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A science that deals withcollecting, organizing,analyzing and
interpreting pertinentdata.
STATISTICS
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Imagine this situation:
You are in a class with just four otherstudents, and the five of you took a 5-
point pop quiz. Today your instructor iswalking around the room, handing backthe quizzes. She stops at your desk and
hands you your paper. Written in boldblack ink on the front is 3/5. How do youreact?
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Are you happy with
your score of 3 ordisappointed?
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How do you decide?
You might calculate yourpercentage correct, realize it is
60%, and be appalled.
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But it is more likely that whendeciding how to react to your
performance, you will wantadditional information.
What additional information
would you like?
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If you are like most students,you will immediately ask your
neighbors, "Whad'ja get?" andthen ask the instructor, "How
did the class do?"
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In other words, the additionalinformation you want is how your
quiz score compares to otherstudents' scores. You thereforeunderstand the importance of
comparing your score to the classdistribution of scores.
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3 Common Measures of
Central Tendency
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Mean of Ungrouped Data
To compute the mean of ungrouped data, simply
add the given observations and divide it by thenumber of observations.
Xi where: Xi sum of all observations
X = ______ n total number
n observations
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Example:
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Mean of Grouped Data
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Example:
C.I. Freq. Xi FiXi
7679 2 77.5 155
8083 5 81.5 407.5
84
87 5 85.5 427.58891 11 89.5 984.5
9295 4 93.5 374
9699 3 97.5 292.5
n = 30 = 2,641
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Median of Ungrouped Data
To get the median of ungrouped
data, arrange the given observationsaccording to magnitude, then
identify the middle value.
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Example
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Note:
In case of an even number of
observations, we expect two middlevalues, what simply need to be done
is to get the average of the two
observations by adding the twoobservations and dividing it by 2.
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Example:
C.I. C.B. Freq. F
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Solution:
Identify the Median class. The median classis computed using the formula n/2. Since
n=30, therefore n/2 = 15.
Locate the computed n/2 in the F
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Identify the value of the differentvariables needed in the formula.
LMe = 87.5n/2 = 15
cfb = 12fMe = 11
c = 4
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Mode of Ungrouped DataTo get the mode of ungrouped data,
identify the observation/s havingthe most number of frequency or
occurrence.
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Example:
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Mode of Grouped Data
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Example:
C.I. C.B. Freq.
7679 75.579.5 2
8083 79.583.5 5
84
87 83.5
87.5 5
88 91 87.5 91.5 11
9295 91.595.5 4
96
99 95.5
99.5 3n = 30
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Solution:
Identify the Modal class. The modalclass is the class having the highestfrequency. Since the highest frequency
is 11, therefore 88 91 is the modalclass.
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Scores of 5 Boys and 5 Girls in Mathematics
Boys Girls
Frederick 70 Grace 82
Russel 95 Irish 80
Murphy 60 Abigail 83
Jerome 80 Sherry 81Tom 100 Kristine 79
Mean: 81 Mean: 81
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Boys
60 70 80 90 100
Girls
60 70 80 90 100
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Measures of Variability or Dispersion
RANGE:The difference between the highest and the
lowest observationR = H L
Boys: R = 100 60
R = 40Girls: R = 83 79
R = 4
Therefore the
girls are more
homogeneous
than the boys in
their math
ability
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Mean Deviation:
The average of the summation of the
absolute deviation of each observationfrom the mean.
MD = | XiX |n
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BOYS Xi lXiXl
Frederick 70 11
Russel 95 14
Murphy 60 21Jerome 80 1
Tom 100 19
Mean: 81 = 405 = 66
M.D = 66 / 5 = 13.2
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GIRLS Xi lXiXl
Grace 82 1
Irish 80 1
Abigail 83 2Sherry 81 0
Kristine 79 2
Mean: 81 = 405 = 6
M.D = 6 / 5 = 1.2
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MD ( boys ) = 13.2MD ( girls ) = 1.2
- based from the computed MeanDeviation, the girls are more
homogeneous than the boys.
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VARIANCE:
The average of the squared deviationfrom the mean.
Population Variance
2 = ( Xi X ) 2
n
Sample Variances 2 = ( Xi X ) 2
n - 1
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BOYS Xi XiX ( XiX ) 2
Frederick 70 -11 121
Russel 95 14 196
Murphy 60 -21 441Jerome 80 -1 1
Tom 100 19 361
Mean: 81 = 405 = 1,120
2 = 1,120 / 5 s2 = 1,120 / 4
= 224 = 280
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GIRLS Xi XiX ( XiX ) 2
Grace 82 1 1
Irish 80 1 1
Abigail 83 2 4Sherry 81 0 0
Kristine 79 2 4
Mean: 81 = 405 = 10
2 = 10 / 5 s2 = 10 / 4
= 2 = 2.5
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BOYS2 = 1,120 / 5 s2 = 1,120 / 4
= 224 = 280
GIRLS
The values of
the Variance
also reveals thatthe score of
boys are more
spread out than
that of the girls.
2 = 10 / 5 s2 = 10 / 4
= 2 = 2.5
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STANDARD DEVIATION:
The square root of the Variance
BOYS
2
= 224 s2
= 280 = 14.97 s = 16.73
GIRLS
2 = 2 s 2 = 2.5
= 1.41 s = 1.58
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Let us pause fora BREAK
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HYPOTHESISTESTING
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HYPOTHESIS TESTING
Inferential Statistics formalized body oftechniques used to make conclusions aboutpopulations based on the results of the study on
the samples.Two areas of Inferential Statistics
Estimation
Point Estimation
Interval Estimation
Hypothesis Testing
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HYPOTHESIS TESTING
Research Problem: How effective is Minoxidil intreating male pattern baldness?
Specific Objectives:
1. To estimate the population proportion of patients whowill show new hair growth after being treated withMinoxidil.
2. To determine whether treatment using Minoxidil is betterthan the existing treatment that is known to stimulate hairgrowth among 40% of patients with male patternbaldness.
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HYPOTHESIS TESTING Hypothesis Testing - is the process of making
an inference or generalization about a populationby using data gathered from a sample of thepopulation
It is an area of statistical inference in which oneevaluates a conjecture about some characteristicof the parent population based upon theinformation contained in the random sample.
Usually the conjecture concerns one of theunknown parameters of the population.
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HYPOTHESIS TESTING
Kinds of Hypothesis:
Scientific Hypothesis is a suggested
explanation or solution to a phenomenon.
Statistical Hypothesis:
Itis a guess or prediction made by a researcher
regarding the possible outcome of the study.It is a claim or a statement about an unknown
parameter.
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HYPOTHESIS TESTING
Examples of Scientific Hypothesis:
When Darwin hypothesized that manevolved from the apes, he was making a
scientific hypothesis.
Similarly when Copernicus hypothesizedthat the earth and the planets in the solarsystem revolved around the sun inconcentric circles with the sun as thecenter.
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HYPOTHESIS TESTING
Examples of Statistical Hypothesis:
1. The correlation between X and Y(in the population)
is equal to zero;
2. There is no significant difference in the mean of
the two groups;
3. The mean IQ of the population is 100;
0XY
BA
100
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HYPOTHESIS TESTING
Two Types of Statistical HypothesisNull hypothesis (H0): It is the hypothesis to be
tested which one hopes to reject. It shows
equality or no significant difference, effect, orrelationship between variables.
denoted by Ho.
the statement being tested.
it represents what the experimenter doubts to be true.
must contain the condition of equality and must be writtenwith the symbol =, , or
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HYPOTHESIS TESTING
For the mean, the null hypothesis will be stated inone of these three possible forms:
Ho: = some value
Ho: some value
Ho: some value
Note: the value of can be obtained from previous studiesor from knowledge of the population
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HYPOTHESIS TESTING
Alternative hypothesis (Ha): It generallyrepresents the idea which the researcher wantsto prove.
denoted by Ha
is the statement that must be true if the nullhypothesis is false
the operational statement of the theory that the
experimenter believes to be true and wishes toprove
is sometimes referred to as the research hypothesis
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HYPOTHESIS TESTING
For the mean, the alternative hypothesis will bestated in only one of three possible forms:
Ha: some value
Ha: > some value
Ha: < some value
Note:
Ha is the opposite of Ho. For example, if Ho is given as
= 37.0, then it follows that the alternative hypothesis isgiven by Ha: 37.0.
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HYPOTHESIS TESTING
Note About Using or in Ho
Even though we sometimes express Ho with the
symbol or as in Ho: 37.0or Ho: 37.0, we conduct the test by assumingthat = 37.0 is true.
We must have a single fixed value for so that wecan work with a single distribution having aspecific mean.
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HYPOTHESIS TESTING
Note About Stating Your OwnHypotheses
If you are conducting a research studyand you want to use a hypothesis test tosupportyour claim, the claim must be
stated in such a way that it becomes thealternative hypothesis, so it cannotcontain the condition of equality.
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HYPOTHESIS TESTING
Example in Stating your Hypothesis
If you believe that your brand of refrigerator
lasts longer than the mean of 14 years forother brands, state the claim that > 14,where is the mean life of your
refrigerators.Ho: = 14 vs. Ha: > 14
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HYPOTHESIS TESTING
In this context of trying to support the goalof the research, the alternative hypothesis issometimes referred to as the research
hypothesis.Also in this context, the null hypothesis is
assumed true for the purpose of conductingthe hypothesis test, but it is hoped that the
conclusion will be rejection of the nullhypothesis so that the research hypothesis issupported.
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HYPOTHESIS TESTING
Research Problem:
Comparative performance in Mathematics ofthe first-born and the last-born children.
H0: There is no significant difference in theperformance in mathematics between the first-born and last-born children.
Ha: There is a significant difference in theperformance in mathematics between the first-born and last-born children.
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HYPOTHESIS TESTING
Research Problem:
Effectiveness of an Instructional Strategy
H0: There is no significant effect of modified workedexample strategy in the problem solving ability ofstudents in physics.
Ha: The modified worked example strategy will have asignificant effect in the problem solving ability of students
in physics.Ha: Students exposed to the modified worked examplesare better problem solvers than those exposed toconventional worked examples.
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HYPOTHESIS TESTING
Research Problem:
Relationship between emotional intelligence ofstudents and their level of math anxiety
H0: There is no significant relationship betweenstudents emotional intelligence and their level ofmath anxiety.
Ha: There is significant relationship betweenstudents emotional intelligence and their level ofmath anxiety.
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HYPOTHESIS TESTING
REMARK:
If the null hypothesis is rejected, thealternative hypothesis is accepted andvice versa. Rejection of the nullhypothesis means it is wrong, whileacceptance of the null hypothesis
does not mean it is true, it simplymeans that we do not have enoughevidence to reject it.
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HYPOTHESIS TESTING
Types of Hypothesis Testing
1. Two-tailed test: It is non-directional test with
the region of rejection lying on both tails of the
normal curve. It is used when the alternativehypothesis uses words such as not equal to,significantly different, etc.
Acceptanceregion Rejection regionRejection region
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HYPOTHESIS TESTING
Example: A teacher wants to know if there issignificant difference in the performance inStatistics between his morning and afternoonclasses.
H0: There is no significant difference in theperformance in Statistics between his morningand afternoon classes.
Ha: There is a significant difference in theperformance in Statistics between his morningand afternoon classes.
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HYPOTHESIS TESTING
2. One-tailed test: It is a directional test with theregion of rejection lying on either left or right tailof the normal curve.
Right directional test. The region of rejection is on theright tail. It used when the alternative hypothesis usescomparatives such as greater than, higher than, betterthan, superiorto, exceeds, etc.
Acceptance region
Region of Rejection
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HYPOTHESIS TESTING
Example:Research Problem: Performance inMathematics of the First-born and Last-born
ChildrenH0: The first born-children perform equallywell in mathematics as the last-born children.
Ha: The first born-children perform better inmathematics than the last-born children.
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HYPOTHESIS TESTING
Left directional test. The region of rejectionis on the left tail. It is used when the alternativehypothesis uses comparatives such as lessthan, smaller than, inferior to, lower than,
below, etc.
Acceptance regionRejection region
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HYPOTHESIS TESTING
Example:Research problem:It is known that in the school canteen, the average waitingtime for a customer to receive and pay for his order is 20
minutes, Additional personnel has been added and nowthe management wants to know if the average waiting timehad been reduced.
H0: The average waiting time had not been reduced or the
average waiting time is equal to 20 minutes.
Ha: The average waiting time had been reduced, or the
average waiting time is less than 20 minutes.
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HYPOTHESIS TESTING
What is a test of Significance?
A test of significance is a problem of decidingbetween the null and the alternative hypotheses onthe basis of the information contained in a randomsample.
The goal will be to reject Ho in favor of Ha, because
the alternative is the hypothesis that the researcherbelieves to be true. If we are successful in rejectingHo, we then declare the results to be significant.
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HYPOTHESIS TESTING
Two Types of Error in Hypothesis Testing:
TYPE 1 ERROR
The mistake of rejecting the null hypothesis when it is
true.It is not a miscalculation or a procedural misstep; it is
an actual error that can occur when a rare eventhappens by chance.
The probability of rejecting the null hypothesis when itis true is called the significance level ( ).
The value of is typically predetermined, and the verycommon choices are = 0.05 and = 0.01.
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HYPOTHESIS TESTING
Examples of Type I Error
1.The mistake of rejecting the null
hypothesis that the mean bodytemperature is 37.0 when that mean isreally 37.0.
2.BFA did not allow the release of aneffective medicine.
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HYPOTHESIS TESTING
Type II Error
The mistake of failing to reject the null
hypothesis when it is false.The symbol (beta) is used to
represent the probability of a type II
error.
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HYPOTHESIS TESTING
Examples of Type II Errors
1.The mistake of failing to reject the null
hypothesis ( = 37.0) when it is actuallyfalse (that is, the mean is not 37.0).
2.BFA allowed the release of an ineffective
drug.
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HYPOTHESIS TESTING
A typical example of testing a statistical hypothesis issummarized in the following table.
Accept H0 Reject H0
H0 is trueCorrectDecision
Type 1 Error
H0 is false Type II ErrorCorrectDecision
HYPOTHESIS TESTING
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HYPOTHESIS TESTING
Controlling Type I and Type II Errors
o The experimenter is free to determine . If the test leads to therejection of Ho, the researcher can then conclude that there issufficient evidence supporting Ha at level of significance.
o Usually, is unknown because its hard to calculate it. The commonsolution to this difficulty is to withholdjudgment if the test leads tothe failure to reject Ho.
o and are inversely related. For a fixed sample size n,as decreases increases.
o In almost all statistical tests, both and can be reduced by
increasing the sample size.o Because of the inverse relationship of and , setting a very small
should also be avoided if the researcher cannot afford a very largerisk of committing a Type II error.
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HYPOTHESIS TESTING
The choice of usually depends on theconsequences associated with making aType I error.
Common Choices
OfConsequences of
Type I Error
0.01 or smaller
0.050.10
Very serious
Moderately seriousNot too serious
HYPOTHESIS TESTING
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HYPOTHESIS TESTING
Level of Confidence
a.) 0.05 level95% sure that the error is only 5%.When a different set of samples is taken from the same
population, the probability of getting a result similar to the presentstudy is 95%.
b.) 0.01 level99% sure that the error is only 1%
Note:
A test is said to be significant if the null hypothesis isrejected at the 0.05 level of significance and is consideredhighly significant if the null hypothesis is rejected at the 0.01level of significance.
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HYPOTHESIS TESTING
Steps in Testing the Hypothesis
1. State the null and alternative hypotheses.
2. Decide on a level of significance, .
3. Determine the testing procedure and methodsof analysis (responsibility of the statistician).
4. Decide on the type of data collected and
choose an appropriate test statistic and testingprocedure.
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HYPOTHESIS TESTING
Steps in Testing the Hypothesis
5. State the decision rule.
6. Collect the data and compute for the value of thetest statistic using the sample data.
7. If decision rule is based on region of rejection:Check if the test statistic falls in the region of
rejection. If yes, reject Ho.
If decision rule is based on p-value: Determinethe p-value. If the p-value is less than or equal to, reject Ho.
8. Interpret results.
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HYPOTHESIS TESTING
The Test Statistic - a statistic computed from the sampledata that is especially sensitive to the differencesbetween Ho and Ha.
1. The test statistic should tend to take on certain values when Hois true and different values when Ha is true.
2. The decision to reject Ho depends on the value of the test statistic
3. A decision rule based on the value of the test statistic:Reject Ho if the computed value of the test statistic falls
in the region of rejection.
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HYPOTHESIS TESTING
Critical Value/s
the value or values that separate the criticalregion from the values of the test statistic that
would not lead to rejection of the null hypothesis.
It depends on the nature of the null hypothesis,the relevant sampling distribution, and the levelof significance.
level of significance (): the smaller is,
the smaller the region of rejection
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HYPOTHESIS TESTING
Test Concerning MeansA. Test for one sample mean
a. When is known and n 30.
z =
b. When unknown and n < 30
t =
n
x
n
s
x
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HYPOTHESIS TESTING
Example 1.
The production manager of a large manufacturingcompany estimates that the mean age of his workers is22.8 years. The treasurer of the firm needs more
accurate employee mean age figure in order to estimatethe cost of an annuity benefit program being consideredfor employees. The treasurer takes a random sample of70 employees and finds that the mean age of thesampled employees is 26.2 years with a standard
deviation of 4.6 years. At 0.05 level of significance, testthe hypothesis that the mean age of the employees is notequal to 22.8 years.
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HYPOTHESIS TESTINGSolution:The steps include:
1. Null Hypothesis : H0 : yearsAlternative Hypothesis: Ha : years
Level of Significance:Test Statistics: Two - tailed Test; n = 70
Critical Region: Reject the null hypothesis if z < -1.96 or z > 1.96, otherwiseaccept it.Note: z is used since our sample size n = 70 is quite large.
Compute:
Decision:Since zc = 6.184 exceeds 1.96, the null hypothesis must be rejected; In otherwords, the difference between = 26.2, and years is too large toattribute it to chance. So we can say that their difference is significant. Hence,the mean age of the employees is not 22.8.
8.228.22
05.0
184.6
70
6.4
8.222.26
cZ
n
xZ
x 8.22
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HYPOTHESIS TESTING
Example 2.
A random sample of 20 drinks from a soft-drink machine has an average content of
21.9 deciliters, with a standard deviation of1.42 deciliters. At .05 level of significance,test the hypothesis that = 22.2 deciliters
against the alternative that < 22.2 andassume that the distribution of the softdrinks contents be normal.
HYPOTHESIS TESTING
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Solution:
Null Hypothesis H0 : = 22.2 deciliters
Alternative Hypothesis H1
: < 22.2 deciliters
Level of Significance: = 0.05
Test Statistics: with df = n-1
Note: The students statistic can be used since our sample size n = 20 issmall and the soft drinks content was assumed to be normally distributed.
Criterion: Reject the null hypothesis if computed t < -1.729 (the tabular valueof t at 20-1 degrees of freedom and otherwise, accept it.
Compute:
Decision: Since computed t = -0.945 is greater than -1.729, we accept H0.Conclude that the mean content of the soft drinks is equal to 22.2 deciliters.In other words, though there is a numerical difference of 0.3, this differencecan be attributed to chance.
n
s
xt
945.0
2042.1
2.229.21
ct
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HYPOTHESIS TESTING
Test of Differences of Two Means
a. When n1 30 or n2 30
z =
b. When n1 < 30 and n2 < 30
t = where
2
2
2
1
2
1
21
nn
xx
21
21
11
nns
xx
p
2
11
21
222
211
nn
snsn
sp =
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HYPOTHESIS TESTING
Example 3.
An instructor wishes to determine which of the twomethods of teaching: A or B, is more effective in teachingcertain concepts in Physics. In a class of 36 students, heused method A and in the other class of 40 students,method B. He gave the same final examination for bothclass and garnered the following results:
Method A Method B
Is the instructor correct in assuming that method A is moreeffective than method B. Use 0.01 level of significance.
781x 70
2x
41 s 62 s
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HYPOTHESIS TESTINGSolution:
Null Hypothesis H0 :Alternative Hypothesis H1 :
Level of Significance:Test Statistics:
Z =
Critical Region: Reject the null hypothesis if Zc > 2.326; otherwise state the differencebetween two sample means is not significant.
Compute:
Zc=
Decision: Since Zc = 6.899 is greater than 2.326 the null hypothesis must be rejected.Conclude that the instructors claim is correct that method A is more effective than method B.
BA
BA
01.0
2
2
2
1
2
1
21
nn
xx
899.6
40
6
36
4
7078
22
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HYPOTHESIS TESTINGSolution:
Null Hypothesis H0 :Alternative Hypothesis H1 :
Level of Significance: with df = n(1) + n(2) - 2Test Statistics:
t = sp =
Criterion: Reject the null hypothesis if tc > 1.68 for 24 + 20 2 = 42 degrees of freedom;otherwise state the difference between two sample means is not significant.
Compute:
sp = t =
Decision: Since tc = 1.684, the null hypothesis must be accepted; in other words, weconclude that the female students of the first section are not taller than the other class.The difference in the mean heights is not significant.
21 21
05.0
21
21
11
nns
xx
p
2
11
21
2
22
2
11
nn
snsn
31.622024
)5.5(120)9.6(12422
675.1
20
1
24
1
31.6
3.1605.163
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CORRELATION
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CORRELATION
Definition:
Correlation is a method used to measure thestrength of relationship between two variables that
tend to vary together in a consistent way. The natureand degree of relationship is indicated by a coefficient,designated by letter r.
By direct causal relations, we mean that ifXand Y
are correlated, then X is partly the cause of Y or Y ispartly the cause ofX.
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CORRELATION
Examples of correlation: There is correlation between
- income and savings
- the extent of fatigue and performance
in speed test.
There is no correlation between- weight and IQ, or
- shoe size and mathematical ability
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CORRELATION
The Scatter Diagram
One can usually and roughly estimate if arelationship exists between two variablesby constructing a scatter diagram. This isdone by plotting the point corresponding to
each observation on a rectangularcoordinate system.
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Scatter Plot Examples
y
x
y
x
y
y
x
x
Linear relationships Curvilinear relationships
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Scatter Plot Examples
y
x
y
x
y
y
x
x
Strong relationships Weak relationships
(continued)
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Scatter Plot Examples
y
x
y
x
No relationship
(continued)
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CORRELATION
Examples:
1. Consider the following marks of five students in Englishand Mathematics. Notice that for each student, there
corresponds two scores (paired observations).
Student English (X) Mathematics (Y)
A 55 69
B 64 85
C 96 99
D 44 52
E 83 89
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CORRELATION
2. The following data are the life spans of nine husbands and wivesrandomly selected from a certain community. Draw a scatter diagramand decide whether a relationship exists between their ages.
Couple Age of Husband (X) Age of Wife (Y)
1 65 902 72 95
3 68 45
4 71 51
5 75 50
6 67 627 76 45
8 73 63
9 71 83
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CORRELATION
Types of Correlation
1. Apositive correlation exists when high values in one variableare associated with high values in the second variable. This isalso true when low values in one variable are associated with low
values in the other. Thus, there is a direct relationship that existsin positive correlated variables. Also, in a positive correlation, thepoints on the scatter diagram closely follow a straight line risingto the right.
Examples:
problem solving ability and reading comprehensionincome and savings
income and expenses
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CORRELATION
Types of Correlation:
2.A negative correlation exists when high values
in one variable are associated with low valuesin the second variable, and vice versa. Here,points on the scatter diagram closely follow astraight line falling to the right.
Example:
pressure and volume (at constant temperature)
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CORRELATION
Types of Correlation
3. A zero correlation exists when scores in one variable
tend to score neither systematically high norsystematically low in the other variable. The points on
the scatter diagram are spread in a random mannerwhen this relationship exists.
Examples:
sex and IQ
athletic ability and mental ability
shoe size andmathematical performance
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CORRELATION
Note:
Correlational descriptions are descriptive and they
may not be sufficient to explain the relationshipbetween two variables.
Correlation coefficient (r) is a numerical measure
of the linear relationship between two variables. Itsvalues range from -1 to +1.
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Correlation Coefficient
The population correlation coefficient (rho) measures the strength of theassociation between the variables
The sample correlation coefficient r isan estimate of and is used to
measure the strength of the linearrelationship in the sampleobservations
(continued)
F f d
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Features of and rUnit free
Range between -1 and 1
The closer to -1, the stronger thenegative linear relationship
The closer to 1, the stronger the positivelinear relationship
The closer to 0, the weaker the linearrelationship
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r = +.3 r = +1
Examples of Approximate r Values
y
x
y
x
y
x
y
x
y
x
r = -1 r = -.6 r = 0
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CORRELATION
Correlational Tests:
1. Pearson Product Moment CorrelationIt measures the degree of relation between two at least
interval scale data.
2.Spearmans Rank Correlation Coefficient It is the measure of the correlation between two ordinal
variables.
3. Phi-CoefficientThe phi coefficient determines the degree of relationship
between two variables which are both nominal dichotomouslike sex (male-female) and marital status (married-unmarried).
4. Point BiserialIt measure correlation between an interval and a nominal
dichotomous data.
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CORRELATION
Interpretation of the Correlation CoefficientOnce the value of r is found significant, the rule of
thumb for assessing the degree of relationship betweenthe two quantitative variables can be interpreted using
the following criteria:r-value Verbal Description
0.00-0.29 Little or weak positive (negative) correlation
0.30-0.49 Low positive (negative) correlation
0.50-0.69 Moderate positive (negative) correlation
0.70-0.89 High positive (negative) correlation
0.90-1.00 Very High or strong positive (negative)
correlation
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CORRELATION
Test of significance for r
When ris calculated on the basis of sample data,
we may get a strong positive or negative correlationpurely by chance, even though there is actually nolinear relationship whatever between the two variablesin the population from which the sample came. The
value we obtain for r is only an estimate of acorresponding parameter, the population correlationcoefficient (). What r measures for a sample, measure s for a population.
CORRELATION
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CORRELATION
1. T-distributionwith n-2 degrees of freedom
This is used to test the significance of r arising fromPearson, Spearman, and Point Biserial.
Note: Reject the null hypothesis of no correlation at
the level of significance, if the computed value oft
exceeds the value of the critical t for one-tailed test orfor a two-tailed test; otherwise we accept the nullhypothesis.
21
2
r
nrt
CORRELATION
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CORRELATION
2. The Inference about the phicoefficient uses
1 nrZ
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CORRELATION
NOTE:
The coefficient of determination, the
square of the coefficient of correlation, r2,is the proportion of the total variation in thedependent variable (y) that can be
attributed to the relationship with theindependent variable (x).
C l l i h C l i C ffi i t
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Calculating the Correlation Coefficient
])yy(][)xx([
)yy)(xx(r
22
where:r = Sample correlation coefficientn = Sample sizex = Value of the independent variabley = Value of the dependent variable
])y()y(n][)x()x(n[
yxxynr
2222
Sample correlation coefficient:
or the algebraic equivalent:
Sample Calculation
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Sample CalculationTree
Height
TrunkDiameter
y X xy y2 x2
35 8 280 1225 64
49 9 441 2401 81
27 7 189 729 49
33 6 198 1089 36
60 13 780 3600 169
21 7 147 441 49
45 11 495 2025 121
51 12 612 2601 144
=321 =73 =3142 =14111 =713
l l l
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0
10
20
30
40
50
60
70
0 2 4 6 8 10 12 14
0.886
](321)][8(14111)(73)[8(713)
(73)(321)8(3142)
]y)()y][n(x)()x[n(
yxxynr
22
2222
Trunk Diameter, x
TreeHeight,y
Sample Calculation(continued)
r = 0.886 strong high positivelinear association between x and y
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Significance Test for Correlation
Hypotheses
H0: = 0 (no correlation)
HA: 0 (correlation exists)
Test statistic
(with n 2 degrees of freedom)
2nr1
rt
2
Solution
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Solution
Is there evidence of a linear relationshipbetween tree height and trunk diameter atthe 0.05 level of significance?
H0:
= 0 (No correlation)
H1: 0 (correlation exists)
= 0.05 , df=8 - 2 = 6
4.68
28
.8861
0.886
2n
r1
rt
22
Solution
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4.68
28
.8861
.886
2n
r1
rt
22
Solution
Conclusion:There is
evidence of alinear relationshipat the 5% level of
significance
Decision:Reject H0
Reject H0Reject H0
/2=.025
-t/2Do not reject H0
0t/2
/2=.025
-2.4469 2.44694.68
d.f. = 8-2 = 6
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Thank You for Listening!
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