Lc07 SL Estimation.ppt_0

Inferences on Inferences on Population ProportionsPopulation Proportions

Use Calculation from Sample to Use Calculation from Sample to Estimate Population ParameterEstimate Population Parameter

Population Sample(select)

Statistic

(calculate)

Parameter(estimate)

(describes)

%63ˆ p?p

Statistic ParameterStatistic Parameter

Describes a Describes a sample.sample.

Always knownAlways known Changes upon Changes upon

repeated repeated sampling.sampling.

Examples:Examples:

Describes a Describes a population.population.

Usually unknownUsually unknown Is fixedIs fixed

Examples:Examples:

pssx ˆ,,, 2 p,,, 2

A Statistic is a Random VariableA Statistic is a Random Variable Upon repeated sampling of the same Upon repeated sampling of the same

population, the value of a statistic population, the value of a statistic changes changes variablevariable..

While we donWhile we don’’ t know what the next t know what the next sample will yield, we do know the sample will yield, we do know the overall pattern over many, many overall pattern over many, many samplings samplings randomrandom..

The distribution of possible values of a The distribution of possible values of a statistic for repeated samples of the statistic for repeated samples of the same size from a population is called same size from a population is called the the sampling distributionsampling distribution of the of the statistic.statistic.

ProportionProportion

We are interested in the distribution ofWe are interested in the distribution of

Note, is cY where c = 1/n is a constant Note, is cY where c = 1/n is a constant and Y is a binomial random variable.and Y is a binomial random variable.

If Y is normally distributed, cY will also be If Y is normally distributed, cY will also be normally distributed. normally distributed.

n

yp ˆ

p̂

If Y is normal cY is normalIf Y is normal cY is normal

For example: If Y is n(µ=100,σ2=4),

then (0.5)Y is n(µ=50, σ2=1)

y cy0.5y

Distribution of Sample Distribution of Sample ProportionsProportions

A normal curve can be used to A normal curve can be used to approximate the distribution of approximate the distribution of sample proportions if:sample proportions if:

The size of the sample or number of The size of the sample or number of repetitions is relatively large (say repetitions is relatively large (say 30 or more).30 or more).

While the sample size is relatively While the sample size is relatively small compared to the population small compared to the population size (say < 10%) size (say < 10%)

Sampling Distribution for Sampling Distribution for

p̂

The sampling distribution of based on large n is approximately normal.

p̂

Sampling Distribution forSampling Distribution forp̂

To completely define the normal distribution of p̂

We need the mean (expected value) and variance.

Sampling Distribution of Sampling Distribution of p̂

pn

ppzp

)1(2/

n

ppzp

)1(2/

So, at most, will be away from p, (1-α)100% of the time. We call this (1-α)100% the level of confidence.

p̂ n

ppz

)1(2/

1-α

Repeated Sampling of Size Repeated Sampling of Size nn

95% of the time our estimate will be within

of the truth.

n

pp )1(96.1

95%

Standard ErrorStandard Error We don’t know the value of p, so we We don’t know the value of p, so we

will usewill use

When we use , we have an When we use , we have an estimate of the standard deviation estimate of the standard deviation for the sampling distribution of .for the sampling distribution of .

We call this estimate the standard We call this estimate the standard error. error.

p̂

p̂

p̂

n

pp )ˆ1(ˆ

Confidence Interval for Confidence Interval for pp

n

pppCI

)ˆ1(ˆ96.1ˆ :pfor %95

n

ppzpCI

)ˆ1(ˆˆ :pfor %100)1( 2/

Example:

Confidence Interval Confidence Interval Based on Normal Based on Normal

DistributionDistributionError) (Standard Value) (Est Pt Z

n

ppzp

)ˆ1(ˆˆ 2/

Standard error is our estimate of the standard deviation for the distribution of the point estimate.

Sample SizeSample Size

(solving for n)

n

pp )ˆ1(ˆz eError Maximum /2

2

22/ )ˆ1(ˆ

e

ppzn

Sample SizeSample Size If you can’t guess a value for :If you can’t guess a value for :

What sample size would you What sample size would you recommend to estimate the recommend to estimate the proportion of exeedences with a 95% proportion of exeedences with a 95% confidence interval and maximum confidence interval and maximum error of 2%?error of 2%?

p̂

2

22/

2

22/

4

)5)(.5(.

e

z

e

zn

Research Update!Research Update! Recent research shows that we get Recent research shows that we get

better coverage for (1-better coverage for (1-α)100% CI’s α)100% CI’s onon

p-hat if we alter the CI formula.p-hat if we alter the CI formula.

95%CI:95%CI:

where where

4

)ˆ1(ˆˆ 2/

n

ppzp

4

2ˆ

n

xp (Agresti/Coull)

Estimation Procedures: An Example

Basic Logic

In estimation procedures, statistics calculated from random samples are used to estimate the value of population parameters.

Example: If we know 42% of a random sample

drawn from a city are Republicans, we can estimate the percentage of all city residents who are Republicans.

Basic Logic Information from

samples is used to estimate information about the population.

Statistics are used to estimate parameters.

POPULATION

SAMPLE

PARAMETER

STATISTIC

Basic Logic Sampling Distribution

is the link between sample and population.

The value of the parameters are unknown but characteristics of the S.D. are defined by theorems.

POPULATION

SAMPLING DISTRIBUTION

SAMPLE

Two Estimation Procedures

A point estimate is a sample statistic used to estimate a population value. A newspaper story reports that 74% of a

sample of randomly selected Americans support capital punishment.

Confidence intervals consist of a range of values. ”between 71% and 77% of Americans

approve of capital punishment.”

Constructing Confidence Intervals For Means Set the alpha (probability that the interval

will be wrong). Setting alpha equal to 0.05, a 95% confidence

level, means the researcher is willing to be wrong 5% of the time.

Find the Z score associated with alpha. If alpha is equal to 0.05, we would place half

(0.025) of this probability in the lower tail and half in the upper tail of the distribution.

Substitute values into equation.

Confidence Intervals For Means

For a random sample of 178 households, average TV viewing was 6 hours/day with s = 3. Alpha = .05. N=178. c.i. = 6.0 ±1.96(3/√177) c.i. = 6.0 ±1.96(3/13.30) c.i. = 6.0 ±1.96(.23) c.i. = 6.0 ± .44


We can estimate that households in this community average 6.0±.44 hours of TV watching each day.

Another way to state the interval: 5.56≤μ≤6.44 We estimate that the population mean is greater

than or equal to 5.56 and less than or equal to 6.44.

This interval has a .05 chance of being wrong.


Even if the statistic is as much as ±1.96 standard deviations from the mean of the sampling distribution the confidence interval will still include the value of μ.

Only rarely (5 times out of 100) will the interval not include μ.

Other confidence levels

Confidence level

Alpha Alpha/2 Z score

90% .10 .05 +/- 1.65

95% .05 .025 +/- 1.96

99% .01 .0050 +/- 2.58

99.9% .001 .0005 +/- 3.29

Constructing Confidence Intervals For Proportions

Procedures: Set alpha. Find the associated Z score. Substitute the sample information into

Formula

Confidence Intervals For Proportions

If 42% of a random sample of 764 from a Midwestern city are Republicans, what % of the entire city are Republicans?

Don’t forget to change the % to a proportion. c.i. = .42 ±1.96 (√.25/764) c.i. = .42 ±1.96 (√.00033) c.i. = .42 ±1.96 (.018) c.i. = .42 ±.04

Confidence Intervals For Proportions

Changing back to %s, we can estimate that 42% ± 4% of city residents are Republicans.

Another way to state the interval: 38%≤Pu≤ 46%

We estimate the population value is greater than or equal to 38% and less than or equal to 46%.

This interval has a .05 chance of being wrong.

SUMMARY

In this situation, identify the following: Population Sample Statistic Parameter

SUMMARY

Population = All residents of the city.

Sample = the 764 people selected for the sample and interviewed.

Statistic = Ps = .42 (or 42%)

Parameter = unknown. The % of all residents of the city who are Republicans.

Estimating Population Parameters: Another Example

Estimating Population Proportion

Example: When surveying 500 people selected at random from the general population we get 200 responses of “yes” when asked if they like broccoli. Estimate the proportion of the population that likes broccoli.

p’ = x/n = 200/500 = 0.40

q’ = 1 - p’ = 0.60

np’ = 200 > 25 and nq’ = 300 > 25 normal aprox. is OK

At the 95% confidence level we have /2 = 0.025 and z/2 = 1.96

The margin for error is E = z/2(.4*.6/500) = 1.96(.022) = 0.043

P(p’- 0.043 < p < p’+ 0.043) = P(.357 < p < .443) = 0.95


There are two user controlled factors that determine the margin of error

E = z/2(p’q’/n)1. The confidence level 1 -

2. The sample size n

The smaller , the greater z/2 and the greater E

The larger a value of n, the smaller E

If the experimenter wants to fix the width of the confidence interval (by setting E to a pre-determined constant) and set the confidence level (by selecting a particular ), then we can use equation (1) above to determine the sample size needed to achieve this level of precision.

(1)


E = z/2(p’q’/n)

Set E and to desired values E and z/2 are constant. Solve equation (1) for n

n = z/22(p’q’)/E2

(1)

In equation (2) we have not yet taken a sample from the population, so we cannot be sure what the proportion of successes might be. The value of p’ that we use in this equation may be an estimate that we make based upon some prior knowledge, or, we may chose p’ = q’ = 0.5, which maximizes n for particular choice of and E

(2)


Returning to our previous example, suppose we choose = 0.05, E = 0.025, and have no prior knowledge about p’

The from equation (2) on the previous slide we obtain

n = z/22(p’q’)/E2

Where

z/2 = 1.96 and

n = (1.96)2 (0.25)/0.0252

n = 1536.4 1537

Estimating the population mean, known

Assume:

•Sample size, n > 30

•Population standard deviation is known

Then from the Central Limit Theorem, we know that the sampling distribution for samples of size n

•Has mean x’ = , the mean of the original population

•Has x’ = /n


Let x’ be the mean of the sample of size n, then we have for a confidence interval 1 - given by

P(x’ – z/2/n < < x’ + z/2/n) = 1 -

Let the margin of error E = z/2/n

Then with a probability of 1 - the population mean lies between x’ – E and x’ + E


x’ mean from sample of size n

E

E

1

2

/2

/2

If the mean were 1 the probability of getting a sample x’ would be /2

If the mean were 2 the probability of getting a sample x’ would be /2

Estimating Population mean with unknown

If the population standard deviation is unknown, the sample will have to provide both an estimate on the population mean and standard deviation.

Estimation of the population mean will be similar to how it was done when is known, but the sample standard deviation will be used instead. Student’s t-distribution will be used to determine the margin of error, and the confidence interval will be somewhat wider than it would be for the same sample size if were known.


Step 1. For a sample of size n, n 30, calculate the sample mean x’, and sample standard deviation s

Recall: the sample variance s2 =(fixi2 – (fixi)2)/(n – 1)

Step 2. Convert to a standard t-score

t = (x’ - )/(s/n)

Where is the (unknown) mean of both the original population and the sampling distribution

Step 3. Select a confidence level 1 - , and determine t/2

Step 4. Find the margin of error E = t/2 (s/n)

Then P(x’ – E x’ + E) = 1 -


Finding t/2

Choose = 0.05 , and assume n = 30 – From table A-3

Degrees of Area in One Tail

freedom 0.005 0.01 0.025 0.05

……………………………………………………….

29 2.756 2.462 2.045 1.699

Then t/2 = 2.045 for n –1 = 29 degrees of freedom


Using the TI calculator to find confidence interval for a statistic with t-distribution

Let n = 106, x’ = 98.2, s = 0.62

Construct 99% Confidence Interval

Step 1: Select STAT > TESTS scroll down to 8: TInterval

Step 2: Select Stats if x’, n, and s are known

Select Data if these values are to be calculated from a list

Step 3: (Stats) Use the arrow key to move to each prompt and enter the values given above. Then Calculate <enter>

Answer: TInterval (98.081, 98.319)

Estimating Population Variance

Requirements:

1. The sample is a simple random sample

2. The population must have normally distributed values

The sample variance has 2 distribution

2 = (n – 1) s2/2

Where

n = sample size

s2 = sample variance

2 = population variance (to be determined)

Estimating Population Standard Deviation

0 2

Confidence Interval for the Population Variance:

(n-1)s2/2R < 2 < (n-1)s2/2

L

2L 2

R

/2/2

2 distribution is skewed right and always positive


Example: Given the following data, find the 95% confidence interval for the population standard deviation

n = 41, x’ = 67200, s = 18277

P{(n-1)s2/2R < 2 < (n-1)s2/2

L} = 0.95

First find 2R and 2

L when each tail of the distribution contains 2.5% of the area under the curve

From Table A-4 for the Chi-Square Distribution

Degrees of Freedom

Area to the right of the Critical Value

0.995 0.99 0.975 0.95 0.10 0.05 0.025

………………………………………………………………………………………….

40 20.707 22.164 24.443 26.509 51.805 55.758 59.342


From the previous slide we have 2R = 59.342 and 2

L = 24.433

And therefore:

(n-1)s2/2R = 40 (18277)2/59.342 = 2.2516 x 108

and

(n-1)s2/2L = 40 (18277)2/24.433 = 5.4688 x 108

Thus we have:

P(2.2516 x 108 < 2 < 5.4688 x 108) = 0.95

and for the standard deviation

P(15,005 < < 23385) = 0.95

Lc07 SL Estimation.ppt_0

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