known):
Determining which distribution to use
Case 2 (is unknown):n very large use zn > 30 use tn < 30 & Normal use tn < 30 & skewed neither
n > 30 use zn < 30 & Normal use zn < 30 & Skewed neither
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1. n = 150 ; x = 100 ; s = 15 skewed distribution2. n = 8 ; x = 100 ; s = 15 normal distribution3. n = 8 ; x = 100 ; s = 15 skewed distribution4. n = 150 ; x = 100 ; σ = 15 skewed distribution5. n = 8 ; x = 100 ; σ = 15 skewed distribution
Determining which distribution to use
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Important Facts about the Student t Distribution
1. Developed by William S. Gosset in 19082. Density function is complex
3. Shape is determined by “n”
4. Has the same general symmetric bell shape as the normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples.
5. The Student t distribution has a mean of t = 0, but the standard deviation varies with the sample size and is always greater than 1
6. Is essentially the normal distribution for large n. For values of n > 30, the differences are so small that we can use the critical z or t value.
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Student tdistributionwith n = 3
Student t Distributions for n = 3 and n = 12
0
Student tdistributionwith n = 12
Standardnormaldistribution
Greater variability than standard normal due to small sample size
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Student t Distribution
If the distribution of a population is essentially normal, then the distribution of
critical values denoted by
t =x - µ
sn
t/ 2
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Book DefinitionDegrees of Freedom (df )
Corresponds to the number of sample values that can vary after certain restrictions have imposed on all data values.
This doesn’t help me, how about you?
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DefinitionDegrees of Freedom (df )
In general, the degrees of freedom of an estimate is equal to the number of independent scores (n) that go into the estimate minus the number of parameters estimated.
In this section
df = n - 1because we are estimating with x
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Table A-3 / Calculators / Excel
Table from website TI – 84 (only) Excel function (tinv)
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Degreesof
freedom
1234567891011121314151617181920212223242526272829
Large (z)
63.6579.9255.8414.6044.0323.7073.5003.3553.2503.1693.1063.0543.0122.9772.9472.9212.8982.8782.8612.8452.8312.8192.8072.7972.7872.7792.7712.7632.7562.575
.005(one tail)
.01(two tails)
31.8216.9654.5413.7473.3653.1432.9982.8962.8212.7642.7182.6812.6502.6252.6022.5842.5672.5522.5402.5282.5182.5082.5002.4922.4852.4792.4732.4672.4622.327
12.7064.3033.1822.7762.5712.4472.3652.3062.2622.2282.2012.1792.1602.1452.1322.1202.1102.1012.0932.0862.0802.0742.0692.0642.0602.0562.0522.0482.0451.960
6.3142.9202.3532.1322.0151.9431.8951.8601.8331.8121.7961.7821.7711.7611.7531.7461.7401.7341.7291.7251.7211.7171.7141.7111.7081.7061.7031.7011.6991.645
3.0781.8861.6381.5331.4761.4401.4151.3971.3831.3721.3631.3561.3501.3451.3411.3371.3331.3301.3281.3251.3231.3211.3201.3181.3161.3151.3141.3131.3111.282
1.000.816.765.741.727.718.711.706.703.700.697.696.694.692.691.690.689.688.688.687.686.686.685.685.684.684.684.683.683.675
.01(one tail)
.02(two tails)
.025(one tail)
.05(two tails)
.05(one tail)
.10(two tails)
.10(one tail)
.20(two tails)
.25(one tail)
.50(two tails)
Table A-3 t Distribution
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Critical z Value vs Critical t Values
See “t distribution pdf.xls”
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Finding t2 for the following Degrees of
Confidence and sample size
1. 1 - n = 12
2. 1 - n = 15
3. 1 - n = 9
4. 1 - n = 20
Find critical value and sketch
Examples:
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Confidence Interval for the Estimate of µ
Based on an Unknown and a Small Simple Random Sample from a Normally Distributed Population
x - E < µ < x + E
where E = t/2 ns
t/2 found in Table A-3
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Using the Normal and t Distribution
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Example: Let’s do an example comparing z and t. Construct confidence interval’s for each using the following data.
n = 16
x = 50
s = 20
= 0.05/2 = 0.025
Now we wouldn’t use a z distribution here due to the small sample but let’s
do it anyway and compare the width of the confidence interval to a confidence interval created using a t distribution
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Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.)
x = 26,227
s = 15,873
= 0.05/2 = 0.025t/2 = 2.201
E = t2 s = (2.201)(15,873) = 10,085.3
n 12
26,227 - 10,085.3 < µ < 26,227 + 10,085.3
x - E < µ < x + E
We are 95% confident that this interval contains the average cost of repairing a Dodge Viper.
$16,141.7 < µ < $36,312.3
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TI-83 Calculator
Finding Confidence intervals using t
1. Press STAT
2. Cursor to TESTS
3. Choose TInterval
4. Choose Input: STATS*
5. Enter s and x and confidence level6. Cursor to calculate
*If your input is raw data, then input your raw data in L1 then use DATA
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Degreesof
freedom
1234567891011121314151617181920212223242526272829
Large (z)
63.6579.9255.8414.6044.0323.7073.5003.3553.2503.1693.1063.0543.0122.9772.9472.9212.8982.8782.8612.8452.8312.8192.8072.7972.7872.7792.7712.7632.7562.575
.005(one tail)
.01(two tails)
31.8216.9654.5413.7473.3653.1432.9982.8962.8212.7642.7182.6812.6502.6252.6022.5842.5672.5522.5402.5282.5182.5082.5002.4922.4852.4792.4732.4672.4622.327
12.7064.3033.1822.7762.5712.4472.3652.3062.2622.2282.2012.1792.1602.1452.1322.1202.1102.1012.0932.0862.0802.0742.0692.0642.0602.0562.0522.0482.0451.960
6.3142.9202.3532.1322.0151.9431.8951.8601.8331.8121.7961.7821.7711.7611.7531.7461.7401.7341.7291.7251.7211.7171.7141.7111.7081.7061.7031.7011.6991.645
3.0781.8861.6381.5331.4761.4401.4151.3971.3831.3721.3631.3561.3501.3451.3411.3371.3331.3301.3281.3251.3231.3211.3201.3181.3161.3151.3141.3131.3111.282
1.000.816.765.741.727.718.711.706.703.700.697.696.694.692.691.690.689.688.688.687.686.686.685.685.684.684.684.683.683.675
.01(one tail)
.02(two tails)
.025(one tail)
.05(two tails)
.05(one tail)
.10(two tails)
.10(one tail)
.20(two tails)
.25(one tail)
.50(two tails)
Table A-3 t Distribution
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6-3
Estimating a population proportion
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Assumptions 1. The sample is a random sample.
2. The conditions for the binomial distribution are satisfied (See Section 4-3.)
3. The normal distribution can be used to approximate the distribution of sample proportions because np 5 and nq 5 are both satisfied.
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q = 1 - p = sample proportion of x failures in a sample size of n
p =ˆ xn sample proportion
ˆ
p = population proportion
(pronounced ‘p-hat’)
of x successes in a sample of size n
Notation for Proportions
ˆ
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DefinitionPoint Estimate
The sample proportion p is the best point estimate of the population
proportion p.
ˆ
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Confidence Interval for Population Proportion
p - E < < + E where
ˆ p ˆ p
z
E =
nˆ ˆp q
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Round-Off Rule for Confidence Interval Estimates of p
Round the confidence interval limits to
three significant digits.
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( )2 ˆp q
Determining Sample Size
zE =
p qnˆ ˆ
(solve for n by algebra)
zn =
ˆE2
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Sample Size for Estimating Proportion p
When an estimate of p is known: ˆ
ˆ( )2 p qn =
ˆE2
z
When no estimate of p is known:
( )2 0.25n =
E2
z
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= [1.645]2 (0.675)(0.325)
n = [z/2 ]2 p q
E2
= 1483.8215= 1484 Americans
Example: We want to determine, with a margin of error of two percentage points, the percentage of Americans
who own their house. Assuming that we want 90% confidence in our results, how many Americans must we survey? An earlier study indicates 67.5% of Americans
own their own home.
To be 90% confident that our sample percentage is within two percentage points of the
true percentage for all Americans, we should
randomly select and survey 1484 households.
0.022
ˆˆ
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Round-Off Rule for Sample Size n
When finding the sample size n, if the result is not a whole number, always increase the value of n to the next larger whole number.
n = 1483.8215 = 1484 (rounded up)
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n = [z/2 ]2 (0.25)
E
= (1.645)2 (0.25)
2
= 1690.9647= 1691
Americans
With no prior information, we need a larger sample to achieve the same results
with 90% confidence and an error of no more than 2%.
Example: We want to determine, with a margin of error of two percentage points, the percentage of Americans
who own their house. Assuming that we want 90% confidence in our results, how many Americans must we
survey? There is no prior information suggesting a possible value for the sample percentage.
0.022
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TI-83 Calculator
Finding Confidence intervals using z (proportions)
1. Press STAT
2. Cursor to TESTS
3. Choose 1-ProbZInt
4. Enter x and n and confidence level5. Cursor to calculate