Chapters 5-8 a Statistical Journey_taming of the Skew Teaching Slides by Dr. DeMoulin & Dr. Kritsonis

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A Statistical Journey:Taming of the Skew

A Discussion Of Chapters 5 – 8

Copyright 2009 by Dr. Donald F. DeMoulinDr.William Allan Kritsonis

Slides May Not Be Altered, Changed, or Modified



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Topics Of This Lesson

1) Probability2) Permutations and Combinations3) Z-Normal Distributions4) Transforming a Raw Score to Standard Deviation Units5) T-Scores and Stanine Scores6) Hypothesis Testing7) Type I and II Errors8) Power, Effect Size, Alpha Level and Sample Size

9) One-Tailed and Two Tailed Tests10) Seven Step Process11) One Sample z-test12) One Sample t-test13) Confidence Interval



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Probability

Chapter 5



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Probabilities have a confined range from zero(0) to one (1)

A probability which equals zero (0) is certainnot to occur while a probability of one (1) is

certain to occur

The closer a probability is to either extreme,the more likely (1) or unlikely (0) theoccurrence

Probability



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Analytical view —the desired outcome divided by

the possible outcomes or the probability of A occurring would be the possible ways for A dividedby the total possible ways (A+B)

For example, if you had 85 red M&Ms and 15 yellowM&Ms in your drawer, the probability of pulling out ared M&M would be Probability (A) = A/A+B)

Probability



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P(A) = A/(A+B)= 85/(85+15)= 85/100= 0.85

Probability



An event is a basic descriptor of an item

For example, the probability of drawing a redM&M out of a bag is referred to as an event

Probability



Events are said to be mutually exclusive if

one event precludes the occurrence of theother

Probability

Mutually Exclusive Events



• The Additive Law simply examines agiven set of mutually exclusive events

• Then, the probability of one event oranother is equal to the sum of their

separate probabilities

• [P(A or B) = P(A) + P(B)]

Additive Law for Mutually Exclusive Events



For example, if we had 30 red M&Ms, 15 yellowM&Ms, and 55 green M&Ms in a sack, whatwould be the probability of drawing an M&Mthat is either red or yellow

First we need to compute the probability of eachoccurrence separately (Analytical View)

P(A) = A/A+B+C = 30/100 = .30P(B) = B/A+B+C = 15/100 = .15P(C) = C/A+B+C = 55/100 = .55




Now we can substitute values in the additiveprocedure to calculate the probability of eitherdrawing a red or yellow M&M from the sack

P(A or B) = P(A) + P(B)

P(A red or B yellow ) = P(A red ) + P(B yellow )= .30 + .15= .45




The Multiplicative Law signifies thatthe probability of a joint occurrenceof two or more independent eventsis the product of their individualprobabilities

[P(A and B) = P(A) x P(B)]

Multiplicative Law for MutuallyExclusive Events



In this sack, we have 30 red M&Ms, 15 yellow M&Ms and 55green M&Ms

What would be the probability of picking a red M&M on the firstdraw, a yellow M&M on the second draw, and a green M&Mon the third draw

P(A and B and C)= P(A) x P(B) x P(C)

= P(A red and B yellow and C green )= P(A red ) x P(B yellow ) x P(C green )= (.30)(.15)(.55)= .0248

Multiplicative Law for MutuallyExclusive Events



Conditional Probability: one eventwill occur given that some otherevent has already occurred

[P(A/B) is read the probability of Agiven that B has already occurred]

Conditional Probability



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Have M&Ms Have No M&Ms TotalHave children 30 5 35

Have no children 15 50 65Total 45 55 100




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What would be the probability of havingM&Ms in a household?

P(A) = A/A+B= 45/100

= .45




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What would be the probability of nothaving M&Ms in a household?

P(B) = B/A+B= 55/100

= .55




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What would be the probability of having M&Ms given that you havechildren

Look at the column under "have M&Ms" and the row "have children“

Here we have 30 parents that have children and have M&Ms in thehousehold out of a total of 35 parents who have children

The probability calculation would be as follows:

P(A/B) = 30/35= .8571= .86




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Figure the probability of having M&Ms in ahousehold given that you have no

children

The procedure is the same:

P(A/B) = 15/65= .23




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The probability of A or B equals theprobability of A plus the probability of Bminus the sum of the probabilities of Aand B or the union ( ) of events A and B

P(A or B) = P(A) + P(B) - P(A and B)

Additive Law for MutuallyUnexclusive Events



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Overlap —one group (A) can also be part of another group (B)



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Children under Number of Probability18 who eat M&Ms Families .

0 15 .331 12 .272 10 .223 5 .114 3 .07 .

Total 45 1.00




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Figure the probability that a randomly selectedfamily has either an odd number of children

or has at least 1 child

First, we must figure the probability of havingan odd number of children (for the sake of argument, zero (0) will be considered an evennumber)

P(A) = A/A+B = (12 + 5)/45 = 17/45 = .38




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Next, we must figure the probability of families having at least one child

P(B) = B/A+B = (12+10+5+3)/45 = 30/45 = .67




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Finally, the solution to the original question of theprobability of randomly selecting a family that haseither an odd number of children, or at least one child

As you can see, the events are mutually unexclusive asa family could fit into both categories

P(A or B) = P(A) + P(B) - P(A and B)= (.38 + .67) - (.27 + .11)= 1.05 - .38= .67




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A factorial (denoted by the sign "!")signifies multiplying a selected integer byeach and every successive integer to zero

For example, 10! (read ten factorial) is thesame as multiplying 10 x 9 x 8 x 7 x 6 x 5 x4 x 3 x 2 x 1**

**zero factorial (0!) equals one

Permutations and Combinations



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A Permutation is concerned about thenumber of possible arrangements thatcan be made regarding a certain order

NPr = N!(N-r)!

N equals the number of items and r equals agrouping, or N items taken r at a time

Permutation



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For example, if a science teacher hassix slides, how many different orderscan be shown if he shows all six at atime

NP r = N! = 6! = 6x5x4x3x2x1 = 720(N-r)! 0! 1

Permutation



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Combinations (denoted NCr) are notconcerned with an order, but are

concerned with the number of differentcombinations that can be made bytaking a certain number at a time

NCr = N!r!(N-r)!

Combination



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How many combinations can be made utilizing a setcontaining the numbers 1, 2, 3, and 4, taking twoat a time

NCr = 4C2 = N! = 4! = 4! = 4!r!(N-r)! 2!(4-2)! 2!(2!) (2X1)(2X1)

= 4X3X2X1 = 24 = 6(2)(2) 4

Combination



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Z-Normal Distribution

Chapter 6



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The z-normal distribution is symmetrical andunimodal (one hump) with a mean of zero and a

standard deviation of one (µ = 0, σ = 1)




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The total area under the z-normal distributionequals one and extends to infinity in eitherdirection (positive and negative)

Total Area = 1.0




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The total area is divided into .5 extending tothe left and .5 extending to the right

.5.5




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If you have an area of .5 for one half of thedistribution, you must have an area of .5 forthe other half as .5 + .5 = 1.00

Total Area = 1.0




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The z-chart in the tables is divided intothree sections:

1) z-score2) the area between the mean and some z-

score3) the area beyond




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z-score

Area betweenmean and z

Areabeyond



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In the z-table, there are no negativenumbers because the area is the same

whether it is positive or negative

Remember, if you have an area of .5 for

one half of the distribution, you musthave an area of .5 for the other half as.5 + .5 = 1.00




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If you look on your chart at a z-score of 1, the "area between" the mean and a

z-score of 1 equals .3413 (34.13%)

Remember, the area between the meanand a z-score of 1 is the same areabetween the mean and a z-score of -1




40



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If you add .3413 and .3413, we get .6826, or 68.26 percent of thescores as previously discussed



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Next, the "area beyond" coincides with the percent of scores that are beyond a certain z-score

For example, the "area between" the mean and -1 orbetween the mean and +1 equals .3413, as we havealready established

Therefore, the "area beyond" a z-score of -1 or +1 mustbe .5000 - .3413 or .1587 (remember the areabetween and beyond must equal .5)




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If we want to know the area beyond z-score of .97, we look under the "area beyond" columnthat corresponds to z-score of .97

Looking at those numbers, we find that area tobe 0.1660

Conversely, the "area between" the mean andz-score of .97 = .3340 (.5000 - .1600 =.3340)




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z = .97 Area Beyond = 0.1660 Area Between Mean and z of .97 = .3340



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Now, if we want to find the area betweentwo z-scores, all we have to do is somesimple addition and subtraction

For instance, we want to know the areabetween a z-score of -1 and a z-score of .97

First we need to locate the area betweenthe mean and a z-score of .97

We need to find the area between z = -1(.3413) and z = .97 (.3340) and add thetwo figures (.3413 + .3340 = .6753)


.6753 z = -1 z = .97



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The process of obtaining a z-score from araw score (X) is simply subtracting the

mean (µ for a population or X- bar for asample) from the raw score (X) anddivide by the standard deviation

z = X-µ or z = X – X-bar

σ s

Transforming a Raw Score to SD Units



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Compute z and locate the area to the left of X = 5(µ = 4, σ = 2)

z = X-µ = 5-4 = 1 = .5σ 2 2

The raw score of 5 is located .5 standard deviationunits above the mean in which .3085 or 30.85percent of the scores lie beyond this figure and .1915or 19.15 percent of the scores lie between the meanand the z-score of .5




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Transforming a Z-Score to a T-Score

The use of a T-score is sometimes preferred over a z-score becauseT-scores eliminate negative numbers

The T- score has a mean of 50 and a standard deviation of 10 (µ = 50, σ =10) with a range from 20 to 80 that parallels with a z-normal distributionof -3 to +3

To compute a T-score, we first must compute a z-score. Then we canplug the z-score into the following formula: T = 10(z) + 50



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Data Summary

X = 26 µ = 20 σ = 3

z = X-µ = 26-20 = 6 = 2σ 3 3

Transform the z-score to aT-score by:

T = 10(z) + 50 = 10(2) + 50 =20 + 50 = 70


z-score = 2

T-Score = 70 z-score = +2



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Data Summary

X = 26 µ = 20 σ = 3

z = X-µ = 26-20 = 6 = 2σ 3 3


Stanine 1 2 3 4 5 6 7 8 9

Stanine = 2(z) + 5

Z-score previously calculated = 2

Stanine = 2(z) + 5 = 2(2) + 5 = 4 + 5 = 9

Stanine of 9 is located at same spotas the T-Score of 70 and a Z-scoreof 2





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It is critical to know that research neverproves anything

It only supports or fails to support yoursuspicion

Hypothesis Testing



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You have basically two different hypotheses

a null hypothesis (Ho) that assumes no significantdifference —or the resulting difference is not greatenough to warrant further attention

an alternative hypothesis (Ha) that assumes asignificant difference —or the resulting difference isgreat enough to warrant further attention

Hypothesis Testing



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You have basically two types of errors - Type I andType II

A Type I error occurs when a researcher rejects anull hypothesis (Ho) when the null hypothesis istrue

A Type II error occurs when a researcher fails toreject a null hypothesis (Ho) when the nullhypothesis is false

Hypothesis Testing

The term power of a statistical test is the



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Testing Errors

The term power of a statistical test is theprobability of action to reject the null hypothesiswhen the null hypothesis is in fact false



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Sample Size Determination



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Power

The Power of a test is the probability that anincorrect null hypothesis is rejected —or the ability toreject a false null hypothesis



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Power

The Power of a test is affected by the Effect Size,Sample Size and Alpha Level



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Effect Size

Effect size can be envisioned as the size of thetreatment effect the researchers wish to detect withprobability power, or the difference between thepopulation mean (µ) and the sample mean (X -bar )when the alternative hypothesis (Ha) is true and adifference can be visually detected



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Effect Size

Effect size has been classified in the following manner:

Small effect size : .25σ (read .25 standard deviations)- difference cannot be visually detected

Medium effect size : .50σ (read .50 standard

deviations) - synonymous with practical significanceLarge effect size : .80σ (read .80 standard deviations)- difference can be visually detected



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Alpha Level

The Alpha Level (or Significance Level) is the P-value that wedecide to accept before we will be confident enough to release

a finding —this is our predetermined acceptance level whichis usually defaulted to .05

Many researchers will not accept a P value greater than .10



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Increase Power

To Increase the Power in a Study Calls for An Increasein Sample size

You Already Have an Established Alpha Level and anEstimated Effect Size so all You Have to Increase Poweris Sample Size



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Increase Power

In an Independent t-test with an alpha level of .05, effect size at.5 and to achieve .80 power with a one-tailed test, you wouldneed a sample size of 50

For .90 power, a sample size of 69 is needed


PER GROUP!!



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Sample Size Chart Example



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Increase Power

In an Independent t-test with an alpha level of .05, effect size at.5 and to achieve .80 power with a two-tailed test, you wouldneed a sample size of 64



PER GROUP!!



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Sample Size Chart Example



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Developing HypothesesDefining a hypothesis

A researcher’s tentative prediction of the

results of the researchFormulated on the basis of knowledge of theunderlying theory or implications from theliterature review

Testing a hypothesis leads to support of thehypothesis or lack thereof



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Developing Hypotheses A good quantitative hypothesis…

is based on sound reasoning

provides a reasonable explanation for thepredicted outcomeclearly and concisely states the expectedrelationships between variablesis testable



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Developing HypothesesTypes of quantitative hypotheses

Research hypotheses state the expectedrelationship between two variables

Non-directional – a statement that no relationship ordifference exists between the variablesDirectional – a statement of the expected direction of therelationship or difference between variables

Null – a statistical statement that no statistically significant relationship or difference exists betweenvariables



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Developing HypothesesNon-Directional Null

There is no relationshipbetween mean math attitudesand mean math achievement

H0: µ = 0H

a: µ ≠ 0

There is no difference in meanmath achievement of studentsusing technology versus those

students who do not

H0: µ 1 - µ 2 = 0Ha: µ 1 - µ 2 ≠ 0



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Developing Hypotheses

Directional Null

Fifth grade boys score significantlyhigher on mean science test scoresthan 5 th grade girls

H0: µ ≤ 0Ha: µ > 0

Students using technology will scoresignificantly lower on mean anxiety

scales than students who do not usetechnology

H0: µ 1 - µ 2 ≥ 0Ha: µ 1 - µ 2 < 0



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One-Tailed and Two Tailed Tests

Critical Value =The Value WhichRepresents theBeginning of theRejection Region

Observed Value =The Calculated ValueThat is Compared tothe Critical Value forStatistical Significance



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Seven-Step ProcessStep 1: Formulate hypothesis, null (Ho) and alternative (Ha), and indicate a test for

difference (non-directional, two-tailed test) or a test for direction (directional,one-tailed test)

Step 2: Establish an alpha level (allowing for power, effect size, and sample size)

Step 3: Determine appropriate sampling distribution

Step 4: Formulate a decision rule

Step 5: Gather data and perform appropriate statistical procedure

Step 6: Summarize procedures based on the decision rule

Step 7: Draw a logical conclusion from results



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Summary

Remember that hypothesis testing does notprove anything

It only supports or fail to support yourhypothesis at that given time with that givensample



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Test of Means forOne Sample

Chapter 8



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Test of Means for One Sample

Review

A null hypothesis (Ho) is written as no difference(µ1 = µ 2) where the population mean of onegroup (µ 1) equals the population mean of another group (µ 2)

When this occurs, no significant difference exists



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Review

The alternative hypothesis (Ha) is written to signify asignificant difference does exist (µ 1 ≠ µ 2), where thepopulation mean of one group (µ 1) does not equal thepopulation mean of another group (µ 2)

Rejecting the null hypothesis, therefore constitutes asignificant difference between groups



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Test of Means for One SampleReview

The alpha level (ά), sometimes referred to as therejection level , the significance level , or the risk factor , is the relative frequency a researcher wishesto make a Type I error

A Type I error is the rejection of the null hypothesiswhen the null hypothesis is actually true

A Type II error is the failure to reject the null hypothesiswhen in fact the null hypothesis is false



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A critical value is a value for an appropriatesampling distribution that will cause a rejection

of the null hypothesis

The critical value is affected by the sample size,alpha level, and type of hypothesis test(directional or non-directional)



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Test of Means for One SampleEach statistical procedure has a table of critical values for

reference

A critical value for one statistical procedure will not be the samecritical value for another statistical procedure

A critical value constitutes the beginning of a critical or

rejection region

A critical region signifies the point(s) designated by the criticalvalue and every point thereafter extending indefinitely



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The point to the left or right of the critical regionon a one-tailed test is known as the ‘safe’

region or the ‘fail to reject’ region

The point in between the critical regions on a two

tailed test is identified as the same — ‘safe’ orthe ‘fail to reject’ region





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A two-tailed testsignifies nodirection, just a

difference, and isconcerned withboth ends of thedistribution since adirection (positiveor negative) is not

important

Ho: =Ha: ≠

T f M f O S l



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Test of Means for One SampleZ-Test

One of the most basic statistical procedures is to test a sampleagainst an established norm

Example

Data collected at a municipal hospital on 100 newborn babiesshow that the average new-born child weighs 7.2 pounds with

a standard deviation of 0.7 pounds

A father wants to know whether his four children who weighed5.0 pounds, 6.8 pounds, 8.6 pounds, and 5.8 pounds at birthwere significantly different from the average newborn

T f M f O S l



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Test of Means for One Sample —Z-Test

In this example, we have one sample, the fourchildren, and an established norm, the average

weight of newborn babies, plus a knownstandard deviation

To solve this mystery, we would perform what isknown as a one-sample z-test

T f M f O S l



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If you recall in an earlier section, we transformeda raw score to a z-score by using the formula

Z = X - µσ

For a one sample z-test a similar formula is usedZ = X- bar - µ

σx ̄

T f M f O S l



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σx ̄ is referred to as the standard error of themean

The Standard Error of the Mean is determined bytaking the Standard Deviation of the sampleand dividing by the square root of the sample

σx ̄ = σ √ n

T t f M f O S l



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Our inquisitional father is interested to see if hisnewborns were indeed significantly different in weightfrom the established average weight of new-born

infants

The seven step process begins with

Step 1: Formulate Ho and Ha

Ho: µ = 7.2Ha: µ ≠ 7.2

T t f M f O S l



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Step 2: Set alpha level (ά) at .05 (purely arbitrary)

Step 3: Sampling Distribution

z-normal distribution for one sample test witha Known σ

T t f M f O S l



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Step 4: Formulate decision rule

This step will be made a bit clearer if we utilize a diagram

Since we are only concerned with a difference, not lighter or heavier than, we are utilizing a non-directional, two-tailed test and the alpha level must be divided equally (since the right half isthe exact mirror of the left half) between each end of the distribution

T t f M f O S l



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The alpha level of .05 becomes .025 for the left end of the tail (negative)and .025 for the right end of thetail (positive)

T t f M f O S l



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In the z-table, we find a z-column , an area betweenmean and z-column , and an area beyondcolumn —we are concerned with the area beyond since that value and every value beyond that pointsignifies the critical region or rejection region

If we look down the column area beyond to .025, wefind a corresponding z-score of 1.96



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Z-critical value

Area Beyond




96


If the calculated or observed value falls at orbeyond 1.96 on the positive end (right half of

the tail) or at or beyond -1.96 on the negativeend (left half of the tail) we will reject thehypothesis

If the calculated or observed value falls between-1.96 and 1.96, we will then fail to reject thenull hypothesis



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We will reject Ho if our z-observed value (z obs ) falls at or beyond our z-critical value (z crit ) of 1.96 or at or beyond our z-critical value of -1.96 otherwise we will fail to reject




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To simplify matters, on all two-tailed tests, we can utilizean absolute value (| |) for our z obs since we are onlyconcerned with a difference

The decision rule now becomes:

We will reject Ho if our |z obs | falls at or beyond our |z crit | of 1.96, otherwise we will fail to reject




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Step 5: Calculation using appropriate statisticalprocedure

Z-test formulaZ = X- bar - µ

σx ̄




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First, we must calculate the standard error of the mean (σ x ̄ ) by the followingformula:

σx ̄ = σ √ n

Here σ equals the established standard deviation and n equals the number inthe sample

Calculation σx ̄ = s = .7 = .7 = .35√ n √ 4 2




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Next we must calculate the average weight (X -bar ) for thesample

Weight (x)5.06.8

8.65.8

ΣX = 26.2

X-bar

= ΣX/N = 26 2 ÷ 4 = 6 55




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Now all we have to do is substitute values intothe z-formula

z = X -bar - µ = 6.55 - 7.2 = -.65 = -1.857 = -1.86

σx ̄ .35 .35

Test of Means for One Sample —



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Step 6: Summarize results

Since the | z obs | of 1.86 does not fall at or beyond the z cri of 1.96, we therefore fail to reject Ho



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Step 7: Draw a logical conclusion based onsummary

Since we failed to reject the Ho, we conclude that there is no difference in the average

weights of the father's four children at birth and the average weight of new-born children at the municipal hospital




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Test of Means for One Sample Z-Test

Example 2

What if the father speculated that the meanweight of his four children at birth wassignificantly less than the average new-bornchild at the municipal hospital

Step 1: Ho: µ = 7.2

Ha: µ < 7 2

Ho: µ ≥ 7.2 Ha: µ < 7.2




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Step 2: Alpha level (ά) equals .05

Step 3: Still a z-normal distribution

Step 4: Decision rule




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To find our critical value, we must again refer to ourz-chart

Since we are not dividing our alpha level and areconcentrating its full effect on the left end of thedistribution, we now look at the area beyond column for .05

Locating .05 we find a corresponding z-value of 1.645and since we are concerned with the left end of thedistribution (negative side) this z-value becomes-1.645



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Z-critical value

Area Beyond



111




112



Reject Ho if z obs falls at or beyond z cri of -1.645,otherwise, fail to reject




113


Step 5: Calculation utilizing appropriatestatistical procedure

We have already calculated the observed valueof -1.86 so we can move to step 6




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Step 6: Summarize results based on decisionrule

Since z obs of -1.86 falls beyond z cri of -1.645, we therefore reject the Ho in favor of the Ha



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Step 7: Conclusion

Since we rejected the Ho, we conclude that the mean weight of the father's four children at birth is significantly less than the mean

average weight of new-born infants at the municipal hospital




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Summary

To justify the use of a one sample z-test, some assumptions are needed

• we must assume that the underlying variable comes from a population thatis normally distributed

• we must assume that the null hypothesis is true

• we must assume that the rules of randomization were followed

• we must know or have access to the population standard deviation




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Test of Means for One Sample t-test

What if the standard deviation is not known or given

If a population standard deviation is not given, then itmust be calculated from a sample

The standard deviation now becomes an estimate and

is denoted as N-1

N-1 becomes the best estimate of the populationstandard deviation




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Test of Means for One Sample t-test

The t-test is used in place of the z-test whenthe standard deviation is not known and mustbe calculated because it functions better (ismore robust) as a measure when the standarddeviation is estimated

Using a t-test makes an adjustment that utilizesthe estimates of the standard deviation (σ) of a sample (N-1) for the population parameter

(µ)




120

Test of Means for One Samplet-test

Example

Over the years, the 'Kids' corporation has sold an average of 3.7 pitchers of lemonadeper day

They want to know if they are faring as well this year

They randomly selected five (5) days for this year and logged the following data set:

Day Pitchers Sold

1 22 33 14 15 5




121

Test of Means for One Samplet-test

Step 1: Formulate Hypothesis

Ho: µ = 3.7Ha: µ ≠ 3.7

Step 2: Alpha level (ά) equals .05




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pt-test

Step 3: Sampling distribution

df is found by N-1 where N equals the totalnumber —N-1 = 5-1 = 4

We have four (4) degrees of freedom for thissample and our appropriate samplingdistribution therefore becomes t 4




123

pt-test


We are performing a non-directional, two-tailed test sowe look in that box for the alpha level of .05(remember, the value has been divided for you)

Finally, we intersect df of 4 and alpha of .05, two-tailedtest and find that our critical value is 2.776



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critical value




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pt-test


Reject Ho if | t obs | falls at or beyond |t crit | of 2.776,otherwise fail to reject

Step 5: Calculation using appropriate statistical

procedure

The t-statistic by utilizes the formula t = X -bar - µest. σx ̄




126

pt-test

Data Set

Sample Day Pitchers Sold X²

1 2 42 3 93 1 14 1 15 5 25

ΣX = 12 ΣX² = 40 N = 5

X-bar = 2.4σ = 1.673




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pt-test

Next, we must calculate the standard error of the mean(σx ̄ ) using the formula below

σx ̄ = s = 1.673 = 1.673 = .747√ n √ 5 2.24

We solve for t by:

t = X -bar - µ = 2.4 – 3.7 = -1.3 = -1.74est. σ x ̄ .747 .747




128

pt-test

Step 6: Based on the calculated information,we offer the following summary:

Since │t obs │of 1.74 does not fall at or beyond |t crit | of 2.776, we therefore fail to reject Ho



129




130

pt-test

Step 7: Conclusion

Since we failed to reject the Ho, we conclude that there is no difference in the amount of pitchers sold per day this year when compared

to last year




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pt-test

What if the 'Kids' corporation suspected that this year's per day pitcher sales were exceeding

last year's per day pitcher sales

Step 1: Formulate Hypothesis

Ho: µ ≤ 3.7Ha: µ > 3.7




132

pt-test

Step 2: ά = .05

Step 3: t 4 as sampling distribution (this doesnot change)





133

pt-test


Referring to the student's t-chart, we find ourcritical value at four (4) degrees of freedomand at an alpha level of .05, one-tailed test to

be 2.132 (t crit = 2.132)



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critical value




135

t-test


Reject Ho if t obs falls at or beyond t crit of 2.132,otherwise, fail to reject



136




137

t-test


Since we have already calculated the t-statistic to be-1.74, we can go to step 6

Step 6: Summary

Since t obs of -1.74 does not fall at or beyond t crit of +2.132 we therefore fail to reject Ho



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t-test

Step 7: Conclusion

Since we failed to reject the Ho, we conclude that this year's per day pitcher sales do not exceed last year's per day pitcher sales




140

t-test

There is another calculation that needs to bedone to construct a confidence interval

A confidence interval generates a range estimatefor the population mean (µ) by generating a

lower limit and an upper limit for possibleparameters




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t-testNo matter what type of hypothesis test you perform (one-tailed

or two-tailed) the critical value in this formula is always designated as though you performed a two-tailed test

In other words, the critical value in any confidence interval is atwo-tailed value

The sample mean will fall outside the constructed populationmean interval when the Ho is rejected on a two-tailed test

A failure to reject signifies the sample mean is within theconstructed population mean interval




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t-test

For the z-test, the formula is:

95% CI = 100(1- ά) = µ ± z cri (σx ̄ )

For the t-test, the formula is:

95% CI = 100(1- ά) = µ ± t crit(σx ̄ )




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t-testThe confidence interval calculations for our z-statistic is:

95% CI = 100(1- ά) = µ ± z cri (σx ̄ )95% = 7.2 ± 1.96(.35)95% = 7.2 ± .68695% = 6.514 ≤ 7.2 ≤ 7.886

In the previous z-test, we failed to reject Ho and therefore the sample mean(X-bar = 6.55) should fall within the confidence interval range

And it does




144

t-test

What the confidence interval relates is that,

based on our random sample, we can have a95% confidence level that the true mean willfall between 6.514 and 7.886




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t-test

Now let us calculate the confidence interval forthe t-statistic just completed

95% CI = 100(1- ά) = µ ± t crit(σx ̄ )95% = 3.7 ± 2.776 (.747)

95% = 3.7 ± 2.776 (.747)95% = 3.7 ± 2.77695% = 0.924 ≤ 3.7 ≤ 6.476




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t-test

Since we failed to reject our Ho in the t-test

problem, our sample mean of 2.4 should fallwithin the confidence interval range

And it does

95% = 0.924 ≤ 3.7 ≤ 6.476

Interval —Ratio Data



Interval Ratio Data

One Sample

Dependent Variable

Known SD frompopulation

Estimated SD fromSample

Documents

Chapters 5-8 a Statistical Journey_taming of the Skew Teaching Slides by Dr. DeMoulin & Dr. Kritsonis