IS 310 Business Statistics CSU Long Beach
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Transcript of IS 310 Business Statistics CSU Long Beach
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IS 310 – Business StatisticsIS 310 – Business Statistics
IS 310
Business Statistic
sCSU
Long Beach
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IS 310 – Business StatisticsIS 310 – Business Statistics
Interval EstimationInterval Estimation
In chapter 7, we studied how to estimate a population In chapter 7, we studied how to estimate a population parameter with a sample statistic. We used a point estimate parameter with a sample statistic. We used a point estimate as a single value. as a single value.
To increase the level of confidence in estimation, we use a To increase the level of confidence in estimation, we use a range of values (rather than a single value) as the estimate range of values (rather than a single value) as the estimate of a population parameter.of a population parameter.
Let’s take a real-life example. Suppose, someone asks you how Let’s take a real-life example. Suppose, someone asks you how long it takes to go from CSULB to LAX. What would be a long it takes to go from CSULB to LAX. What would be a more reliable estimate – 30 minutes or between 25 and 40 more reliable estimate – 30 minutes or between 25 and 40 minutes? minutes?
If you use 30 minutes as estimate, you are using a point If you use 30 minutes as estimate, you are using a point estimate. On the other hand, if you use between 25 and 40 estimate. On the other hand, if you use between 25 and 40 minutes as estimate, you are using an interval estimationminutes as estimate, you are using an interval estimation..
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IS 310 – Business StatisticsIS 310 – Business Statistics
Interval EstimationInterval Estimation
Interval estimation uses a range of values.Interval estimation uses a range of values.
The width of the range indicates the level of The width of the range indicates the level of confidence. The narrower the range, lower the confidence. The narrower the range, lower the confidence. Wider the range, higher the confidence. Wider the range, higher the confidence.confidence.
Once a confidence level is specified, the interval Once a confidence level is specified, the interval estimate can be calculated using the formula, 8.1, estimate can be calculated using the formula, 8.1, in the book. If one wants 95% confidence, the in the book. If one wants 95% confidence, the interval estimate is known as 95% Confidence interval estimate is known as 95% Confidence Interval. For a 90% confidence, the interval is Interval. For a 90% confidence, the interval is known as 90% Confidence Interval.known as 90% Confidence Interval.
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IS 310 – Business StatisticsIS 310 – Business Statistics
Interval EstimationInterval Estimation
In this chapter, we will cover the following:In this chapter, we will cover the following:
Interval Estimate for a Population Mean Interval Estimate for a Population Mean (known (known σσ))
Interval Estimate for a Population Mean Interval Estimate for a Population Mean (unknown (unknown σσ))
Determining Size of a SampleDetermining Size of a Sample
Interval Estimate for a Population Proportion Interval Estimate for a Population Proportion
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The general form of an interval estimate of aThe general form of an interval estimate of a population mean ispopulation mean is
The general form of an interval estimate of aThe general form of an interval estimate of a population mean ispopulation mean is
Margin of Errorx Margin of Errorx
Margin of Error and the Interval EstimateMargin of Error and the Interval Estimate
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Margin of ErrorMargin of Error
The Margin of Error implies bothThe Margin of Error implies both
Confidence level andConfidence level and
Amount of error Amount of error
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Interval Estimate ofInterval Estimate of
Interval Estimate of a Population Mean:Interval Estimate of a Population Mean: Known Known
x zn
/2x zn
/2
where: is the sample meanwhere: is the sample mean 1 -1 - is the confidence coefficient is the confidence coefficient zz/2 /2 is the is the zz value providing an area of value providing an area of /2 in the upper tail of the standard /2 in the upper tail of the standard
normal probability distributionnormal probability distribution is the population standard deviationis the population standard deviation nn is the sample size is the sample size
xx
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Interval Estimate for Population Mean Interval Estimate for Population Mean (Known (Known σσ))
Example:Example: Lloyd’s Department Store wants to determine Lloyd’s Department Store wants to determine
a 95% confidence interval on the average a 95% confidence interval on the average amount spent by all of its customers amount spent by all of its customers (population mean).(population mean).
Lloyd took a sample of 100 customers and Lloyd took a sample of 100 customers and found that the average amount spent is found that the average amount spent is $82 (this is sample mean or x‾ ). Lloyd $82 (this is sample mean or x‾ ). Lloyd assumes that the population standard assumes that the population standard deviation ( deviation ( σσ ) is $20. ) is $20.
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IS 310 – Business StatisticsIS 310 – Business Statistics
Interval Estimation for Population MeanInterval Estimation for Population Mean(Known (Known σσ ) )
Using Formula 8.1, the 95% confidence interval is:Using Formula 8.1, the 95% confidence interval is: __ x ± z (x ± z (σσ/√ n) = 0.05/√ n) = 0.05 0.0250.025
82 ± 1.96 (20/√100)82 ± 1.96 (20/√100)
82 ± 3.9282 ± 3.92
78.08 to 85.9278.08 to 85.92
We are 95% sure that the average amount spent by all We are 95% sure that the average amount spent by all Lloyd customers is between $78.08 and $85.92.Lloyd customers is between $78.08 and $85.92.
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Sample ProblemSample Problem
Problem # 5 (10-Page 306; 11-Page 315)Problem # 5 (10-Page 306; 11-Page 315)
__
Given: x = 24.80 n = 49 Given: x = 24.80 n = 49 σσ = 5 confidence level = 95% = = 5 confidence level = 95% = 0.05 or 5%0.05 or 5%
a.a. Margin of Error = z x (Margin of Error = z x (σσ / √n) / √n)
/2/2
= 1.96 (5/√49) = 1.4= 1.96 (5/√49) = 1.4
b.b. 95% Confidence Interval is:95% Confidence Interval is:
_ _
x ± Margin of Error 24.8 ± 1.4 23.4, 26.2x ± Margin of Error 24.8 ± 1.4 23.4, 26.2
It means that we are 95% confident that the average amount It means that we are 95% confident that the average amount spent by all customers for dinner at the Atlanta restaurant is spent by all customers for dinner at the Atlanta restaurant is between $23.40 and $26.20between $23.40 and $26.20
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Interval Estimation of a Population Mean:Interval Estimation of a Population Mean: Unknown Unknown
If an estimate of the population standard If an estimate of the population standard deviation deviation cannot be developed prior to cannot be developed prior to sampling, we use the sample standard sampling, we use the sample standard deviation deviation ss to estimate to estimate . . This is the This is the unknown unknown case. case.
In this case, the interval estimate for In this case, the interval estimate for is based is based on the on the tt distribution. distribution.
(We’ll assume for now that the population is (We’ll assume for now that the population is normally distributed.)normally distributed.)
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The The t t distribution distribution is a family of similar probability is a family of similar probability distributions.distributions. The The t t distribution distribution is a family of similar probability is a family of similar probability distributions.distributions.
tt Distribution Distribution
A specific A specific tt distribution depends on a parameter distribution depends on a parameter known as the known as the degrees of freedomdegrees of freedom.. A specific A specific tt distribution depends on a parameter distribution depends on a parameter known as the known as the degrees of freedomdegrees of freedom..
Degrees of freedom refer to the sample size minus 1.Degrees of freedom refer to the sample size minus 1. Degrees of freedom refer to the sample size minus 1.Degrees of freedom refer to the sample size minus 1.
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tt Distribution Distribution
A A tt distribution with more degrees of freedom has distribution with more degrees of freedom has less dispersion.less dispersion. A A tt distribution with more degrees of freedom has distribution with more degrees of freedom has less dispersion.less dispersion.
As the number of degrees of freedom increases, theAs the number of degrees of freedom increases, the difference between the difference between the tt distribution and the distribution and the standard normal probability distribution becomesstandard normal probability distribution becomes smaller and smaller.smaller and smaller.
As the number of degrees of freedom increases, theAs the number of degrees of freedom increases, the difference between the difference between the tt distribution and the distribution and the standard normal probability distribution becomesstandard normal probability distribution becomes smaller and smaller.smaller and smaller.
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tt Distribution Distribution
StandardStandardnormalnormal
distributiondistribution
tt distributiondistribution(20 degrees(20 degreesof freedom)of freedom)
tt distributiondistribution(10 degrees(10 degrees
of of freedom)freedom)
00zz, , tt
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tt Distribution Distribution
Degrees Area in Upper Tail
of Freedom .20 .10 .05 .025 .01 .005
. . . . . . .
50 .849 1.299 1.676 2.009 2.403 2.678
60 .848 1.296 1.671 2.000 2.390 2.660
80 .846 1.292 1.664 1.990 2.374 2.639
100 .845 1.290 1.660 1.984 2.364 2.626
.842 1.282 1.645 1.960 2.326 2.576
Standard Standard normalnormalzz values values
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Interval EstimateInterval Estimate
x tsn
/2x tsn
/2
where: 1 -where: 1 - = the confidence coefficient = the confidence coefficient
tt/2 /2 == the the tt value providing an area of value providing an area of /2/2 in the upper tail of a in the upper tail of a t t distribution distribution with with nn - 1 degrees of freedom - 1 degrees of freedom ss = the sample standard deviation = the sample standard deviation
Interval Estimation of a Population Mean:Interval Estimation of a Population Mean: Unknown Unknown
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Interval Estimate for Population Mean Interval Estimate for Population Mean (Unknown (Unknown σσ))
Example:Example: We want to estimate the mean credit card debt We want to estimate the mean credit card debt
of all customers in the U.S. with a 95% of all customers in the U.S. with a 95% confidence.confidence.
A sample of 70 households provided the credit A sample of 70 households provided the credit card balances shown in Table 8.3. The sample card balances shown in Table 8.3. The sample standard deviation (s) is $4007.standard deviation (s) is $4007.
Using formula 8.2,Using formula 8.2, __ x ± t (s/√n) = 0.05x ± t (s/√n) = 0.05 0.0250.025 9312 ± 1.995 (4007/√70)9312 ± 1.995 (4007/√70)
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Interval Estimate (Continued)Interval Estimate (Continued)
9312 ± 9559312 ± 955 8357 to 10,2678357 to 10,267
We are 95% confident that the average credit We are 95% confident that the average credit card balance of all customers in the U.S. are card balance of all customers in the U.S. are between $8,357 and $10,267. between $8,357 and $10,267.
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Summary of Interval Estimation Summary of Interval Estimation ProceduresProcedures
for a Population Meanfor a Population Mean
Can theCan thepopulation standardpopulation standard
deviation deviation be assumed be assumed known ?known ?
Use the sampleUse the samplestandard deviationstandard deviation
ss to estimate to estimate ss
UseUse
YesYes NoNo
/ 2
sx t
n / 2
sx t
nUseUse
/ 2x zn
/ 2x z
n
KnownKnownCaseCase
UnknownUnknownCaseCase
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Sample ProblemSample Problem
Problem # 16 (10-Page 315; 11-Page 324)Problem # 16 (10-Page 315; 11-Page 324)
__
Given: x = 49 n = 100 s = 8.5 Confidence Level = 90%Given: x = 49 n = 100 s = 8.5 Confidence Level = 90%
a.a. At 90% confidence, the Margin of Error is:At 90% confidence, the Margin of Error is:
t . (s/√n) 1.660 x (8.5/10) = 1.411 t . (s/√n) 1.660 x (8.5/10) = 1.411
/2 /2
b.b. The 90% Confidence Interval is:The 90% Confidence Interval is:
_ _
x ± 1.411 49 ± 1.411 47.59, 50.41x ± 1.411 49 ± 1.411 47.59, 50.41
It means that we are 90% confident that the average hours It means that we are 90% confident that the average hours of flying for Continental pilots is between 47.59 and 50.41of flying for Continental pilots is between 47.59 and 50.41
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Let Let EE = the desired margin of error. = the desired margin of error. Let Let EE = the desired margin of error. = the desired margin of error.
EE is the amount added to and subtracted from the is the amount added to and subtracted from the point estimate to obtain an interval estimate.point estimate to obtain an interval estimate. EE is the amount added to and subtracted from the is the amount added to and subtracted from the point estimate to obtain an interval estimate.point estimate to obtain an interval estimate.
Sample Size for an Interval EstimateSample Size for an Interval Estimateof a Population Meanof a Population Mean
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Sample Size for an Interval EstimateSample Size for an Interval Estimateof a Population Meanof a Population Mean
E zn
/2E zn
/2
nz
E
( )/ 22 2
2n
z
E
( )/ 22 2
2
Margin of ErrorMargin of Error
Necessary Sample SizeNecessary Sample Size
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Determination of Sample SizeDetermination of Sample Size
Example Problem (10-Page 317; 11-Page 326):Example Problem (10-Page 317; 11-Page 326): We want to know the average daily rental rate of all We want to know the average daily rental rate of all
midsize automobiles in the U.S. We want our estimate midsize automobiles in the U.S. We want our estimate with a margin of error of $2 and a 95% level of confidence.with a margin of error of $2 and a 95% level of confidence.
What should be the size of our sample?What should be the size of our sample? Given for this problem: E =2, = 0.05, Given for this problem: E =2, = 0.05, σσ = 9.65 = 9.65
Using Formula 8.3,Using Formula 8.3, 2 2 22 2 2 n = (z ) (n = (z ) (σσ ) / E ) / E 0.0250.025 = 89.43= 89.43 The sample size needs to be at least 90.The sample size needs to be at least 90.
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Sample ProblemSample Problem
Problem #26 (10-Page 318; 11-Page 327)Problem #26 (10-Page 318; 11-Page 327)
Given: µ = 2.41 Given: µ = 2.41 σσ = 0.15 Margin of Error = 0.07 = 0.15 Margin of Error = 0.07
Confidence Level = 95%Confidence Level = 95%
Margin of Error = z . (Margin of Error = z . (σσ /√n) /√n)
/2 2 2 2/2 2 2 2
0.07 = 1.96 (0.15/√n) n = (1.96) (0.15) / 0.07)0.07 = 1.96 (0.15/√n) n = (1.96) (0.15) / 0.07)
= 17.63 or 18= 17.63 or 18
A sample size of 18 is recommended for The Cincinnati A sample size of 18 is recommended for The Cincinnati Enquirer for its studyEnquirer for its study
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The general form of an interval estimate of aThe general form of an interval estimate of a population proportion ispopulation proportion is
The general form of an interval estimate of aThe general form of an interval estimate of a population proportion ispopulation proportion is
Margin of Errorp Margin of Errorp
Interval EstimationInterval Estimationof a Population Proportionof a Population Proportion
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Interval EstimationInterval Estimationof a Population Proportionof a Population Proportion
The sampling distribution of plays a key role inThe sampling distribution of plays a key role in computing the margin of error for this intervalcomputing the margin of error for this interval estimate.estimate.
The sampling distribution of plays a key role inThe sampling distribution of plays a key role in computing the margin of error for this intervalcomputing the margin of error for this interval estimate.estimate.
pp
The sampling distribution ofThe sampling distribution of can be approximated can be approximated by a normal distribution whenever by a normal distribution whenever npnp >> 5 and 5 and nn(1 – (1 – pp) ) >> 5. 5.
The sampling distribution ofThe sampling distribution of can be approximated can be approximated by a normal distribution whenever by a normal distribution whenever npnp >> 5 and 5 and nn(1 – (1 – pp) ) >> 5. 5.
pp
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/2/2 /2/2
Interval EstimationInterval Estimationof a Population Proportionof a Population Proportion
Normal Approximation of Sampling Distribution Normal Approximation of Sampling Distribution of of
pp
Samplingdistribution of
Samplingdistribution of pp
(1 )p
p p
n
(1 )p
p p
n
pppp
/ 2 pz / 2 pz / 2 pz / 2 pz
1 - of all values1 - of all valuespp
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Interval EstimateInterval Estimate
Interval EstimationInterval Estimationof a Population Proportionof a Population Proportion
p zp pn
/
( )2
1p z
p pn
/
( )2
1
where: 1 -where: 1 - is the confidence coefficient is the confidence coefficient
zz/2 /2 is the is the zz value providing an area of value providing an area of
/2 in the upper tail of the standard/2 in the upper tail of the standard
normal probability distributionnormal probability distribution
is the sample proportionis the sample proportionpp
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Interval Estimate for a Population Interval Estimate for a Population ProportionProportion
Example (10-Page 320; 11-Page 329):Example (10-Page 320; 11-Page 329): We want to estimate the proportion of all women golfers in the We want to estimate the proportion of all women golfers in the
U.S. who are satisfied with the availability of tee times with a U.S. who are satisfied with the availability of tee times with a 95% confidence level.95% confidence level.
We take a sample of 900 women golfers and find that 396 are We take a sample of 900 women golfers and find that 396 are satisfied with the tee times.satisfied with the tee times.
__ Given: p = 396/900 = 0.44 = 0.05Given: p = 396/900 = 0.44 = 0.05 The 95% confidence interval is:The 95% confidence interval is: _ _ __ _ _ p ± z √ [p (1 – p)]/n = 0.44 ± 1.96√ [0.44 (1 – 0.44)]/900p ± z √ [p (1 – p)]/n = 0.44 ± 1.96√ [0.44 (1 – 0.44)]/900 0.44 ± 0.0324 0.4076 to 0.4724 or 40.76% to 47.24%0.44 ± 0.0324 0.4076 to 0.4724 or 40.76% to 47.24% We are 95% confident that the proportion of all women golfers We are 95% confident that the proportion of all women golfers
who are satisfied with tee times is between 40.76% and who are satisfied with tee times is between 40.76% and 47.24%.47.24%.
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Sample ProblemSample Problem
#38 (10-Page 323; 11-Page 332))#38 (10-Page 323; 11-Page 332))
a.a. Point estimate of the proportion of companies that Point estimate of the proportion of companies that
fell short of estimates = 29/162 = 0.179fell short of estimates = 29/162 = 0.179
_ _
b.b. Given : p = 104/162 = 0.642 _ _Given : p = 104/162 = 0.642 _ _
Margin of error = z √ [(p (1 – p ) / n)]Margin of error = z √ [(p (1 – p ) / n)]
0.025 0.025
= 1.96 √ [(0.642) (1 – 0.642) / = 1.96 √ [(0.642) (1 – 0.642) / 162]162]
= 0.0738= 0.0738
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Sample Problem (contd)Sample Problem (contd)
95% confidence interval:95% confidence interval: 0.642 ± 0.07380.642 ± 0.0738 0.5682 and 0.71580.5682 and 0.7158
We are 95% confident that the proportion of companies We are 95% confident that the proportion of companies that beat estimates of their profits is between 56.82% that beat estimates of their profits is between 56.82% and 71.58%. and 71.58%.
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c. Required sample size, n = [(1.96) (0.642) c. Required sample size, n = [(1.96) (0.642) (0.358)]/(0.05) (0.358)]/(0.05)
= 353.18 use 354= 353.18 use 354
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End of Chapter 8End of Chapter 8