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A term too broad to define, Statistics is an important subject studied by

almost all commerce graduates and undergraduates across the globe.

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Estimation & Confidence Interval Interval Estimation for Population Mean: s

Known Interval Estimation for Population Mean: s Unknown Determining the Sample Size

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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%.

An Interval Estimate states the range within which a population parameter probably lies.

A point estimate is a single value (statistic) used to estimate a population value (parameter).

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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.

95% of the sample means for a specified sample size will lie within 1.96 standard deviations of the hypothesized population mean.

For a 95% confidence interval about 95% of the similarly constructed intervals will contain the parameter being estimated.

Interpretation of Interval Estimation

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A point estimator cannot be expected to provide the exact value of the population parameter.

An interval estimate can be computed by adding and subtracting a margin of error to the point estimate. Point Estimate +/- Margin of Error

The purpose of an interval estimate is to provide information about how close the point estimate is to the value of the parameter.

Margin of Error and the Interval Estimate

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The general form of an interval estimate of a population mean is Margin of Errorx

Margin of Error and the Interval Estimate (Continued)

Point Estimate

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Interval Estimation of a Population Mean:s Known (Continued)

In order to develop an interval estimate of a population mean, the margin of error must be computed using either:• the population standard deviation s , or• the sample standard deviation s

s is rarely known exactly, but often a good estimate can be obtained based on historical data or other information.

We refer to such cases as the s known case.

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There is a 1 - probability that the value of asample mean will provide a margin of error of or less.

z x s/2

/2 /21 - of all valuesx

Sampling distribution of x

x

z x s/2z x s/2

Interval Estimation of a Population Mean:s Known (Continued)

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/2 /21 - of all valuesx

Sampling distribution of x

x

z x s/2z x s/2[------------------------- -------------------------]

[------------------------- -------------------------]

[------------------------- -------------------------]

xx

x

intervaldoes not

include

intervalincludes

intervalincludes

Interval Estimate of a Population Mean:s Known (Continued)

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Interval Estimate of

Interval Estimate of a Population Mean:s Known (Continued)

where: is the sample mean 1 - is the confidence coefficient z/2 is the z value providing an area of /2 in the upper tail of the standard

normal probability distribution s is the population standard deviation n is the sample size

x

x zn

s

/2 Margin of Error

Point Estimation ofPopulation Mean

14 Slide

SD

Interval Estimate of Population Mean:s Known

Example: Discount SoundsDiscount Sounds has 260 retail outlets

throughout the United States. The firmis evaluating a potential location for anew outlet, based in part, on the meanannual income of the individuals inthe 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.

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The margin of error is:

s

/ 2

4,5001.96 1,47036zn

Thus, at 95% confidence, the margin of error is $1,470.

SD

Interval Estimate of Population Mean:s Known

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

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Interval estimate of is:

Interval Estimate of Population Mean:s Known S

D

We are 95% confident that the interval contains thepopulation mean. Note that the sample mean was = $31,100.

$31,100 + $1,470or

$29,630 to $32,570x z

n

s/2

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Interval Estimation of a Population Mean:s Unknown

If an estimate of the population standard deviation s cannot be developed prior to sampling, we use the sample standard deviation s to estimate s . This is the s unknown case.

In this case, the interval estimate for is based on the t distribution.

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The t distribution is a family of similar probability distributions.

t Distribution

A specific t distribution depends on a parameter known as the degrees of freedom.

Degrees of freedom refer to the number of independent pieces of information that go into the computation of s.

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t Distribution (Continued)

A t distribution with more degrees of freedom has less dispersion.

As the number of degrees of freedom increases, the difference between the t distribution and the standard normal probability distribution becomes smaller and smaller.

20 Slide

t Distribution (Continued)

Standardnormal

distribution

t distribution(20 degreesof freedom)

t distribution(10 degrees

of freedom)

0z, t

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Student’s t Table

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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

Student’s t Table

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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

Student’s t Table

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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 valuest0

/ 2

/ 2

Student’s t Table

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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 valuest0

Assume:n = 3df = n - 1 = 2 = .10 / 2 =.05

Student’s t Table

/ 2

/ 2

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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 valuest0

Assume:n = 3df = n - 1 = 2 = .10 / 2 =.05

Student’s t Table

/ 2

/ 2

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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 valuest0

Assume:n = 3df = n - 1 = 2 = .10 =.05

.05

Student’s t Table

/ 2

/ 2

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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

t0

Assume:n = 3df = n - 1 = 2 = .10 / 2 =.05

2.920t values

.05

Student’s t Table

/ 2

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Interval Estimate

x t sn

/2

where: 1 - = the confidence coefficient t/2 = the t value providing an area of /2

in the upper tail of a t distribution with n - 1 degrees of freedom s = the sample standard deviation

Interval Estimation of a Population Mean:s Unknown

Margin of Error

Point Estimation ofPopulation Mean

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A reporter for a student newspaper is writing anarticle on the cost of off-campushousing. A sample of 16efficiency apartments within ahalf-mile of campus resulted ina sample mean of $650 per month and a samplestandard deviation of $55.

Interval Estimation of a Population Mean:s Unknown

Example: Apartment Rents

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Let us provide a 95% confidence interval estimate of the mean rent permonth for the population of efficiency apartments within ahalf-mile of campus.

Interval Estimation of a Population Mean:s Unknown (Example Continued)

Example: Apartment Rents

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At 95% confidence, = .05, and /2 = .025.

Degrees Area in Upper Tailof Freedom .20 .100 .050 .025 .010 .005

15 .866 1.341 1.753 2.131 2.602 2.94716 .865 1.337 1.746 2.120 2.583 2.92117 .863 1.333 1.740 2.110 2.567 2.89818 .862 1.330 1.734 2.101 2.520 2.87819 .861 1.328 1.729 2.093 2.539 2.861. . . . . . .

In the t distribution table we see that t.025 = 2.131.t.025 is based on n - 1 = 16 - 1 = 15 degrees of freedom.

Interval Estimation of a Population Mean:s Unknown (Example Continued)

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x t sn

.025

We are 95% confident that the mean rent per monthfor the population of efficiency apartments within ahalf-mile of campus is between $620.70 and $679.30.

Interval Estimate

Interval Estimation of a Population Mean:s Unknown (Example Continued)

55650 2.131 650 29.3016

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Summary of Interval Estimation Procedures

for a Population MeanCan the

population standard deviation s be assumed

known ?Use the sample

standard deviations to estimate s

Use

Yes No

/ 2sx tn

Use/ 2x z

ns

s KnownCase

s UnknownCase &

Small Sample

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If s is unknown and n>30, the standard deviation of the sample, designated by s, is used to approximate the population standard deviation.

Interval EstimationSummary

If the population standard deviation (s ) is known or the sample (n) is n≥30 we use the z distribution.

nszX

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nstX

The value of t for a given confidence level depends upon its degrees of freedom.

If the population standard deviation (s ) 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 EstimationSummary

Confidence Interval Estimate for Mean (s Known)

More Examples

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Thinking Challenge Example

You’re a Q/C inspector for Gallo. The s for 2-liter 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 2-liter bottles?

2 liter

To find the Z:α =1-90% =0.1 and α/2 =0.1/2 =0.051 – 0.05 = 0.95 and from table, the Z is:(1.64 + 1.65)/ 2 = 1.645

39 Slide

Confidence Interval Solution

X Zn

X Zn

-

-

s s

/ /

. . . . . .

. .

2 2

199 1645 05100

199 1645 05100

1982 1998

We are 90% confident that interval estimate of the true mean amount in 2-liter bottles is between 1.98 and 1.99.

40 Slide

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?

Confidence Interval of (s Unknown and n ³ 30)

Example

41 Slide

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.

Example & Solution (Continued)

42 Slide

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.

Example & Solution (Continued)

Confidence Interval Estimate for Mean (s Unknown, and n<30)

More Examples

44 Slide

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?

45 Slide

Solution

`X = 3.7 S = 0.38987 n = 6, df = n - 1 = 6 - 1 = 5 S / Ön = 3.8987 / Ö6 = 0 .1592 t.05,5 = 2.0150 3.7 - (2.015)(0.1592) 3.7 + (2.015)

(0.1592) 3.385 4.015

We are 90% confident that the interval estimate of the population mean task time is between 3.4 and 4.0 minutes.

46 Slide

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.

47 Slide

Excel Solution--SWStat

Data Area

48 Slide

Excel Solution—SWStat (Continued)(SWStat Statistics Intervals &

Tests)

49 Slide

Excel Solution—SWStat (Continued)

$227,099 ≤ µ ≤ $297,502

50 Slide

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.

Confidence Interval Excel Application(Another Problem)

51 Slide

The DataSubscriber 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 3610 28 22 32 34 3311 33 23 29 35 4612 35 24 33 36 37

52 Slide

Excel Solution--SWStat

53 Slide

Excel Solution (Continued)SWStat

54 Slide

Let E = the desired margin of error.

We said E is the amount added to and subtracted from the point estimate to obtain an interval estimate.

Sample Size for an Interval Estimateof a Population Mean

x t sn

/2 E = Margin of Error

Interval Estimate

of the mean

55 Slide

Sample Size for an Interval Estimateof a Population Mean (Continued)

E zn

s

/2

n zE

( )/ s2

2 2

2

Margin of Error

Necessary Sample Size

Margin of Error

56 Slide

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 inthe marketing area of the new location.

Suppose that Discount Sounds’ management teamwants an estimate of the population mean such thatthere is a 0.95 probability that the sampling error is $500or less.

How large a sample size is needed to meet the required precision? S

D

Sample Size for an Interval Estimateof a Population Mean--Example

57 Slide

At 95% confidence, z.025 = 1.96. Recall that s = 4,500.

zns

/2 500

2 2

2(1.96) (4,500) 311.17 312(500)n

Sample Size for an Interval Estimateof a Population Mean--Solution S

D

A sample of size 312 is needed to reach a desired precision of + $500 at 95% confidence.

n zE

( )/ s2

2 2

2

E =

Given

58 Slide

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 error--margin of error. 3- The variation of the population.

Sample Size for an Interval Estimateof a Population Mean (Continued)

n zE

( )/ s2

2 2

2

59 Slide

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 sample is required?

n = [(2.58)(20)/5]2 = 106.5024 » 107

n zE

( )/ s2

2 2

2

Thinking Challenge Sample Size Example 1

60 Slide

What sample size is needed to be 90% confident of being correct within 5? A pilot study suggested that the standard deviation is 45.

Thinking Challenge Sample Size Example 1 (Continued)

61 Slide

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.

Thinking Challenge Sample Size Example 1 (Continued)

62 Slide

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?

63 Slide

Thinking Challenge Sample Size Example 2 (Solution)

64 Slide

Interval Estimationof a Population Proportion

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65 Slide

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