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ES Chapters 14 & 16: Introduction to Statistical Inferences

Level ofConfidence

1

MaximumError

E

SampleSize

n

En

z

2

ES Chapter Goals• Learn the basic concepts of estimation

• Consider questions about a population mean using two methods that assume the population standard deviation is known

• Consider: what value or interval of values can we use to estimate a population mean?

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ES The Nature of Estimation• Discuss estimation more precisely

• What makes a statistic good ?

• Assume the population standard deviation, , is known throughout this chapter

• Concentrate on learning the procedures for making statistical inferences about a population mean

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ES Point Estimate for a ParameterPoint Estimate for a Parameter: The value of the corresponding statistic

Note: The quality of an estimation procedure is enhanced if the sample statistic is both less variable and unbiased

How good is the point estimate? Is it high? Or low? Would another sample yield the same result?

Example: is a point estimate (single number value) for the mean of the sampled population

x 14 7.

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ES Unbiased StatisticUnbiased Statistic: A sample statistic whose sampling distribution has a mean value equal to the value of the population parameter being estimated. A statistic that is not unbiased is a biased statistic.

Example: The figures on the next slide illustrate the concept of being unbiased and the effect of variability on a point estimate

Assume A is the parameter being estimated

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ES Illustrations

A

Positive biasOver estimateLow variability

A

UnbiasedOn target estimate

A

Negative biasUnder estimateHigh variability

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ES Notes

2. Sample means vary from sample to sample. We don’t expect the sample mean to be exactly equal the population mean .

3. We do expect the sample mean to be close to the population mean

4. Since closeness is measured in standard deviations, we expect the sample mean to be within 2 standard deviations of the population mean

x

1. The sample mean, ,is an unbiased statistic because the mean value of the sampling distribution is equal to the population mean:

x

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ES Important DefinitionsInterval Estimate: An interval bounded by two values and used to estimate the value of a population parameter. The values that bound this interval are statistics calculated from the sample that is being used as the basis for the estimation.

Level of Confidence 1 - : The probability that the sample to be selected yields an interval that includes the parameter being estimated

Confidence Interval: An interval estimate with a specified level of confidence

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• To construct a confidence interval for a population mean , use the CLT

• Use the point estimate as the central value of an intervalx

Summary

• The level of confidence for the resulting interval is approximately 95%, or 0.95

• We can be more accurate in determining the level of confidence

x

• Since the sample mean ought to be within 2 standard deviations of the population mean (95% of the time), we can find the bounds to an interval centered at :

x xx x 2 2( ) ( ) to

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ES Illustration

x x x2( )x x 2( )

Distribution of x

• The interval is an approximate 95% confidence interval for the population mean based on this x

x xx x 2 2 to

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ES Estimation of Mean ( Known)

• Formalize the interval estimation process as it applies to estimating the population mean based on a random sample

• Assume the population standard deviation is known

• The assumptions are the conditions that need to exist in order to correctly apply a statistical procedure

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ES The Assumption...

Assumption satisfied by:

1. Knowing that the sampled population is normally distributed, or

2. Using a large enough random sample (CLT)Note: The CLT may be applied to smaller samples (for examplen = 15) when there is evidence to suggest a unimodal distribution that is approximately symmetric. If there is evidence of skewness, the sample size needs to be much larger.

x

The assumption for estimating the mean using a known :The sampling distribution of has a normal distribution

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2. z: confidence coefficient, the number of multiples of the standard error needed to construct an interval estimate of the correct width to have a level of confidence 1-

The 1- Confidence Interval of

• A 1- confidence interval for is found by

0 z

1 / 2 / 2

- z z

xn

xn

to z z

Notes:1. is the point estimate and the center point of the confidence

intervalx

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3. : standard error of the meanThe standard deviation of the distribution of / n

x

Notes Continued

4. : maximum error of estimate EOne-half the width of the confidence interval (the product of the confidence coefficient and the standard error)

n( / )z

5. : lower confidence limit (LCL) : upper confidence limit (UCL)

x

x n( / )z

n( / )z

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ES The Confidence IntervalA Five-Step Model:

1. Describe the population parameter of concern2. Specify the confidence interval criteria

a. Check the assumptionsb. Identify the probability distribution and the formula to be usedc. Determine the level of confidence, 1 -

3. Collect and present sample information4. Determine the confidence interval

a. Determine the confidence coefficientb. Find the maximum error of estimatec. Find the lower and upper confidence limits

5. State the confidence interval

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2. Specify the confidence interval criteria

a. Check the assumptionsThe weights are normally distributed, the distribution of is normal

b. Identify the probability distribution and formula to be usedUse the standard normal variable z with = 0.27

c. Determine the level of confidence, 1 - The question asks for 95% confidence: 1 - = 0.95 x

Example Example: The weights of full boxes of a certain kind of cereal are normally

distributed with a standard deviation of 0.27 oz. A sample of 18 randomly selected boxes produced a mean weight of 9.87 oz. Find a 95% confidence interval for the true mean weight of a box of this

cereal.Solution:

1. Describe the population parameter of concernThe mean, , weight of all boxes of this cereal

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3. Collect and present informationThe sample information is given in the statement of the problem

Given: n x 18 9 87; .

Solution Continued

1.15 1.28 1.65 1.96 2.33 2.580.75 0.80 0.90 0.95 0.98 0.991

4. Determine the confidence interval

a. Determine the confidence coefficient

The confidence coefficient is found using Table A or C:

z

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ES Solution Continuedb. Find the maximum error of estimate

Use the maximum error part of the formula for a CI

5. State the confidence interval

9.75 to 10.00 is a 95% confidence interval for the true mean weight, , ofcereal boxes

En

1960 2718

01247. . .= z

c. Find the lower and upper confidence limitsUse the sample mean and the maximum error:

xn

9 87 01247 9 87 012479 7453 9 9947

9 75 10 00

to

to to to

. . . .. .

. .

z xn

+ z

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ES Example Example: A random sample of the test scores of 100 applicants for clerk-

typist positions at a large insurance company showed a mean score of 72.6. Determine a 99% confidence interval for the mean score of all applicants at the insurance company. Assume the standard deviation of test scores is 10.5.

Solution:1. Parameter of concern

The mean test score, , of all applicants at the insurance company

2. Confidence interval criteria

a. Assumptions: The distribution of the variable, test score, is not known. However, the sample size is large enough (n = 100) so that the CLT applies

b. Probability distribution: standard normal variable z with = 10.5

c. The level of confidence: 99%, or 1 - = 0.99

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3. Sample informationGiven: n = 100 and = 72.6x

Solution Continued

5. Confidence intervalWith 99% confidence we say, “The mean test score is between 69.9 and 75.3”, or “69.9 to 75.3 is a 99% confidence interval for the true mean test score”

Note: The confidence is in the process. 99% confidence means: if we conduct the experiment over and over, and construct lots of confidence intervals, then 99% of the confidence intervals will contain the true mean value .

72 6 2 709 69 891 72 6 2 709 75 309. . . . . . to

4. The confidence intervala. Confidence coefficient:

b. Maximum error:

c. The lower and upper limits:z(0.005) .2 58 z

E n ( / ) ( . )( . / ) . 2 58 10 5 100 2 709z

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ES Sample Size• Problem: Find the sample size necessary in order to

obtain a specified maximum error and level of confidence (assume the standard deviation is known)

En

z

Solve this expression for n:

nE

2z

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ES ExampleExample: Find the sample size necessary to estimate a population

mean to within 0.5 with 95% confidence if the standard deviation is 6.2

Note: When solving for sample size n, always round up to the next largest integer (Why?)

Solution:

Therefore, n = 591

n

( . )( . )0.

. .196 6 25

[24 304] 590 6842

2

n E 2z