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Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-1
Business Statistics: A Decision-Making Approach
8th Edition
Chapter 8Estimating Single
Population Parameters
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 8-2
Chapter Goals
After completing this chapter, you should be able to: Distinguish between a point estimate and a confidence
interval estimate Construct and interpret a confidence interval estimate for a
single population mean using both the z and t distributions Determine the required sample size to estimate a single
population mean or a proportion within a specified margin of error
Form and interpret a confidence interval estimate for a single population proportion
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Overview of the Chapter
Builds upon the material from Chapter 1 and 7 Introduces using sample statistics to estimate population
parameters Because gaining access to population parameters can be
expensive, time consuming and sometimes not feasible Confidence Intervals for the Population Mean, μ
when Population Standard Deviation σ is Known when Population Standard Deviation σ is Unknown
Confidence Intervals for the Population Proportion, p Determining the Required Sample Size for means and
proportions
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Estimation Process
(mean, μ, is unknown)
Population
Random Sample
(point estimate)
Mean x = 50
Sample
I am 95% confident that μ is between 40 & 60.
confidenceinterval
Confidence Level
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Point Estimate
Suppose a poll indicate that 62% (sample mean) of the people favor limiting property taxes to 1% of the market value of the property.
The 62% is the point estimate of the true population of people who favor the property-tax limitation.
EPA Automobile Mileage Test Result (point estimate) A point estimate is a single number, used to estimate
an unknown population parameter
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Confidence Interval
The point estimate is not likely to exactly equal the population parameter because of sampling error.
Probability of “sample mean = population mean” is zero With sample mean, it is impossible to determining how
far the sample mean is from the population mean. To overcome this problem, “confidence interval” can be
used as the most common procedure. Stated in terms of level of confidence: Never 100% sure
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Confidence Level
Confidence Level (not same as critical value, α) Describes how strongly we believe that a particular
sampling method will produce a confidence interval that includes the true population parameter.
A percentage (less than 100%) Most common: 90% (α = 0.1), 95% (α = 0.05), 99%
Suppose confidence level = 95% In the long run, 95% of all the confidence intervals will
contain the unknown true parameter
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General Formula
The general formula for the confidence interval is:
Point Estimate (Critical Value)(Standard Error)
z-value (or t value) based on the level of confidence desired
n
σzx
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How to Find the Critical Value
The Central Limit Theorem states that the sampling distribution of a statistic will be normal or nearly normal, if any of the following conditions apply.
n > 30: will give a sampling distribution that is nearly normal The sampling distribution of the mean is normally distributed
because the population distribution is normally distributed.
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How to Find the Critical Value
When one of these conditions is satisfied, the critical value can be expressed as a z score or as a t score. To find the critical value, follow these steps.
Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 0.95 = 0.05 Find the critical probability (p*): p* = 1 – α/2 (because there are lower
confidence limit and upper confidence limit) =1 - 0.05/2 = 0.975 To express the critical value as a z score, find the z score having a
cumulative probability equal to the critical probability (p*). See the example 8-1 on page 335
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t score as the critical value
When the population standard deviation is unknown or when the sample size is small, the t score is preferred.
Find the degrees of freedom (DF). When estimating a mean score or a proportion from a single sample, DF is equal to the sample size minus one. For other applications, the degrees of freedom may be calculated differently.
The critical t score (t*) is the t score having degrees of freedom equal to DF and a cumulative probability equal to the critical probability (p*).
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From Chapter 7
The standard deviation of the possible sample means computed from all random samples of size n is equal to the population standard deviation divided by the square root of the sample size:
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n
σσx Also called the
standard errorn
σzx
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Margin of Error
Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval
n
σzx
n
σze
Example: Margin of error for estimating μ, σ known:
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Point and Interval Estimates
So, a confidence interval provides additional information about variability within a range of z-values
The interval incorporates the sampling error
Point Estimate
Lower
Confidence
Limit
Upper
Confidence
Limit
Width of confidence interval
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Using Analysis ToolPak
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Using Analysis ToolPak
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Margin of Error
Confidence Interval
119.9 (mean) ± 2.59
117.31 ------ 122.49
Download “Confidence Interval Example” Excel file
Don’t even worry about p* = 1 - α/2
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Using Analysis ToolPak(large sample: use normal (z) distribution automatically )
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Using Analysis ToolPak (small sample: use (t) distribution automatically)
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Factors Affecting Margin of Error
Data variation, σ : e as σ
Sample size, n : e as n
Level of confidence, 1 - : e if 1 -
Video Lecture: Confidence Intervals
n
σze
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Confidence Intervals
Population Mean
σ Unknown
ConfidenceIntervals
PopulationProportion
σ Known
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Confidence Interval for μ(σ Known)
Assumptions Population standard deviation σ is known If population is not normal, use larger sample n > 30
(Central Limit Theorem)
Confidence interval estimate
n
σzx
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Finding the Critical Value
Consider a 95% confidence interval:
n
σzx
-z = -1.96 z = 1.96
.95/22/)(1
.0252
α .025
2
α
Point EstimateLower Confidence Limit
UpperConfidence Limit
z units:
x units: Point Estimate
0
1.96z
n
σzx x
0.95/2 = 0.475 find on z table
Normsinv(0.5 – 0.475)=1.959963…
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Common Levels of Confidence
Commonly used confidence levels are 90%, 95%, and 99%
Confidence Level
Critical value, z
1.28
1.645
1.96
2.33
2.58
3.08
3.27
80%
90%
95%
98%
99%
99.8%
99.9%
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Computing a Confidence Interval Estimate for the Mean (s known)
1. Select a random sample of size n
2. Specify the confidence level
3. Compute the sample mean
4. Determine the standard error
5. Determine the critical value (z) from the normal table
6. Compute the confidence interval estimate
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Example
A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is 0.35 ohms.
Determine a 95% (α = 0.05) confidence interval for the true mean resistance of the population.
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2.4068 1.9932
.2068 2.20
)11(0.35/ 1.96 2.20
n
σz x
Example Solution
Solution:(continued)
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Interpretation
We are 95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms Although the true mean may or may not be in this interval, 95% of
intervals formed in this manner will contain the true mean
An incorrect interpretation is that there is 95% probability that this
interval contains the true population mean.
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In most real world situations, population mean and StdDev are NOT KNOWN.
When the population standard deviation is unknown or when the sample size is small, the t score is preferred.
When the sample size is large (n > 30), it doesn't make much difference. Both approaches yield similar results.
So we use the t distribution instead of the normal distribution.
Confidence Interval for μ(σ Unknown)
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Assumptions Population standard deviation is unknown
Sample size is small If population is not normal, use large sample n > 30
Use Student’s t Distribution Confidence Interval Estimate
Confidence Interval for μ(σ Unknown)
n
stx
(continued)
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The t is a family of distributions The t value depends on degrees of freedom (d.f.)
For example, if n = 28, then the d.f. is 27.
As the d.f. increase, the t distribution approaches the normal distribution (see the t distribution simulation one the class website)
d.f. = n – 1 Only n-1 independent pieces of data information left in the
sample because the sample mean has already been obtained
Student’s t Distribution
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Student’s t Distribution
t0
t (df = 5)
t (df = 13)t-distributions are bell-shaped and symmetric, but have ‘fatter’ tails than the normal
Standard Normal
(t with df = )
Note: t compared to z as n increases
As n the estimate of s becomes better so t
converges to z
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Student’s t Table
Confidence Level
df 0.50 0.80 0.90
1 1.000 3.078 6.314
2 0.817 1.886 2.920
3 0.765 1.638 2.353
t0 2.920The body of the table contains t values, not probabilities
Let: n = 3 df = n - 1 = 2 confidence level: 90%
/2 = 0.05
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t Distribution Values
With comparison to the z value
Confidence t t t z Level (10 d.f.) (20 d.f.) (30 d.f.) ____
0.80 1.372 1.325 1.310 1.28
0.90 1.812 1.725 1.697 1.64
0.95 2.228 2.086 2.042 1.96
0.99 3.169 2.845 2.750 2.58
Note: t compared to z as n increases
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Example
A random sample of n = 25 has x = 50 and s = 8. Form a 95% confidence interval for μ
d.f. = n – 1 = 24, so
The confidence interval is
2.0639t0.025,241n,/2 t
25
8(2.0639)50
n
stx
46.698 53.302
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TINV
NORMSINV(p) gives the z-value that puts probability (area) p to the left of that value of z.
TINV(p,DF) gives the t-value that puts one-half the probability (area) to the right with DF degrees of freedom.
Download and then review “TDIST Vs. TINV” Only need to review “TINV”
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TINV Example
For 95% confidence intervals we use α = .05, so that we are looking t.025.
Suppose d.f. is 17 t.025,17 = t value that puts .025 to the right of t with 17
degrees of freedom. Since TINV splits α = .05 to .025, this value is =TINV(.05,17).
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In-class Practice Example
The file Excel file Coffee on the class website contains a random sample of 144 German coffee drinkers and measures the annual coffee consumption in kilograms for each sampled coffee drinker. A marketing research firm wants to use this information to develop an advertising campaign to increase German coffee consumption.
Develop a 95% , 90% and 75% confidence interval estimates for the mean annual coffee consumption of German coffee drinkers.
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Determining Required Sample Size
Wishful situation High confidence level, low margin of error, and small sample size
In reality, conflict among three…. For a given sample size, a high confidence level will tend to
generate a large margin of error For a given confidence level, a small sample size will result in an
increased margin of error Reducing of margin of error requires either reducing the confidence
level or increasing the sample size, or both
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Determining Required Sample Size
How large a sample size do I really need? Sampling budget constraint Cost of selecting each item in the sample
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Determining Required Sample SizeWhen σ is known
The required sample size can be found to reach a desired margin of error (e) and
level of confidence (1 - ) Required sample size, σ known:
2
222
e
σz
e
σzn
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Required Sample Size Example
If = 45 (known), what sample size is needed to be 90% confident of being correct within ± 5?
(Always round up)
219.195
(45)1.645
e
σzn
2
22
2
22
So the required sample size is n = 220
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If σ is unknown (most real situation)
If σ is unknown, three possible approaches
1. Use a value for σ that is expected to be at least as large as the true σ
2. Select a pilot sample (smaller than anticipated sample size) and then estimate σ with the pilot sample standard deviation, s
3. Use the range of the population to estimate the population’s Std Dev. As we know, µ ± 3σ contains virtually all of the data. Range = max – min.
Thus, R = (µ + 3σ) – (µ - 3σ) = 6σ. Therefore, σ = R/6 (or R/4 for a more conservative estimate, producing a larger sample size)
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Example when σ is unknown
Jackson’s Convenience Stores Using a pilot sample approach Available on the class website
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Confidence Intervals for the Population Proportion, π
An interval estimate for the population proportion ( π ) can be calculated by adding an allowance for uncertainty to the sample proportion ( p ). For example, estimation of the proportion of
customers who are satisfied with the service provided by your company
Sample proportion, p = x/n
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Confidence Intervals for the Population Proportion, π
Recall that the distribution of the sample proportion is approximately normal if the sample size is large, with standard deviation
We will estimate this with sample data:
(continued)
n
p)p(1sp
n
π)π(1σπ
See Chpt. 7!!
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Confidence Interval Endpoints
Upper and lower confidence limits for the population proportion are calculated with the formula
where z is the standard normal value for the level of confidence desired p is the sample proportion n is the sample size
n
p)p(1zp
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Example
A random sample of 100 people
shows that 25 are left-handed.
Form a 95% confidence interval for
the true proportion of left-handers
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Example
A random sample of 100 people shows that 25 are left-handed. Form a 95% confidence interval for the true proportion of left-handers.
1.
2.
3.
0.0433 /1000.25(0.75)p)/np(1S
0.2525/100 p
p
0.3349 0.1651
(0.0433) 1.96 0.25
(continued)
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Interpretation
We are 95% confident that the true percentage of left-handers in the population is between
16.51% and 33.49%
Although this range may or may not contain the true proportion, 95% of intervals formed from samples of size 100 in this manner will contain the true proportion.
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Changing the sample size
Increases in the sample size reduce the width of the confidence interval.
Example: If the sample size in the above example is
doubled to 200, and if 50 are left-handed in the sample, then the interval is still centered at 0.25, but the width shrinks to
0.19 0.31
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Finding the Required Sample Size
for Proportion Problems
n
π)π(1ze
Solve for n:
Define the margin of error:
2
2
e
π)(1πzn
π can be estimated with a pilot sample, if necessary (or conservatively use π = 0.50 – worst possible variation thus the largest sample size)
Will be in % units
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What sample size...?
How large a sample would be necessary to estimate the true proportion defective in a large population within 3%, with 95% confidence?
(Assume a pilot sample yields p = 0.12)
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What sample size...?
Solution:For 95% confidence, use Z = 1.96e = 0.03p = 0.12, so use this to estimate π
So use n = 451
450.74(0.03)
.12)0(0.12)(1(1.96)
e
π)(1πzn
2
2
2
2
(continued)
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Using PHStat
PHStat can be used for confidence intervals for the mean or proportion
Two options for the mean: known and unknown population standard deviation
Required sample size can also be found Download from the textbook website
The link available on the class website
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PHStat Interval Options
options
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PHStat Sample Size Options
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Using PHStat (for μ, σ unknown)
A random sample of n = 25 has x = 50 and s = 8. Form a 95% confidence interval for μ
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Using PHStat (sample size for proportion)
How large a sample would be necessary to estimate the true proportion defective in a large population within 3%, with 95% confidence?
(Assume a pilot sample yields p = 0.12)
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Chapter Summary
Discussed point estimates Introduced interval estimates Discussed confidence interval estimation for
the mean [σ known] Discussed confidence interval estimation for
the mean [σ unknown] Addressed determining sample size for mean
and proportion problems Discussed confidence interval estimation for
the proportion
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