Stats Lecture 07 Confidence Intervals

8/3/2019 Stats Lecture 07 Confidence Intervals

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Confidence Intervals

For Means and Proportions

Quantitative Methods for Economics

Dr. Katherine Sauer

Metropolitan State College of Denver


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We have seen that the sample mean can vary greatly from thepopulation mean.

We have also seen that as the sample size increases, the variation

decreases.


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Sometimes we dont know the population mean.

- we cant know if our sample mean is close to the

population mean

We can estimate an interval estimate for the mean.

x margin of error


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I. Confidence Intervals for the Population Mean

When the population mean is know, an interval that contains 95%

of sample means is

x 96.1

This can also be written as where x 96.1

This interval has a fixed center point of and total width of 2that contains 95% of all sample means.


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If we dont know the population mean, we have to estimate it.

xx 96.1

In this case, the center is no longer fixed.

- it will move according to the value of each new point

estimate of the sample mean


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Observations:

An interval estimate will contain the population mean if

the sample mean is one of the 95% of sample meanswithin the interval.

- sample means 1, 2, and 4 from the diagram

Each of the sample means that is within the interval willresult in an interval that contains the population mean

somewhere within the interval.

An interval estimate will not contain the population mean

if the sample mean lies in either tail.- sample means 3 and 5 from the diagram


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A. How To Construct a Confidence Interval

A Confidence Interval ispoint estimate + (critical Z value)(standard error)

For a 95% confidence interval:

5% is divided into two tails 0.025 each

Z0.025 = 1.96 and the margin of error is

For a 99% confidence interval:

1% is divided into two tails 0.005 each

Z0.005 = 2.5758 and the margin of error is

x 96.1

x 5758.2


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Denote the area in the two tails as .

The area between the two tails is (1 - ).

In general, if we denote the area in each tail as /2 then the

corresponding critical Z value is Z/2 and the margin of

error isxZ 2/

xZx 2/

(1 - )(100%) is called the level of confidence.

The (1 - )(100%) confidence interval is given by:


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Example: An importer of Herbs and Spices claims that the average

weight of packets of saffron is 20 grams. A random sample of 36

packets is selected. The sample average is calculated to be 19.35

grams. The populations standard deviation is 1.8 grams.

Calculate the 99% confidence interval for the population average

weight.

Step 1: find

3.036

8.1

nx

2/,, Zx x

we are given that =19.35 grams

we can calculate

x

x


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For the critical Z:

= 1% /2 = 0.5%

so Z/2 = 2.5758

Step 2: construct the intervalx

Zx 2/

19.35 + (2.5758)(0.3)

19.35 + 0.77274

18.5773 to 20.1227

We are 99% sure that the population mean is somewhere

between 18.5773 and 20.1227.


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The manufacturer makes the claim that the average weight of a

packet is 20 grams.

Does our confidence interval support this claim?

yes, 20 lies between the upper and lower limits of our

confidence interval


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B. One-Sided Confidence Intervals

Sometimes we wish to know the lower limit, above which we are

(1)(100%) confident the population mean lies.

Sometimes we wish to know the upper limit, below which we are(1)(100%) confident the population mean lies.

xZx

2/ xZx 2/


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Ex: Suppose a property investor claims that the average rental

income per room in student housing is at most $5000 per year. The

average rent paid by a sample of 36 students was calculated to be$5200. The sample standard deviation is $735.

Calculate a 90% confidence interval for the true mean annual rental

income.

Step 1: find2/

,,

Zsxx

5.12236

735

n

ssx

we are given that =5200

we can calculate

x

xs


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For the critical Z:

= 10% /2 = 5%

so Z/2 = 1.6449

Step 2: construct the intervalxsZx 2/

5200 + (1.6449)(122.5)

5200 + 201.50

4998.50 to 5401.50

We are 90% sure that the true population mean is somewhere

between $4998.50 and $5401.50.


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The investor makes the claim that the average rent is at most

$5000.

Does our confidence interval support this claim?

no, $5000 is $401 below the upper bound of our

confidence interval


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II. Confidence Intervals for Proportions

The (1)(100%) confidence interval for a population proportion

is given by

n

ppZp

)1(2/

center of interval margin of error


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Example: In a poll of 200 voters, 88 stated that they will vote for

the Green party candidate. Construct a 95% confidence interval

for the true proportion of support for the Green party candidate.

Step 1: find2/

,,

Znp

n = 200

p = 88/200 = 0.44

Z0.025 = 1.96

0351.0200

)44.01(44.0)1(

n

pp


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So, the confidence interval is:

0.44 + (1.96)(0.0351)

0.44 + 0.068796

Meaning, we are 95% confident that the population proportion

is between 0.371204 and 0.508796.

That is, we are 95% sure that between 37.1% and 50.8% of the

voters will vote for the Green party.

The margin of error is 6.88 percentage points.

What would you say about the usefulness of this information?


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III. The Precision of Confidence Intervals

The margin of error tells us about the precision of the interval.

nZ

2/

It depends on Z, , and n.


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The population standard deviation is fixed for a given

population.

- The margin of error (width of the interval) will begreater when the population standard deviation is large.

The greater the level of confidence, the larger Z is and the wider

the interval.- A more precise interval can be obtained by reducing

the confidence level.

The larger the sample size, the smaller the margin of error.

Sample size has a major effect on the precision of the confidence

interval.


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How to calculate the sample size needed that will result in a

confidence interval with a specified precision:

2

2/

Zn ))1((

2

2/ ppZ

n


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Example: An importer of Herbs and Spices claims that the

average weight of packets of saffron is 20 grams. The

populations standard deviation is 1.8 grams.

Calculate the sample sized needed to construct a 99% confidence

interval with a margin of error of + 0.5.

2

2/

Zn

986.855.0

)8.1)(5758.2(2

For 99%, Z/2 = 2.5758

A sample size of 86 would allow us to construct a 99%

confidence interval with a margin of error of 0.5.


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Example: In a poll of 200 voters, 88 stated that they will vote

for the Green party candidate.

Calculate the sample sized needed to construct a 95% confidence

interval with a margin of error of + 0.01.

For 95%, Z/2 = 1.96

p=88/200 = 0.44

A sample size of 9466 would allow us to construct a 95%

confidence interval with a margin of error of 0.01.

))1((

2

2/ ppZn

7.9465))44.01(44.0(

01.0

96.12


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IV. Confidence Intervals for Differences between Means and

Proportions

Ex: Is the average commuting time by bus faster than by car?

Ex: Is the percentage of people likely to vote for candidate A

larger than the percentage of people likely to vote for candidate B?


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The distribution for the difference between two normalindependent random variables is normal with a mean equal to the

difference of the means and the variance equal to the sum of the

variances.

If and

1

2

111,~n

NX

2

2

222,~n

NX

then (for large n)

2

2

2

1

2

12121,~

nnNXX


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The confidence interval for differences between population

means is:

point estimate + (critical value)(standard error)

The (1)(100%) confidence interval for 12:

2

22

1

21

2/21nn

Zxx

2

2

2

1

2

12/21

n

s

n

sZxx

Or, if the standard deviation is unknown:

(for large n)


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For sample proportions that are normally distributed, thedifference between the proportions will also be normally

distributed.

If and

1

112

1111

)1(,~

nNp

pp

then

2

22

1

112

21212121

)1()1(,~

nnNpp pppp

2

222

2222

)1(,~

nNp

pp


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The confidence interval for differences between population

proportions is:

point estimate + (critical value)(standard error)

The (1)(100%) confidence interval for p1p2:

2

22

1

112/21

)1()1(

n

pp

n

ppZpp


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How to interpret:

Let the confidence interval for the difference between means be

A < 12 < B

Case 1: If A and B are both positive, then we can be (1-)(100%)

confident that 1 is greater than 2 by an amount ranging from A to

B.

Case 2: If A and B are both negative, then we can be (1-)(100%)

confident that 1 is less than 2 by an amount ranging from A to B.

Case 3: If A and B have different signs, then the interval contains

12 = 0. We can be (1-)(100%) confident that A is equal to B.


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Example: Designers of rowing equipment investigate the

difference between the mean weights (kg)of male and female

rowing teams. Random samples of male and female rowers are

selected.

Calculate the 95% confidence interval for the difference in mean

weights between male and female rowers.

Note: we dont know the population mean or standard deviation

2

2

2

1

2

12/21

n

s

n

sZxx

ll d h diff i d


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Well need the difference in means, Z, s, and n.

difference in means = 60.552.6 = 7.9

Z/2=1.96

n1 = 42 n2 = 30

s1 = 6.8 s2 = 4.530

5.4

42

8.696.19.7

22

7.9 + 1.96(1.3326)

7.9 + 2.6119

We are 95% confident that the difference between the mean

weights of male and female rowers is between 5.2881kg and

10.5119kg.


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Both the lower and upper limits are positive.

1 > 2

The average weight for male rowers is greater than the

average weight for female rowers.

Lower Limit: 1 > 2 by 5.2881

Upper Limit: 1 > 2 by 10.5991

We are 95% confident that the mean weight of male rowers

exceeds the mean weight of female rowers by 5.2881kg to

10.5991kg.

We cannot assume that the weight of any individual male rower

exceeds the weight of any individual female rower.

variance for individual variance for means2

n

x

2

2


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Example: Here are the results for polls taken in 2 locations.

Calculate the 90% confidence interval for the difference in

proportions who will vote for the Green party in area A and Area

B.

2

22

1

112/21 )1()1(

npp

nppZpp


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Well need p, the difference in p, Z, and n.

p1 = 88/200 = 0.44 p2 = 54/160 = 0.3375

difference in proportions = 0.440.3375 = 0.1025

n1 = 200 n2 = 160

Z/2=1.6449

160

)3375.01(3375.0

200

)44.01(44.06449.11025.0

0.1025 + 1.6449(0.0513)0.1025 + 0.0843

We are 90% confident that the difference between the two

proportions is between 0.0182 and 0.1868.


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Both the lower and upper limits are positive.1 > 2

The proportion of voters supporting the Green party in area

A exceeds the proportion of voters supporting the Green

party in area B.

Lower Limit: 1 > 2 by 0.0182

Upper Limit: 1 > 2 by 0.1868

We are 90% confident that 1 > 2.


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Concepts:

Interpreting confidence intervals and their precision

Skills:

Construct a 90%, 95% and 99% confidence interval for means

and proportions

Calculate the sample size needed in order to construct a

confidence interval of a specified margin of error

- means and proportions

Construct a 90%, 95% and 99% confidence interval for the

difference between means or proportions.

Stats Lecture 07 Confidence Intervals

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