Unformatted text preview: Business Statistics (BUSA 3101)
Dr. Lari H. Arjomand
[email protected] Slide 1 Chapter 9 Estimation & Confidence Interval
s Interval Estimation for Population Mean: σ Known s Interval Estimation for Population Mean: σ Unknown
Determining the Sample Size s Slide 2 A point estimate is a single value (statistic) used to estimate a population value (parameter). An Interval Estimate states the range within which a population parameter probably lies. A confidence interval is a range of values within which the population parameter is expected to occur. The two confidence intervals that are used extensively are the 95% and the 99%. Slide 3 Interpretation of Interval Estimation For a 95% confidence interval about 95% of the similarly constructed intervals will contain the parameter being estimated. 95% of the sample means for a specified sample size will lie within 1.96 standard deviations of the hypothesized population mean. For the 99% confidence interval, 99% of the sample means for a specified sample size will lie within 2.58 standard deviations of the hypothesized population mean. Slide 4 Margin of Error and the Interval Estimate A point estimator cannot be expected to provide the
A point estimator cannot be expected to provide the exact value of the population parameter.
exact value of the population parameter. An interval estimate can be computed by adding and
An interval estimate can be computed by adding and subtracting a margin of error to the point estimate.
subtracting a margin of error to the point estimate. Point Estimate +/− Margin of Error The purpose of an interval estimate is to provide
The purpose of an interval estimate is to provide information about how close the point estimate is to
information about how close the point estimate is to the value of the parameter.
the value of the parameter. Slide 5 Margin of Error and the Interval Estimate (Continued) The general form of an interval estimate of a
The general form of an interval estimate of a population mean is
population mean is x ± Margin of Error Point Estimate Slide 6 Interval Estimation of a Population Mean:
σ Known (Continued)
s In order to develop an interval estimate of a population mean, the margin of error must be computed using either:
• the population standard deviation σ , or
• the sample standard deviation s s σ is rarely known exactly, but often a good estimate can be obtained based on historical data or other information. s We refer to such cases as the σ known case. Slide 7 Interval Estimation of a Population Mean:
σ Known (Continued) There is a 1 − α probability that the value of a
zα / 2 σ x
sample mean will provide a margin of error of or less.
Sampling distribution of x α/2 1 α of all
x values zα / 2 σ x µ α/2
x zα / 2 σ x Slide 8 Interval Estimate of a Population Mean:
σ Known (Continued)
Sampling distribution of x α/2
interval
does not
include µ 1 α of all
x values zα / 2 σ x µ x
zα / 2 σ x x ]
[ x ] [[ x ] α/2 interval
includes µ Slide 9 Interval Estimate of a Population Mean:
σ Known (Continued)
s Interval Estimate of µ Point Estimation of
Population Mean x ± zα / 2 σ
n Margin of Error where: is the sample mean
x 1 α is the confidence coefficient zα/2 is the z value providing an area of α/2 in the upper tail of the standard normal probability distribution σ is the population standard deviation n is the sample size Slide 10 Interval Estimate of Population Mean:
σ Known
Example: Discount Sounds
Discount Sounds has 260 retail outlets throughout the United States. The firm
is evaluating a potential location for a
new outlet, based in part, on the mean
annual income of the individuals in
the marketing area of the new location.
A sample of size n = 36 was taken;
the sample mean income is $31,100. The population standard deviation is estimated to be $4,500,
and the confidence coefficient to be used in the interval estimate is 0.95. s D S Slide 11 Interval Estimate of Population Mean:
σ Known D S The margin of error is: zα / 2 σ 4,500 = 1.96 = 1, 470
n 36 Thus, at 95% confidence, the margin of error is $1,470. Note: To find the Z from the table do the following:
α/2 = .05/2=.025 and 1.025 = .975 and from table Z is 1.96 Slide 12 Interval Estimate of Population Mean:
σ Known D S Interval estimate of µ is: x ± zα / 2 σ
n $31,100 + $1,470
or
$29,630 to $32,570 We are 95% confident that the interval contains the
population mean. Note that the sample mean was = $31,100. Slide 13 Interval Estimation of a Population Mean:
σ Unknown
s If an estimate of the population standard deviation σ cannot be developed prior to sampling, we use the sample standard deviation s to estimate σ . s This is the σ unknown case. s In this case, the interval estimate for µ is based on the t distribution. Slide 14 t Distribution The t distribution is a family of similar probability
The t distribution is a family of similar probability distributions.
distributions. A specific t distribution depends on a parameter
A specific t distribution depends on a parameter known as the degrees of freedom.
known as the degrees of freedom. Degrees of freedom refer to the number of Degrees of freedom refer to the number of independent pieces of information that go into the
independent pieces of information that go into the computation of ss.
computation of . Slide 15 t Distribution (Continued) A t distribution with more degrees of freedom has
A t distribution with more degrees of freedom has less dispersion.
less dispersion. As the number of degrees of freedom increases, the
As the number of degrees of freedom increases, the difference between the t distribution and the
difference between the t distribution and the standard normal probability distribution becomes
standard normal probability distribution becomes smaller and smaller.
smaller and smaller. Slide 16 t Distribution (Continued)
t distribution
(20 degrees
of freedom) Standard
normal
distribution t distribution
(10 degrees of freedom) 0 z, t Slide 17 Student’s t Table Slide 18 Student’s t Table Upper Tail Area
df .25 .10 .05 1 1.000 3.078 6.314
2 0.817 1.886 2.920
3 0.765 1.638 2.353 Slide 19 Student’s t Table Upper Tail Area
df .25 .10 .05 1 1.000 3.078 6.314
2 0.817 1.886 2.920
3 0.765 1.638 2.353 t values Slide 20 Student’s t Table Upper Tail Area
df .25 .10 α /2 .05 1 1.000 3.078 6.314
2 0.817 1.886 2.920 α /2 3 0.765 1.638 2.353 t values 0 t Slide 21 Student’s t Table Upper Tail Area
df .25 .10 Assume:
n=3
df = n  1 = 2
α = .10
.10
α / 2 =.05 α /2 .05 1 1.000 3.078 6.314
2 0.817 1.886 2.920 α /2 3 0.765 1.638 2.353 t values 0 t Slide 22 Student’s t Table Upper Tail Area
df .25 .10 Assume:
n=3
df = n  1 = 2
α = .10
.10
α / 2 =.05 α /2 .05 1 1.000 3.078 6.314
2 0.817 1.886 2.920 α /2 3 0.765 1.638 2.353 t values 0 t Slide 23 Student’s t Table Upper Tail Area
df .25 .10 Assume:
n=3
df = n  1 = 2
α = .10
.10
=.05
α / 2 =.05 α /2 .05 1 1.000 3.078 6.314
2 0.817 1.886 2.920 .05 3 0.765 1.638 2.353 t values 0 t Slide 24 Student’s t Table Upper Tail Area
df .25 .10 Assume:
n=3
df = n  1 = 2
α = .10
.10
α / 2 =.05 α /2 .05 1 1.000 3.078 6.314
2 0.817 1.886 2.920 .05 3 0.765 1.638 2.353 t values 0 2.920 t Slide 25 Interval Estimation of a Population Mean:
σ Unknown
s Interval Estimate Point Estimation of
Population Mean x ± tα / 2 s
n Margin of Error where: 1 α = the confidence coefficient tα/2 = the t value providing an area of α/2 in the upper tail of a t distribution with n 1 degrees of freedom s = the sample standard deviation Slide 26 Interval Estimation of a Population Mean:
σ Unknown
s Example: Apartment Rents
A reporter for a student newspaper is writing an
article on the cost of offcampus
housing. A sample of 16
efficiency apartments within a
halfmile of campus resulted in
a sample mean of $650 per month and a sample
standard deviation of $55. Slide 27 Interval Estimation of a Population Mean:
σ Unknown (Example Continued)
s Example: Apartment Rents
Let us provide a 95% confidence interval estimate of the mean rent per
month for the population of efficiency apartments within a
halfmile of campus. Slide 28 Interval Estimation of a Population Mean:
σ Unknown (Example Continued)
At 95% confidence, α = .05, and α/2 = .025.
t.025 is based on n − 1 = 16 − 1 = 15 degrees of freedom.
In the t distribution table we see that t.025 = 2.131.
D e gre e s Are a in Uppe r Tail of Fre e dom .20 .100 .050 .025 .010 .005 15 .866 1.341 1.753 2.131 2.602 2.947 16 .865 1.337 1.746 2.120 2.583 2.921 17 .863 1.333 1.740 2.110 2.567 2.898 18 .862 1.330 1.734 2.101 2.520 2.878 19 .861 1.328 1.729 2.093 2.539 2.861 . . . . . . . Slide 29 Interval Estimation of a Population Mean:
σ Unknown (Example Continued) s Interval Estimate x ± t.025 s
n 55
650 ± 2.131
= 650 ± 29.30
16 We are 95% confident that the mean rent per month
for the population of efficiency apartments within a
halfmile of campus is between $620.70 and $679.30. Slide 30 Summary of Interval Estimation Procedures
for a Population Mean
Can the
population standard deviation σ be assumed known ? Yes Use the sample
standard deviation
s to estimate σ σ Known
Case
Use σ x ± zα / 2 n No σ Unknown
Case & Small Sample Use s x ± tα / 2 n Slide 31 If σ is unknown and n>30, the standard deviation of the sample, designated by s, is used to approximate the population standard deviation. Interval Estimation
Summary If the population standard X±z s
n deviation (σ ) is known or the sample (n) is n≥30 we use the z distribution. Slide 32 If the population standard deviation (σ ) is unknown, and the underlying population is approximately normal, and the sample size is less than 30 (n<30) we use the t distribution. Interval Estimation
Summary X ±t s
n The value of t for a given confidence level depends upon its degrees of freedom. Slide 33 Mo re Example s Confidence Interval Estimate for Mean (σ Known) Thinking Challenge Example You’re a Q/C inspector for Gallo. The σ for 2liter bottles is 0.05 liters. A random sample of 100 bottles showed that sample mean = 1.99 liters. What is the 90% confidence interval estimate of the true mean amount in 2liter bottles? To find the Z:
α =190% =0.1 and α/2 =0.1/2 =0.05
1 – 0.05 = 0.95 and from table, the Z is:
(1.64 + 1.65)/ 2 = 1.645 2 liter Slide 35 Confidence Interval Solution σ
σ
X − Zα / 2 ⋅
≤ µ ≤ X + Zα / 2 ⋅
n
n
.05
.05
1.99 − 1.645 ⋅
≤ µ ≤ 1.99 + 1.645 ⋅
100
100
1.982 ≤ µ ≤ 1.998
We are 90% confident that interval estimate of the true mean amount in 2liter bottles is between 1.98 and 1.99. Slide 36 Confidence Interval of µ
(σ Unknown and n ≥ 30)
Example
Example The School of Business Dean at CSU wants to estimate the mean number of hours worked per week by business students. A sample of 49 students showed a mean of 24 hours with a standard deviation of 4 hours. What is the point estimate of the mean number of hours worked per week by students? The point estimate is 24 hours (sample mean). What is the 95% confidence interval for the average number of hours worked per week by the students? Slide 37 Example & Solution (Continued) Using the formula, we have 24 ± 1.96(4/7) or we have 22.88 to 25.12.
What are the 95% confidence limits?
The endpoints of the confidence interval are the confidence limits. The lower confidence limit is 22.88 and the upper confidence limit is 25.12. What degree of confidence is being used?
The degree of confidence (level of confidence) is 0.95. Slide 38 Example & Solution (Continued) Interpret the findings. If we had time to select 100 samples of size 49 from the population of the number of hours worked per week by business students at CSU and compute the sample means and 95% confidence intervals, the population mean of the number of hours worked by the students per week would be found in about 95 out of the 100 confidence intervals. Either a confidence interval contains the population mean or it does not. In this example, about 5 out of the 100 confidence intervals would not contain the population mean. Slide 39 Mo re Example s Confidence Interval Estimate for Mean (σ Unknown, and n<30) Thinking Challenge Example You’re a time study analyst in manufacturing. You’ve recorded the following task times (min.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1.
What is the 90% confidence interval estimate of the population mean task time? Slide 41 Solution
X = 3.7 S = 0.38987 n = 6, df = n 1 = 6 1 = 5 S / √n = 3.8987 / √6 = 0 .1592 t.05,5 = 2.0150 3.7 (2.015)(0.1592) ≤ 3.385 ≤ µ ≤ 3.7 + (2.015)(0.1592) 3.7 + (2.015)(0.1592) µ ≤ 4.015 We are 90% confident that the interval estimate of the population mean task time is between 3.4 and 4.0 minutes. Slide 42 Confidence Interval Excel Application Given the price of ten (10) houses in one of the subdivisions located in Henry County, use Excel to construct a 95% confidence interval for the population mean. Data: $230,000, $240,000, $310,000, $198,000, $257,000, $345,000, $315,000, $260,000, $198,000, $270,000. Slide 43 Excel SolutionSWStat Data Area Slide 44 Excel Solution—SWStat (Continued)
(SWStat Statistics Intervals & Tests) Slide 45 Excel Solution—SWStat (Continued) $227,099 ≤ µ ≤ $297,502
$227,099
$297,502 Slide 46 Confidence Interval Excel Application
(Another Problem)
s s A random sample of 36 magazine subscribers is taken to estimate the mean age of all subscribers. Use Excel to construct a 90% confidence interval estimate of the mean age of all of this magazine’s subscribers. See next slide for the data. Slide 47 The Data
Subscriber Age Subscriber Age Subscriber Age 1 39 13 40 25 38 2 27 14 35 26 51 3 38 15 35 27 26 4 33 16 41 28 39 5 40 17 34 29 35 6 35 18 46 30 37 7 51 19 44 31 33 8 36 20 44 32 41 9 47 21 43 33 36 10 28 22 32 34 33 11 33 23 29 35 46 12 35 24 33 36 37
Slide 48 Excel SolutionSWStat Slide 49 Excel Solution (Continued)
SWStat Slide 50 Sample Size for an Interval Estimate
of a Population Mean Let E = the desired margin of error.
Let E = the desired margin of error. We said E is the amount added to and subtracted We said E is the amount added to and subtracted from the point estimate to obtain an interval estimate.
from the point estimate to obtain an interval estimate. x ± tα / 2 s
n Interval Estimate
of the mean E = Margin of Error Slide 51 Sample Size for an Interval Estimate
of a Population Mean (Continued)
s Margin of Error E = zα / 2
s σ
n Margin of Error Necessary Sample Size ( zα / 2 ) 2 σ 2
n=
E2 Slide 52 Sample Size for an Interval Estimate
of a Population MeanExample Recall that Discount Sounds is evaluating a potential location for a new retail outlet, based in part, on the mean annual income of the individuals in
the marketing area of the new location. Suppose that Discount Sounds’ management team
wants an estimate of the population mean such that
there is a 0.95 probability that the sampling error is $500
or less. How large a sample size is needed to meet the required precision? D S Slide 53 Sample Size for an Interval Estimate
of a Population MeanSolution
E = zα / 2 σ
= 500
n D S Given
Given At 95% confidence, z.025 = 1.96. Recall that σ = 4,500.
= 4,500. ( zα / 2 ) 2 σ 2
n=
E2
(1.96)2 (4, 500)2
n=
= 311.17 = 312
2
(500) A sample of size 312 is needed to reach a desired precision of + $500 at 95% confidence. Slide 54 SampleSize for an Interval Estimate
of a Population Mean (Continued) There are 3 factors that determine the size of a sample, none of which has any direct relationship to the size of the population. They are: 1 The degree of confidence selected. 2 The maximum allowable errormargin of error. 3 The variation of the population. ( zα / 2 ) 2 σ 2
n=
E2 Slide 55 Thinking Challenge Sample Size Example 1 A consumer group would like to estimate the mean monthly electric bill for a single family house in July. Based on similar studies the standard deviation is estimated to be $20.00. A 99% level of confidence is desired, with an accuracy of ± $5.00. How large a $5.00
sample is required? n = [(2.58)(20)/5]2 = 106.5024 ≈ 107 ( zα / 2 ) 2 σ 2
n=
E2 Slide 56 Thinking Challenge Sample Size Example 1 (Continued) What sample size is needed to be 90% confident of being correct within ± 5? A pilot study suggested that the standard deviation is 45. Slide 57 Thinking Challenge Sample Size Example 1 (Continued) What sample size is needed to be 90% confident of being correct within ± 5? A pilot study suggested that the standard deviation is 45. Note that in this example both the degree of confidence and population standard deviation are changed hence, the sample size is changed too. Slide 58 Thinking Challenge Sample Size Example 2 You work in Human Resources at Merrill Lynch. You plan to survey employees to find their average medical expenses. You want to be 95% confident that the sample mean is within ± $50. A pilot study showed that sample standard deviation was about $400. What sample size do you use? Slide 59 Thinking Challenge Sample Size Example 2 (Solution) Slide 60 Interval Estimation
of a Population Proportion
OPTIONAL
READINGS Slide 61 End of Chapter 9 Slide 62 ...
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 Fall '09
 Statistics, interval estimate

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