lec5

Quantitative Methods for Decision Making

Dr. Raziuddin Siddiqui

January 28, 2012

Lecture 5Dr. Raziuddin Siddiqui Quantitative Methods for Decision Making

Inferential Statistics for Decision Making

Distributions of random variable:

symmetric distribution

uniform distribution

normal distribution

skewed distribution

Dr. Raziuddin Siddiqui Quantitative Methods for Decision Making

normal distribution

skewed distribution

normal distribution

skewed distribution

normal distribution

skewed distribution

random variable X

standardized variable Z

z = x−µσ

for a sample of size n we have z = x̄−µσ√n

wherex̄ is the mean of sample of the random variable Xµ is the mean of the population.σ is the standard deviation of the population.

random variable X

z = x−µσ

random variable X

z = x−µσ

random variable X

z = x−µσ

Key Fact

Properties:

the total area under z-curve is always 1.

the mean of variable z is always 0.

the standard deviation of z is always 1.

Key Fact

Properties:

Key Fact

Properties:

µ + σ

|µ + 2σ

|µ− σ

|µ− 2σ

|µ + 3σ

|µ− 3σ

A normal curve with (µX , σX )

A normal curve with (µZ = 0, σZ = 1)

Example

Finding the Area to the Left of a Specified z-ScoreDetermine the area under the standard normal curve that liesto the left of z = 1.23Solution:P(Z < 1.23) = 0.8907

Example

Finding the Area to the Right of a Specified z-ScoreDetermine the area under the standard normal curve that liesto the right of z = 0.76Solution:P(Z > 0.76) = 1− P(Z < 0.76) = 1− 0.7764 = 0.2236

Example

Finding the Area between Two Specified z-ScoresDetermine the area under the standard normal curve that liesbetween z = −0.68 and z = 1.82Solution:

P(−0.68 < Z < 1.82) =P(Z < 1.82)− P(Z < −0.68)

=0.9656− 0.2483

=0.7173

Example

Finding the z-Score Having a Specified Area to Its LeftDetermine the z-score having an area of 0.04 to its left underthe standard normal curveSolution:P(Z < z) = 0.04⇒ z = −1.75

Definition

The symbol zα is used to denote the z-score that has an areaof α (alpha) to its right under the standard normal curve.Read ”zα” as ”z sub α”.

Example

Finding zαDetermine z0.025 and z0.05

Solution:1− 0.025 = 0.9750P(Z > z0.025) = 0.9750⇒ z0.025 = 1.961− 0.05 = 0.9500P(Z > z0.05) = 0.9500⇒ z0.05 = 1.645

Class Activity

Finding zαDetermine z0.005 and z0.01

Example

Intelligence quotients (IQs) measured on the Stanford Revisionof the Binet-Simon Intelligence Scale are normally distributedwith a mean of 100 and a standard deviation of 16. Determinethe percentage of people who have IQs between 115 and 140.Solution:µ = 100 and σ = 16by using the transformation rulez = x−µ

for x = 115 we have z = 115−10016

= 0.94for x = 140 we have z = 140−100

16= 2.50

Example(Cont’d)

Solution:

P(115 < X < 140) =P(0.96 < Z < 0.2.50)

=P(Z < 2.50)− P(Z < 0.94)

=0.9938− 0.8264

=0.174

Interpretation: 16.74% of all people have IQs between 115and 140. Equivalently, the probability is 0.1674 that arandomly selected person will have an IQ between 115 and140.

Class Activity

According to the National Health and Nutrition ExaminationSurvey, the serum (noncellular portion of blood) totalcholesterol level of U.S. females 20 years old or older isnormally distributed with a mean of 206 mg/dL (milligramsper deciliter) and a standard deviation of 44.7 mg/dL.

a. Determine the percentage of U.S. females 20 years old orolder who have a serum total cholesterol level between150 mg/dL and 250 mg/dL.

b. Determine the percentage of U.S. females 20 years old orolder who have a serum total cholesterol level below 220mg/dL.

c. Obtain and interpret the quartiles for serum totalcholesterol level of U.S. females 20 years old or older.

Class Activity

Definition

The 68.26-95.44-99.74 RuleAny normally distributed variable has the following properties.

68.26% of all possible observations lie within onestandard deviation to either side of the mean, that is,between µ− σ and µ + σ.

95.44% of all possible observations lie within twostandard deviations to either side of the mean, that is,between µ− 2σ and µ + 2σ.

99.74% of all possible observations lie within threestandard deviations to either side of the mean, that is,between µ− 3σ and µ + 3σ.

Definition

Class Activity

Apply 68.26-95.44-99.74 rule to the previous example.

Confidence interval for one population mean

Definition

Confidence interval (CI): An interval of numbersobtained from a point estimate of a parameter.

Confidence level: The confidence we have that theparameter lies in the confidence interval (i.e., that theconfidence interval contains the parameter).

Confidence-interval estimate: The confidence leveland confidence interval.

Definition

Procedure

Assumptions:

1. Simple random sample

2. Normal population or large sample

3. σ known

Procedure(Cont’d)

1: For a confidence level of 1− α, use Table to find zα/2

2. The confidence interval for µ is from

x̄ − zα/2 ·σ√n

to x̄ + zα/2 ·σ√n

where zα/2 is found in step 1, n is the sample size, and x̄is computed from the sample data.

3. Interpret the confidence interval.

Procedure(Cont’d)

x̄ − zα/2 ·σ√n

Procedure(Cont’d)

x̄ − zα/2 ·σ√n

Key Fact

When to Use the One-Mean z-Interval Procedure

For small samples-say, of size less than 15–the z-intervalprocedure should be used only when the variable underconsideration is normally distributed or very close to beingso.

For samples of moderate size–say, between 15 and 30–thez-interval procedure can be used unless the data containoutliers or the variable under consideration is far frombeing normally distributed.

Key Fact

When to Use the One-Mean z-Interval Procedure

For small samples-say, of size less than 15–the z-intervalprocedure should be used only when the variable underconsideration is normally distributed or very close to beingso.

For samples of moderate size–say, between 15 and 30–thez-interval procedure can be used unless the data containoutliers or the variable under consideration is far frombeing normally distributed.

Key Fact(Cont’d)

For large samples–say, of size 30 or more–the z-intervalprocedure can be used essentially without restriction.However, if outliers are present and their removal is notjustified, you should compare the confidence intervalsobtained with and without the outliers to see what effectthe outliers have. If the effect is substantial, use adifferent procedure or take another sample, if possible.

If outliers are present but their removal is justified andresults in a data set for which the z-interval procedure isappropriate (as previously stated), the procedure can beused.

Key Fact(Cont’d)

For large samples–say, of size 30 or more–the z-intervalprocedure can be used essentially without restriction.However, if outliers are present and their removal is notjustified, you should compare the confidence intervalsobtained with and without the outliers to see what effectthe outliers have. If the effect is substantial, use adifferent procedure or take another sample, if possible.

If outliers are present but their removal is justified andresults in a data set for which the z-interval procedure isappropriate (as previously stated), the procedure can beused.

Key Fact

A Fundamental Principle of Data AnalysisBefore performing a statistical-inference procedure, examinethe sample data. If any of the conditions required for using theprocedure appear to be violated, do not apply the procedure.Instead use a different, more appropriate procedure, if oneexists.

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