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OutlineState what is estimated
Point estimation
Interval estimation
Compute sample size
Tests of Hypotheses
Null Hypothesis and Tests of Hypotheses
Hypotheses concern one mean
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Why Statistics?The purpose of most statistical investigations is to generalize from information contained in random samples about the populations from which the samples were obtained.
How: estimation and tests of Hypothesis
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ThinkingSuppose you’re interested in the average amount of money that students in this class (the population) have on them. How would you find out?
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Introduction to Estimation
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Statistical Methods
StatisticalMethods
DescriptiveStatistics
InferentialStatistics
EstimationHypothesis
Testing
StatisticalMethods
DescriptiveStatistics
InferentialStatistics
EstimationHypothesis
Testing
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Estimation Process
Mean, , is unknown
PopulationPopulation Random SampleRandom SampleI am 95%
confident that is between
40 & 60.
Mean X= 50
Sample
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Estimation Methods
Estimation
PointEstimation
IntervalEstimation
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Point estimationWe want to know the mean of a population. However, it is unavailable.
Hence, we choose a sample data, and calculate from the choosing sample data. Then estimate that the population is also the same mean.
Point estimation: concerns the choosing of a statistic, that is, a single number calculated from sample data for which we have some expectation, or assurance, that is reasonably close the parameter it is supposed to estimate.
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Point Estimation1. Provides Single Value
Based on Observations from 1 Sample
2. Gives No Information about How Close Value Is to the Unknown Population Parameter
3. Example: Sample MeanX = 3 Is Point Estimate of Unknown Population Mean
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Point estimation of a meanParameter: Population mean
Data: A random sample
Estimator:
Estimate of standard error: S
n
X
1 2, , , nX X X
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EXScientists need to be able to detect small amounts of contaminants in the environment. Sample data is listed as follows:
2.4 2.9 2.7 2.6 2.9 2.0 2.8 2.2 2.4 2.4 2.0 2.5
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EX of point estimationCompute the point estimator and estimate its standard deviation (also called the estimated standard error of ).
Solution:
X
X
29.82.483
12x
2 2 22
2 1 1
( )75.08 (29.8) /12
0.097881 1 12 1
n n
i ii i
x x x nxs
n n
n
Hence the estimated standard deviation is 0.09788 /12 0.090s
n
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Bias or UnbiasQuestion: is the estimation good enough?
EX.In previous ex, as a check on the current capabilities, the measurements were made on test specimens spiked with a known concentration 1.25 ug/l of lead. That is the readings should average 1.25 if there is no background lead in the samples. There appears to be either a bias due to laboratory procedure or some lead already in the samples before they were spiked.
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Unbiased estimatorLet be the parameter of interest and
be a statistic.
A statistic is said to be an unbiased estimator, or its value an unbiased estimate, if and only if the mean of the sampling distribution of the estimator equals , whatever the value of .
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More efficient unbiased estimatorEstimator is not unique: for example: it can be shown that for a random sample of size n=2, the mean as well as the weighted mean where a, b are positive constants, are unbiased estimates of the mean of the population.
A statistics is said to be a more efficient unbiased estimator of the parameter than the statistics if
1. and are both unbiased estimators of 2. the variance of the sampling distribution of the first
estimator is no larger than that of the second and is smaller for at least one value of .
1 2
12
1 2
2
X X
1 2aX bX
a b
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Finding Sample Sizes
I don’t want to sample too much or too little!
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The error The error between the estimator and the quantity it is supposed to estimate is: X
/
X
n
is a random variable having approximately the standard normal distribution
We could assert with probability that the inequality
/ 2 / 2/
Xz z
n
1
/ 2( ) 1 / 2P Z z Remind that
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Maximum error of estimateThe error will be less than
with probability .1
| |X
/ 2E zn
Specially, / 2 1.96, 0.05z when
/ 2 2.575, 0.01z when
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EXAn industrial engineer intends to use the mean of a random sample of size n=150 to estimate the average mechanical aptitude of assembly line workers in a large industry. If, on the basis of experience, the engineer can assume that for such data, what can he assert with probability 0.99 about the maximum size of his error.
6.2
6.22.575 1.3
150E
Thus, the engineer can assert with probability 0.99 that his error is at most 1.3
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Determine of sample sizeSuppose that we want to use the mean of a large random sample to estimate the mean of a population, and want to be able to assert with probability that the error will be at most some prescribed quantity E. As before, we get
2/ 2( )z
nE
1
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EXA research worker want to determine the average time it takes a mechanic to rotate the tires of a car, and she wants to be able to assert with 95% confidence that the mean of her sample is off by at most 0.5 minute. If she can presume from past experience that minutes, how large a sample will she have to take?
1.6
21.96 1.6( ) 39.3
0.5n
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ContinuousThe method discussed so far requires be known or it can be approximated with the sample standard deviation s, thus requiring that n be large. Another approach: if it is reasonable to assume that we are sampling from a normal population, we get
is a random variable having the t distribution with n-1 degree of freedom.
/
Xt
S n
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EXThe first example, n=12, 0.09788 /12 0.090
s
n
0.01 2.718t For n=11 degrees of freedom
0.01 2.718 0.09 0.2s
tn
2.483 1.25 1.233 0.2X
Thus, one can assert with 98% confidence that the maximum error is within 0.2
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Estimation Methods
Estimation
PointEstimation
IntervalEstimation
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Interval Estimation
1. Provides Range of Values Based on Observations from 1 Sample
2. Gives Information about Closeness to Unknown Population Parameter
Stated in terms of Probability
3. Example: Unknown Population Mean Lies Between 50 & 70 with 95% Confidence
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Interval Estimation
Sample statistic Sample statistic
(point estimate)(point estimate)
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Interval Estimation
Confidence Confidence intervalinterval
Sample statistic Sample statistic
(point estimate)(point estimate)
Confidence Confidence limit (lower)limit (lower)
Confidence Confidence limit (upper)limit (upper)
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Interval Estimation
Confidence Confidence intervalinterval
Sample statistic Sample statistic
(point estimate)(point estimate)
Confidence Confidence limit (lower)limit (lower)
Confidence Confidence limit (upper)limit (upper)
A A probabilityprobability that the population parameter that the population parameter falls somewhere within the interval.falls somewhere within the interval.
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Confidence Interval Estimates
ConfidenceIntervals
ProportionMean Variance
UnknownKnown
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Confidence Interval Mean ( Known)
AssumptionsPopulation standard deviation is known
Population is normally distributed
If not normal, can be approximated by normal distribution (n 30)
Confidence Interval Estimate
/ 2 / 2X Z X Zn n
/ 2 / 2X Z X Zn n
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Interval EstimationInterval estimation: with intervals for which we can assert with a reasonable degree of certainty that they will contain the parameter under consideration.
For a large random sample (n > 30) from a population with the unknown mean and the known variance. When the observed value becomes available, we obtainx
/ 2 / 2x z x zn n
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Confidence intervalWe can claim with confidence that the interval
(1 )100%
/ 2 / 2[ , ]x z x zn n
Contains
It is customary to refer to an interval of this kind as a confidence interval for having the degree of confidence
(1 )100%
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Confidence Interval Estimates
ConfidenceIntervals
ProportionMean Variance
UnknownKnown
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Solution for Small Samples1. Assumptions
Population of X Is Normally Distributed
Use Student’s t Distribution1.Define variable
2.T has the Student distribution with n -1 degrees of freedom (When X is normally distributed)
• There’s a different Student distribution for different degrees of freedom
• As n gets large, Student distribution approximates a normal distribution with mean = 0 and sigma = 1
/
XT
s n
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Small sample (n<30)Small sample and we assume to get sampling from a normal distribution population.
We get the confidence interval formula
(1 )100%
/ 2 / 2
s sx t x t
n n
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EX.The mean weight loss of n=16 grinding balls after a certain length of time in mill slurry is 3.42 grams with a standard deviation of 0.68 gram. Construct a 99% confidence interval for the true mean weight loss of such grinding balls under the stated condition.
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Confidence Depends on Interval (z)
90% Samples90% Samples
95% Samples95% Samples
99% Samples99% Samples
+1.65+1.65x x +2.58+2.58xx
xx__
XX
+1.96+1.96xx
-2.58-2.58xx -1.65-1.65xx
-1.96-1.96xx
XX= = ± Z ± Zxx
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1. Probability that the Unknown Population Parameter Falls Within Interval
2. Denoted (1 - Is Probability That Parameter Is Not Within Interval
3. Typical Values Are 99%, 95%, 90%
Confidence Level
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Intervals & Confidence Level
x =
1 - /2/2
X_
x_
x =
1 - /2/2
X_
x_Sampling Sampling
Distribution Distribution of Meanof Mean
Intervals derived from Intervals derived from many samplesmany samples
Intervals Intervals extend from extend from X - ZX - ZXX to to
X + ZX + ZXX
(1 - (1 - ) % of ) % of intervals intervals contain contain . .
% do not.% do not.
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Confidence interval & levelIt is useful to think confidence intervals as a range of "plausible" values for the parameter. confidence interval is different from interpreting the confidence level
suppose we've taken a random sample of 10 ice-cream cones, and determined that a 95% confidence interval for the mean caloric contents of a single scoop of ice-cream is (260,310). Interpret the confidence level: If we repeatedly took samples of size 10 and then formed confidence intervals, we would expect 95% of them to contain the true (but unknown) mean. Interpret this particular confidence interval: we are 95% confident that the true mean caloric content lies between 260 and 310.
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Confidence interval & levelThe wider the confidence interval you are willing to accept, the more certain you can be that the whole population answers would be within that range. For example, if you asked a sample of 1000 people in a city which brand of cola they preferred, and 60% said Brand A, you can be very certain that between 40 and 80% of all the people in the city actually do prefer that brand, but you cannot be so sure that between 59 and 61% of the people in the city prefer the brand.
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1. Data DispersionMeasured by
2. Sample Size
3. Level of Confidence (1 - )
Affects Z
Factors Affecting Interval Width
Intervals Extend fromIntervals Extend from
X - ZX - ZXX to toX + ZX + ZXX
© 1984-1994 T/Maker Co.
/x n
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Statistical Methods
StatisticalMethods
DescriptiveStatistics
InferentialStatistics
EstimationHypothesis
Testing
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Hypothesis Testing
PopulationPopulation
I believe the population mean age is 50 (hypothesis).
I believe the population mean age is 50 (hypothesis).
Reject hypothesis! Not close.
Reject hypothesis! Not close.
MeanMean
X X = 20= 20
Random Random samplesample
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What’s a Hypothesis?
A belief about a population parameter
Parameter is Population mean, proportion, variance
Must be statedbefore analysis
I believe the mean GPA I believe the mean GPA of this class is 3.8!of this class is 3.8!
© 1984-1994 T/Maker Co.
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Tests of HypothesesSuppose that a consumer protection agency wants to test a paint manufacturer’s claim that the average drying time of his new “fast-drying” paint is 20 minutes. It instructs a member of its research staff to paint each of 36 boards using a different 1-gallon can of the paint, with the intention of rejecting the claim if the mean of the drying time exceeds 20.75 minutes. Otherwise, it will accept the claim.Question: Is it a infallible criterion for accepting or rejecting the claim?
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EX cont.Assuming that it is known from past experience that the
standard deviation 2.4
Let us investigate the possibility that the sample may exceed 20.75 minutes even though the true mean is 20 minutes
20 20.75
x Minutes
Reject the claim
that 20 Accept the claim
that 20
0.0304
20.75 201.875
2.4 / 36z
The probability of erroneously rejecting the hypothesis that 20
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Another possibilityThe procedure fails to detect that 20
21 20.75
0.2660
20
Accept the claim
that
Reject
20
the claim
that
20.75 210.625
2.4 / 36z
Suppose that the true mean of drying time is 21
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Type of errorsH: the hypothesis. Ex.
Accept H Reject H
H is true Correct decision Type I error
H is false Type II error Correct decision
20
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Type I error: If hypotheses is true but rejected. Denoted by the letter
EX.
Type II error: If hypotheses is false but not rejected. Denoted by the letter
EX.
0.0304
0.2660
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7.4 Null HypothesesQuestion: Can we formulate minutes, where can take on more than one possible value?
Null Hypotheses : (Pronounced H-nought) Usually require that we hypothesize the opposite of what we hope to prove.
EX. If we want to show that one method of teaching computer programming is more efficient than another, we hypothesize that the two methods are equally effective.
The null hypothesis proposes something initially presumed true. It is rejected only when it becomes evidently false.
20
0H
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Null Hypothesis
1. What is tested
2. has serious outcome if incorrect decision made
3. Designated H0
4. Specified as H0: Some Numeric Value
Specified with = Sign Even if , or Example, H0: 3
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Alternative Hypothesis
1. Opposite of Null Hypothesis
2. Always Has Inequality Sign: ,, or
3. Designated Ha
4. Specified Ha: < Some Value
Example, Ha: < 3
will lead to two-sided tests
<, > will lead to one-sided tests
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AlternativeOne-sided alternative: In the drying time example, the null hypothesis is minutes and the alternative hypothesis is
Two-sided alternative: where is the value assumed under the null hypothesis.
20
20
0 0
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Selecting the null hypothesis Guideline for selecting the null hypothesis: When the goal of an experiment is to establish an assertion, the negation of the assertion should be taken as the null hypothesis. The assertion becomes the alternative hypothesis.
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Identifying Hypotheses Steps
1. Example Problem: Test That the Population Mean Is Not 3
2. StepsState the Question Statistically ( 3)
State the Opposite Statistically ( = 3)Must Be Mutually Exclusive & Exhaustive
Select the Alternative Hypothesis ( 3)Has the , <, or > Sign
State the Null Hypothesis ( = 3)
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State the question statistically: = 12
State the opposite statistically: 12
Select the alternative hypothesis: Ha: 12
State the null hypothesis: H0: = 12
Is the population average amount of TV Is the population average amount of TV viewing 12 hours?viewing 12 hours?
What Are the Hypotheses?
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State the question statistically: 12
State the opposite statistically: = 12
Select the alternative hypothesis: Ha: 12
State the null hypothesis: H0: = 12
Is the population average amount of TV Is the population average amount of TV viewing viewing differentdifferent from 12 hours? from 12 hours?
What Are the Hypotheses?
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State the question statistically: 20
State the opposite statistically: 20
Select the alternative hypothesis: Ha: 20
State the null hypothesis: H0: 20
Is the average cost per hat less than or Is the average cost per hat less than or equal to $20?equal to $20?
What Are the Hypotheses?
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State the question statistically: 25
State the opposite statistically: 25
Select the alternative hypothesis: Ha: 25
State the null hypothesis: H0: 25
Is the average amount spent in the Is the average amount spent in the bookstore greater than $25?bookstore greater than $25?
What Are the Hypotheses?
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Decision Making Risks
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Errors in making decision1. Type I Error
Reject True Null HypothesisHas Serious ConsequencesProbability of Type I Error Is (Alpha)
Called Level of Significance
2. Type II ErrorDo Not Reject False Null HypothesisProbability of Type II Error Is (Beta)
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Jury Trial H0 Test
Actual Situation Actual Situation
Verdict Innocent Guilty Decision H0 True H0 False
Innocent Correct Error Do Not Reject
H0 1 - Type II
Error ()
Guilty Error Correct Reject H0
Type I Error ()
Power (1 - )
Jury Trial H0 Test
Actual Situation Actual Situation
Verdict Innocent Guilty Decision H0 True H0 False
Innocent Correct Error Do Not Reject
H0 1 - Type II
Error ()
Guilty Error Correct Reject H0
Type I Error ()
Power (1 - )
Decision ResultsHH00: Innocent: Innocent
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Jury Trial H0 Test
Actual Situation Actual Situation
Verdict Innocent Guilty Decision H0 True H0 False
Innocent Correct Error
Accept H0
1 - Type II Error
()
Guilty Error Correct Reject H0
Type I Error ()
Power (1 - )
Jury Trial H0 Test
Actual Situation Actual Situation
Verdict Innocent Guilty Decision H0 True H0 False
Innocent Correct Error
Accept H0
1 - Type II Error
()
Guilty Error Correct Reject H0
Type I Error ()
Power (1 - )
Decision ResultsHH00: Innocent: Innocent
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& have an inverse relationship
You can’t reduce both errors simultaneously!
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Factors Affecting 1. True Value of Population Parameter
Increases When Difference With Hypothesized Parameter Decreases
2. Significance Level, Increases When Decreases
3. Population Standard Deviation, Increases When Increases
4. Sample Size, n
Increases When n Decreases
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Hypothesis Testing
PopulationPopulation
I believe the population mean age is 50 (hypothesis).
I believe the population mean age is 50 (hypothesis).
Reject hypothesis! Not close.
Reject hypothesis! Not close.
MeanMean
X X = 20= 20
Random Random samplesample
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Basic Idea
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Basic Idea
Sample Mean = 50 Sample Mean = 50
HH00HH00
Sampling DistributionSampling Distribution
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Basic Idea
Sample Mean = 50 Sample Mean = 50
Sampling DistributionSampling Distribution
It is unlikely It is unlikely that we would that we would get a sample get a sample mean of this mean of this value ...value ...
20202020HH00HH00
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Basic Idea
Sample Mean = 50 Sample Mean = 50
Sampling DistributionSampling Distribution
It is unlikely It is unlikely that we would that we would get a sample get a sample mean of this mean of this value ...value ...
... if in fact this were... if in fact this were the population mean the population mean
20202020HH00HH00
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Basic Idea
Sample Mean = 50 Sample Mean = 50
Sampling DistributionSampling Distribution
It is unlikely It is unlikely that we would that we would get a sample get a sample mean of this mean of this value ...value ...
... if in fact this were... if in fact this were the population mean the population mean
... therefore, ... therefore, we reject the we reject the hypothesis hypothesis
that that = 50.= 50.
20202020HH00HH00
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Hypothesis testing1. We formulate a null hypothesis and an appropriate alternative hypothesis which we accept when the null hypothesis must be rejected.
2. We specify the probability of a Type I error. If possible, desired, or necessary, we may also specify the probabilities of Type II errors for particular alternatives.
3. Based on the sampling distribution of an appropriate statistic, we construct a criterion for testing the null hypothesis against the given alternative.
4. We calculate from the data the value of the statistic on which the decision is to be based.
5. We decide whether to reject the null hypothesis or whether to fail to reject it.
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Level of Significance1. Probability
2. Defines Unlikely Values of Sample Statistic if Null Hypothesis Is True
Called Rejection Region of Sampling Distribution
3. Designated (alpha)
4. Selected by Researcher at Start
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Level of significanceThe probability of a Type I errorType I error is also called the level of significance. Usually, we set .
Step 2 can often be performed even when the null hypothesis specifies a range of values for the parameter.
Ex. The null hypothesis:Then we can claim that
0.05 0.01or
20
0.0304
In general, we can only specify the maximum probability of a Type I error, and by again the worst possibility.
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Criterion One-sided criterion (one-sided test or one-tailed test): ex. One-sided alternative
Two-sided criterion (two-sided test or two-tailed test): ex. two-sided alternative
20
4
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Rejection Region (One-Tail Test)
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Sampling DistributionSampling Distribution
1 - 1 -
Level of ConfidenceLevel of Confidence
Rejection Region (One-Tail Test)
H0
valueCriticalvalue
Sample Statistic
RejectionRegion
NonrejectionRegion
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HoValueCritical
Value
Sample Statistic
RejectionRegion
NonrejectionRegion
HoValueCritical
Value
Sample Statistic
RejectionRegion
NonrejectionRegion
Sampling DistributionSampling Distribution
1 - 1 -
Level of ConfidenceLevel of Confidence
Observed sample statisticObserved sample statistic
Rejection Region (One-Tail Test)
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Rejection Region (One-Tail Test)
HoValueCritical
Value
Sample Statistic
RejectionRegion
NonrejectionRegion
HoValueCritical
Value
Sample Statistic
RejectionRegion
NonrejectionRegion
Sampling DistributionSampling Distribution
1 - 1 -
Level of ConfidenceLevel of Confidence
Observed sample statisticObserved sample statistic
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Rejection Region (Two-Tailed Test)
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Sampling DistributionSampling Distribution
1 - 1 -
Level of ConfidenceLevel of Confidence
Rejection Region (Two-Tailed Test)
HoValue Critical
ValueCriticalValue
1/2 1/2
Sample Statistic
RejectionRegion
RejectionRegion
NonrejectionRegion
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Sampling DistributionSampling Distribution
1 - 1 -
Level of ConfidenceLevel of Confidence
Rejection Region (Two-Tailed Test)
HoValue Critical
ValueCriticalValue
1/2 1/2
Sample Statistic
RejectionRegion
RejectionRegion
NonrejectionRegion
Observed sample statisticObserved sample statistic
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Sampling DistributionSampling Distribution
1 - 1 -
Level of ConfidenceLevel of Confidence
Rejection Region (Two-Tailed Test)
HoValue Critical
ValueCriticalValue
1/2 1/2
Sample Statistic
RejectionRegion
RejectionRegion
NonrejectionRegion
Observed sample statisticObserved sample statistic
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Sampling DistributionSampling Distribution
1 - 1 -
Level of ConfidenceLevel of Confidence
Rejection Region (Two-Tailed Test)
HoValue Critical
ValueCriticalValue
1/2 1/2
Sample Statistic
RejectionRegion
RejectionRegion
NonrejectionRegion
Observed sample statisticObserved sample statistic
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H0 Testing Steps
Set up critical values
Collect data
Compute test statistic
Make statistical decision
Express decision
State HState H00
State HState Haa
Choose Choose
Choose Choose nn
Choose testChoose test
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One Population Tests
OnePopulation
Z Test(1 & 2tail)
t Test(1 & 2tail)
Z Test(1 & 2tail)
Mean Proportion Variance
2 Test(1 & 2tail)
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7.5 Hypothesis concerning one mean
Suppose we want to test on the basis of n=35 determinations and at the 0.05 level of significance whether the thermal conductivity of a certain kind of cement brick is 0.340, as has been claimed. We can expect that the variability of such determinations is given by 0.010
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Solution1) Null hypothesis:
alternative hypothesis:
2) Level of significance:
3) Criterion:
0.340
/
XZ
n
z
0.05
0.340
/ 2z/ 2z
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Criterion Region for testing (Normal population and known)
Alternative hypothesis
Reject null hypothesis if
0
0
0
0
Z z
Z z
/ 2 / 2Z z or Z z
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Critical values
One-sided alternatives
Two-sided alternatives
-1.645
1.645
-1.96
1.96
-2.33
2.33
-2.575
2.575
0.05
0.01
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EX. Cont.3) Criterion: Reject the null hypothesis if
1.96 1.96Z or Z
/
XZ
n
where
4) Calculations:0.343 0.34
1.770.01/ 35
z
5) Decision: The null hypothesis cannot be rejected.
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P-valueP value (or tail probability): the probability of getting difference between and
greater than or equal to that actually observed.
EX. In above example
1.771.77
( )P Z z
x 0
2(1 0.9616) 0.078
½ P-value½ P-value
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P-Value1. Probability of Obtaining a Test Statistic More Extreme (or than Actual Sample Value Given H0 Is True
2. Called Observed Level of SignificanceSmallest Value of H0 Can Be Rejected
3. Used to Make Rejection DecisionIf p-Value , Do Not Reject H0
If p-Value < , Reject H0
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P valueP value for a given test statistic and null hypothesis: The P value is the probability of obtaining a value for the test statistic that is as extreme or more extreme than the value actually observed. Probability is calculated under the null hypothesis.
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EXA process for producing vinyl floor covering has been stable for a long period of time, and the surface hardness measurement of the flooring produced has a normal distribution with mean 4.5 and standard deviation 1.5. A second shift has been hired and trained and their production needs to be monitored. Consider testing the hypothesis
0 1: 4.5 : 4.5H versus H
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A random sample of hardness measurements is made of n=25 vinyl specimens produced by the second shift. Calculate the P value when using the test statistic
If
/
XZ
n
3.9X
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SolutionThe observed value of the test statistic is
3.9 4.52.0
1.5 / 25z
Since the alternative hypothesis is two-sided, we must consider large negative value for Z as well as large positive values
( 2.0) ( 2.0) 0.0228P Z P Z
0.0456Consequently, the P value is
0.05 The small P value suggests the mean of the second shift is not at the target value of 4.5
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P valueTo understand P values, you have to understand fixed level testing.
With fixed level testing, a null hypothesis is proposed (usually, specifying no treatment effect) along with a level for the test, usually 0.05. All possible outcomes of the experiment are listed in order to identify extreme outcomes that would occur less than 5% of the time in aggregate if the null hypothesis were true.
This set of values is known as the critical region. They are critical because if any of them are observed, something extreme has occurred. Data are now collected and if any one of those extreme outcomes occur the results are said to be significant at the 0.05 level. The null hypothesis is rejected at the 0.05 level of significance and one star (*) is printed somewhere in a table. Some investigators note extreme outcomes that would occur less than 1% of the time and print two stars (**) if any of those are observed.
Many researchers quickly realized the limitations of reporting only whether a result achieved the 0.05 level of significance. Was a result just barely significant or wildly so? Would data that were significant at the 0.05 level be significant at the 0.01 level? At the 0.001 level? Even if the result are wildly statistically significant, is the effect large enough to be of any practical importance?
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P valueObserved significance level (or P value)--the smallest fixed level at which the null hypothesis can be rejected. If your personal fixed level is greater than or equal to the P value, you would reject the null hypothesis.
If your personal fixed level is less than to the P value, you would fail to reject the null hypothesis.
For example, if a P value is 0.027, the results are significant for all fixed levels greater than 0.027 (such as 0.05) and not significant for all fixed levels less than 0.027 (such as 0.01). A person who uses the 0.05 level would reject the null hypothesis while a person who uses the 0.01 level would fail to reject it.
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P-Value Thinking Challenge
You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)?
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p-Value Solution*
Z0-2.65
p-Value.004
Z0-2.65
p-Value.004
Z value of Z value of sample statisticsample statistic
From Z table: From Z table: lookup 2.65lookup 2.65
.4960.4960
Use Use alternative alternative hypothesis hypothesis to find to find directiondirection
.5000.5000-- .4960.4960
.0040.0040
p-Value is P(Z p-Value is P(Z -2.65) = .004. -2.65) = .004.p-Value < (p-Value < ( = .01). Reject H = .01). Reject H00..
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One Population Tests
OnePopulation
Z Test(1 & 2tail)
t Test(1 & 2tail)
Z Test(1 & 2tail)
Mean Proportion Variance
2 Test(1 & 2tail)
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t Test for Mean ( Unknown)1. Assumptions
Population Is Normally Distributed
If Not Normal, Only Slightly Skewed & Large Sample (n 30) Taken
2. Parametric Test Procedure
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Statistic for small sample
/
XT
S n
The test of null hypothesis on the statistic
0
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Criterion Region for testing (Statistic for small sample )
Alternative hypothesis
Reject null hypothesis if
0
0
0
0
T t
T t
/ 2 / 2T t or T t
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t0 t0
Two-Tailed t TestFinding Critical t Values
Given: n = 3; Given: n = 3; = .10 = .10
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t0 t0
/2 = .05/2 = .05
/2 = .05/2 = .05
Given: n = 3; Given: n = 3; = .10 = .10
Two-Tailed t TestFinding Critical t Values
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t0 t0
/2 = .05/2 = .05
/2 = .05/2 = .05
Given: n = 3; Given: n = 3; = .10 = .10
df = n - 1 = 2df = n - 1 = 2
Two-Tailed t TestFinding Critical t Values
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v t.10 t.05 t.025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182
v t.10 t.05 t.025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182t0 t0
Critical Values of t Table Critical Values of t Table (Portion)(Portion)
/2 = /2 = .05.05
/2 = .05/2 = .05
Given: n = 3; Given: n = 3; = .10 = .10
df = n - 1 = df = n - 1 = 22
Two-Tailed t TestFinding Critical t Values
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v t.10 t.05 t.025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182
v t.10 t.05 t.025
1 3.078 6.314 12.706
2 1.886 2.920 4.303
3 1.638 2.353 3.182t0 2.920-2.920 t0 2.920-2.920
Critical Values of t Table Critical Values of t Table (Portion)(Portion)
/2 = .05/2 = .05
/2 = .05/2 = .05
Given: n = 3; Given: n = 3; = .10 = .10
df = n - 1 = 2df = n - 1 = 2
Two-Tailed t TestFinding Critical t Values
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One-Tailed t TestYou’re a marketing analyst
for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3. At the .05 level, is there evidence that the average bear sales per store is more than 5 ($ 00)?
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One-Tailed t Test Solution*
H0:
Ha:
=
df =
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
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One-Tailed t Test Solution*
H0: = 5
Ha: > 5
=
df =
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
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One-Tailed t Test Solution*
H0: = 5
Ha: > 5
= .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
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117/139t0 1.8331
.05
Reject
t0 1.8331
.05
Reject
One-Tailed t Test Solution*
H0: = 5
Ha: > 5
= .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
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118/139t0 1.8331
.05
Reject
t0 1.8331
.05
Reject
One-Tailed t Test Solution*
H0: = 5
Ha: > 5
= .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
tX
Sn
6 4 5
3 37310
131..
.tX
Sn
6 4 5
3 37310
131..
.
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119/139t0 1.8331
.05
Reject
t0 1.8331
.05
Reject
One-Tailed t Test Solution*
H0: = 5
Ha: > 5
= .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
tX
Sn
6 4 5
3 37310
131..
.tX
Sn
6 4 5
3 37310
131..
.
Do not reject at Do not reject at = .05 = .05
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120/139t0 1.8331
.05
Reject
t0 1.8331
.05
Reject
One-Tailed t Test Solution*
H0: = 5
Ha: > 5
= .05
df = 10 - 1 = 9
Critical Value(s):
Test Statistic: Test Statistic:
Decision:Decision:
Conclusion:Conclusion:
tX
Sn
6 4 5
3 37310
131..
.tX
Sn
6 4 5
3 37310
131..
.
Do not reject at Do not reject at = .05 = .05
There is no evidence There is no evidence average is more than 5average is more than 5