2INTRODUCTIONOne of the ways of estimating the true value of population parameters is to test the validity of the claim(assertion or statement) made about this value using simple statisticse.g. a pharmaceutical company claims the efficacy of a medicine against a disease that 95 % of all persons suffering from the disease get cured
3INTRODUCTIONTo test such claims sample data is collected and analyzed. On the basis of sample findings the hypothesized value of population parameter is either accepted or rejectedThe process that enables a decision maker to test the validity (or significance ) of the claim by analyzing the difference between the value of sample statistic and the corresponding hypothesized population parameter value is called hypothesis testing
4HypothesisHypotheses are tentative, intelligent guesses as to the solution of the problem.A hypothesis is an assumption about the population parameter.A parameter is a characteristic of the population, like its mean or variance.Goal:Make statement(s) regarding unknown population parameter values based on sample data
5Types of Hypotheses Null that no statistically significant difference exists between the parameter and the statistic being comparedAlternativelogical opposite of the null hypothesisthat a statistically significant difference does exist between the parameter and the statistic being compared.H0 always contains =HA matches the question asked; it is what we’re trying to prove.
6Various types of H0 and HA Case Null Hypothesis Alternate Hypothesis H 0 H A 1. µ = µ0 µ ≠ µ0 2 . µ ≤ µ0 µ > µ0 3 . µ ≥ µ0 µ < µ0 µ0 is Hypothesised mean
7Always the null hypothesis is tested, i. e Always the null hypothesis is tested, i.e., we want to either accept or reject the null hypothesis because we have information only for the null hypothesis.Ho is taken as true and tested for its possible rejectionEither H0 or HA must be true, making the other false.H0 and H1 are:Mutually exclusiveOnly one can be true.ExhaustiveTogether they cover all possibilities, so one or the other must be true.
8FORMATS OF HYPOTHESIS Directional Hypothesis. A hypothesis is a statement to be tested about the true value of population parameterA hypothesis whether there exists any significant difference between two or more populations with respect to any of their common parameter can also be tested. In this case the hypothesis can be stated in the form if- then statement .ConsiderIf inflation rate has increased then wholesale index will also increaseIf employees are healthy then they will take less sick leave
9FORMATS OF HYPOTHESIS Directional Hypothesis. If terms such as positive, negative , more than etc are used to make a statement , such a hypothesis is called directional hypothesis as it indicates the direction of the relationshipNon Directional Hypothesis.It indicates a relationship but offers no direction of relationship ( or difference)e.g.There is a relationship between age and job satisfactionThere is difference between average pulse rate of men and women
10Hypothesis about other Parameters Hypotheses about other parameters such as population proportions and and population variances are also possible. For exampleH0: p %H1: p < 40%H0: sH1: s2 > 50
11RATIONALE FOR HYPOTHESIS TESTING Since statistics are random values , therefore their sampling distributions show the tendency of variationConsequently we do not expect the sample statistic value to be equal to hypothesized valueIf value of sample statistic differ significantly from the hypothesized parameter value then the question arises whether the hypothesized parameter value is correct or not
12RATIONALE FOR HYPOTHESIS TESTING The probability level at which the decision maker concludes that observed difference between the value of the test statistic and hypothesized value cannot be due to chance is called the level of significance
13PROCEDURE FOR HYPOTHESIS TESTING STEP 1 –STATE THE NULL HYPOTHESIS (Ho) AND ALTERNATIVE HYPOTHESIS(H1)Null hypothesis Ho represents the claim or statement made about the value or range of value of the population parameter. The letter H stands for hypothesis and zero implies no difference between sample statistic and parameter value
14PROCEDURE FOR HYPOTHESIS TESTING Null hypothesis is always expressed in the form of mathematical statement which includes (≤, ≥,=)making a claim regarding the specific value of the population parameterHo: μ (≤, ≥,=) μo where μ is population mean and μo represents a hypothesized value of μAlternate hypothesis H1 is the counter claim made against the value of particular population parameter. That is alternate hypothesis must be true when null hypothesis is found to be false.
15PROCEDURE FOR HYPOTHESIS TESTING In other words alternative hypothesis states that specific population parameter is not equal to the value stated in the null hypothesis H1:μ ≠ μo Consequently H1 :μ<μo or μ>μo
16ERRORS IN HYPOTHESIS TESTING Ideally the hypothesis testing procedure should lead to acceptance of the null hypothesis Ho when it is true and the rejection of Ho when it is not trueHowever the correct decision is not always possibleSince the decision to reject or accept a hypothesis is based on sample date there is a possibility of an incorrect decision or error
17ERRORS IN HYPOTHESIS TESTING There are two types of errorSTATE OF NATURESample decisionHo is trueHo is falseAccept HoCorrect Decision=1- Type II error.P(Type II)=Reject HoType I error.p(Type I)= Correct DecisionPower=1-
18ERRORS IN HYPOTHESIS TESTING TYPE 1 ERRORThis is the probability of rejecting the null hypothesis when it is true and some alternative hypothesis is wrongThe probability of making a Type I error is denoted by the symbol αIt is represented by the area under the sampling distribution curve over the region of rejectionThe probability of making a Type I error is referred to as level of significanceThe complement 1-α of the probability of Type I error measure the probability level of not rejecting a true null hypothesis. It is also referred to as confidence level
19ERRORS IN HYPOTHESIS TESTING TYPE II ERRORThis is the probability of accepting the null hypothesis when it is false and some alternative hypothesis is trueThe probability of Type II error varies with the actual values of the population parameter being tested when null hypothesis Ho is false
20The Power of a TestThe power of a statistical hypothesis test is the probability of rejecting the null hypothesis when the null hypothesis is false.Power = (1 - )
21The Power FunctionThe probability of a type II error, and the power of a test, depends on the actual value of the unknown population parameter. The relationship between the population mean and the power of the test is called the power function.Value of m b Power = (1 - b)769854321.PowerfaOn-Tildst:=,Power
22Factors Affecting the Power Function The power depends on the distance between the value of the parameter under the null hypothesis and the true value of the parameter in question: the greater this distance, the greater the power.The power depends on the population standard deviation: the smaller the population standard deviation, the greater the power.The power depends on the sample size used: the larger the sample, the greater the power.The power depends on the level of significance of the test: the smaller the level of significance,, the smaller the power.
23To Note: Would like a and b to be as small as possible. a and b are inversely related.Usually set a (and don’t worry too much about b.)Most common values for a and b are 0.01 and 0.05.
24PROCEDURE FOR HYPOTHESIS TESTING STEP 2-STATE THE LEVEL OF SIGNIFICANCESelect an appropriate significance level(Type I error), it is denoted by α. It is the risk a decision maker takes of rejecting the null hypothesis when it is trueTraditionally α = 0.05 for consumer research projects, α = 0.01 for quality assurance and α=0.10 for political polling
25Significance level gives the confidence with which a researcher rejects or retain the null hypothesis. If we take 5% level of significance then there are about 5 chances out of 100 that we would reject the Null hypothesis when it should be accepted or that we are 95% confident about the right decision
26PROCEDURE FOR HYPOTHESIS TESTING STEP 3- ESTABLISH CRITICAL OR REJECTION REGIONThe area under the sampling distribution curve of test statistic is divided into two regions. These regions are called acceptance region and the rejection region ( critical region)
27Acceptance region and critical regions for two tailed test in hypothesis testing
28One-sided hypothesis where the critical region lies in the right tail
29One-sided hypothesis where the critical region lies in the left tail
30Given the same level of significance the two tailed test is more conservative, i.e., it is more rigorous than the one- tailed test because the rejection point is farther out in the tail. It is more difficult to reject H0 with a two-tailed test than with a one-tailed test.
311-Tailed and 2-Tailed Tests The tails of a statistical test are determined by the need for an action. If action is to be taken if a parameter is greater than some value a, then the alternative hypothesis is that the parameter is greater than a, and the test is a right-tailed test. H0: 50H1: 50If action is to be taken if a parameter is less than some value a, then the alternative hypothesis is that the parameter is less than a, and the test is a left-tailed test. H0: 50H1: 50If action is to be taken if a parameter is either greater than or less than some value a, then the alternative hypothesis is that the parameter is not equal to a, and the test is a two-tailed test. H0: 50H1: 50
32PROCEDURE FOR HYPOTHESIS TESTING STEP 4 –SELECT THE TEST OF SIGNIFICANCE OR TEST STATISTICThe tests of significance or test statistic are divided into two categories: parametric and non parametricParametric tests are more powerful because their data is derived from interval and ratio measurementsNon parametric tests are used to test hypothesis with nominal and ordinal data
33Univariate Techniques Parametric TestOne Sample•t test•Z testTwo or More SamplesIndependent•One way ANOVARelated•Paired t testNonparametric Test•Frequency•Chi-square•K-S•Runs•Binomial•Chi-Square•Median•Sign
34How to Select a Test Which does the test involve? one sample, two samplesk samplesIf two or k samples,are the individual cases independent or related?Is the measurement scale nominal, ordinal, interval, or ratio?
35Testing Population Means Cases in which the test statistic is Zs is known and the population is normal.s is known and the sample size is at least 30. (The population need not be normal)
36Testing Population Means Cases in which the test statistic is ts is unknown but the sample standard deviation is known and the population is normal.
37PROCEDURE FOR HYPOTHESIS TESTING STEP 5 – FORMULATE A DECISION RULE TO ACCEPT NULL HYPOTHESISCompare the calculated value of test statistic with the critical value ( also called standard value of test statistic). The decision rule for null hypothesis are as followsAccept Ho if the test statistic value falls within the area of acceptanceReject otherwise
38PROCEDURE FOR HYPOTHESIS TESTING The p-Value approachThe p-value is the probability of obtaining a value of the test statistic as extreme as, or more extreme than, the actual value obtained, when the null hypothesis is true.The p-value is the smallest level of significance, , at which the null hypothesis may be rejected using the obtained value of the test statistic.Policy: When the p-value is less than a , reject H0.
40EXAMPLE 1An automatic bottling machine fills cola into two liter (2000 cc) bottles. A consumer advocate wants to test the null hypothesis that the average amount filled by the machine into a bottle is at least 2000 cc. A random sample of 40 bottles coming out of the machine was selected and the exact content of the selected bottles are recorded. The sample mean was cc. The population standard deviation is known from past experience to be 1.30 cc.Test the null hypothesis at the 5% significance level.
41H0: 2000 H1: 2000 n = 40 For = 0.05, the critical value of z is The test statistic is: Do not reject H0 if: [z -1.645] Reject H0 if: z ]
43Example : p-value approach n = 40For = 0.05, the critical valueof z isThe test statistic is:Do not reject H0 if: [p-value 0.05]Reject H0 if: p-value 0.0]
44Examples :2As part of a survey to determine the extent of required in-cabin storage capacity, a researcher needs to test the null hypothesis that the average weight of carry-on baggage per person is 0 = 12 pounds, versus the alternative hypothesis that the average weight is not 12 pounds. The analyst wants to test the null hypothesis at = 0.05.H0: = 12H1: 12For = 0.05, critical values of z are ±1.96The test statistic is:Do not reject H0 if: [-1.96 z 1.96]Reject H0 if: [z <-1.96] or z 1.96]Lower RejectionRegionUpper Rejection.87654321.025.95Nonrejectionz1.96-1.96The Standard Normal Distribution
45Lower RejectionRegionUpper Rejection.87654321.025.95Nonrejectionz1.96-1.96The Standard Normal DistributionSince the test statistic falls in the upper rejection region, H0 is rejected, and we may conclude that the average amount of carry-on baggage is more than 12 pounds.
46Examples :3An insurance company believes that, over the last few years, the average liability insurance per board seat in companies defined as “small companies” has been $ Using = 0.01, test this hypothesis using Growth Resources, Inc. survey data.H0: = 2000H1: 2000For = 0.01, critical values of z are ±2.576The test statistic is:Do not reject H0 if: [ z 2.576]Reject H0 if: [z <-2.576] or z 2.576]
47Examples 3 : ContinuedLower RejectionRegionUpper Rejection.87654321.005.99Nonrejectionz2.576-2.576The Standard Normal DistributionSince the test statistic falls in the upper rejection region, H0 is rejected, and we may conclude that the average insurance liability per board seat in “small companies” is more than $2000.
48Examples :4The average time it takes a computer to perform a certain task is believed to be 3.24 seconds. It was decided to test the statistical hypothesis that the average performance time of the task using the new algorithm is the same, against the alternative that the average performance time is no longer the same, at the 0.05 level of significance.H0: = 3.24H1: 3.24For = 0.05, critical values of z are ±1.96The test statistic is:Do not reject H0 if: [-1.96 z 1.96]Reject H0 if: [z < -1.96] or z 1.96]
49Examples 4 : ContinuedLower RejectionRegionUpper Rejection.87654321.025.95Nonrejectionz1.96-1.96The Standard Normal DistributionSince the test statistic falls in the nonrejection region, H0 is not rejected, and we may conclude that the average performance time has not changed from 3.24 seconds.
50Examples :5According to the Japanese National Land Agency, average land prices in central Tokyo soared 49% in the first six months of An international real estate investment company wants to test this claim against the alternative that the average price did not rise by 49%, at a 0.01 level of significance.H0: = 49H1: 49n = 18For = 0.01 and (18-1) = 17 df ,critical values of t are ±2.898The test statistic is:Do not reject H0 if: [ t 2.898]Reject H0 if: [t < ] or t 2.898]
51Examples 5: ContinuedLower RejectionRegionUpper Rejection.87654321.005.99Nonrejectiont2.898-2.898The t DistributionSince the test statistic falls in the rejection region, H0 is rejected, and we may conclude that the average price has not risen by 49%. Since the test statistic is in the lower rejection region, we may conclude that the average price has risen by less than 49%.
52Examples :6An investment analyst for Goldman Sachs and Company wanted to test the hypothesis made by British securities experts that 70% of all foreign investors in the British market were American. The analyst gathered a random sample of 210 accounts of foreign investors in London and found that 130 were owned by U.S. citizens. At the = 0.05 level of significance, is there evidence to reject the claim of the British securities experts?H0: p = 0.70H1: p 0.70n = 210For = 0.05 critical values of z are ±1.96The test statistic is:Do not reject H0 if: [-1.96 z 1.96]Reject H0 if: [z < -1.96] or z 1.96]
53Examples :7A floodlight is said to last an average of 65 hours. A competitor believes that the average life of the floodlight is less than that stated by the manufacturer and sets out to prove that the manufacturer’s claim is false. A random sample of 21 floodlight elements is chosen and shows that the sample average is 62.5 hours and the sample standard deviation is 3. Using =0.01, determine whether there is evidence to conclude that the manufacturer’s claim is false.H0: 65H1: 65n = 21For = 0.01 an (21-1) = 20 df, thecritical valueThe test statistic is:Do not reject H0 if: [t -2.528]Reject H0 if: z ]
54Examples 7 : Continued0.95-2.5285-.4321tf(t)NonrejectionRegionRejectionCritical Point for a Left-Tailed Test-3.82Since the test statistic falls in the rejection region, H0 is rejected, and we may conclude that the manufacturer’s claim is false, that the average floodlight life is less than 65 hours.
55Examples :8“After looking at 1349 hotels nationwide, we’ve found 13 that meet our standards.” This statement by the Small Luxury Hotels Association implies that the proportion of all hotels in the United States that meet the association’s standards is 13/1349= The management of a hotel that was denied acceptance to the association wanted to prove that the standards are not as stringent as claimed and that, in fact, the proportion of all hotels in the United States that would qualify is higher than The management hired an independent research agency, which visited a random sample of 600 hotels nationwide and found that 7 of them satisfied the exact standards set by the association. Is there evidence to conclude that the population proportion of all hotels in the country satisfying the standards set by the Small Luxury hotels Association is greater than ?H0: p H1: p n = 600For = the critical value 1.282The test statistic is:Do not reject H0 if: [z 1.282]Reject H0 if: z ]
56Examples 8 : ContinuedSince the test statistic falls in the nonrejection region, H0 is not rejected, and we may not conclude that proportion of all hotels in the country that meet the association’s standards is greater than0.901.2825-.4321zf(z)NonrejectionRegionRejectionCritical Point for a Right-Tailed Test0.519
57EXAMPLE 9A packaging device is set to fill detergent powder packets with a mean weight of 3 kg, with a standard deviation of 0.21 kg. The weight of packets can be assumed to be normally distributed. The weight of packets is known to drift upwards over a period of time due to machine fault which is no tolerable. A sample of 100 packets is taken and weighed. This sample has a mean weight of 5.03 kg. can we conclude that the mean weight produced by the machine has increased? Use a 5 % level of significance
58EXAMPLE 9: ContinuedLet null hypothesis Ho that mean weight has increasedHo : μ ≥ 5 and H1 ; μ < 5Given n=100,x=5.3 kg, σ=0.21, α= 5%Z= (x- μ) / σ√n = /0.21/√100=1.428Since calculated value z(cal) = is less than its critical value z(α) =1.645 at α = 0.05, the null hypothesis Ho is acceptedHence we conclude that mean weight is likely to be more than 5 kg
59HYPOTHESIS TESTING FOR DIFFERENCE BETWEEN TWO POPULATION MEANS Idea is to test whether there is significant difference between the means of these populationsLet μ1 and μ2 be mean of two populationσ1 and σ2 be std deviation of two population
60HYPOTHESIS TESTING FOR POPULATION PARMETERS WITH LARGE SAMPLES Z statistic = [(x1-x2) – (μ1-μ2)]/ σx1 –x2= [(x1-x2) – (μ1-μ2)]/ √σ₁²/n1 + σ₂²/n2The null and alternate hypothesis are stated as Null hypothesisHo :μ1 – μ2 = 0
62HYPOTHESIS TESTING FOR POPULATION PARMETERS WITH LARGE SAMPLES Decision RuleReject Ho at a specified level of significance whenOne Tailed TestZ(cal) >z(α) [or when z<-z(α)]Two tailed testZ(cal)>z(α/2) or z(cal) <-z(α/2)
63EXAMPLE 10A firm believes that the tyres produced by process A on an average last longer than tyres produced by process B. To test this belief, random supplies of tyres produced by the two processes were tested and the result areIs there evidence at a 5 % level of significance that the firm is correct in its belief?ProcessSample sizeAverage lifetime(km)Standard deviation(km)A50224001000B21800
64EXAMPLE 10Let us take the null hypothesis that there is no significant difference in the average life of tyres produced by processes A and B Ho : μ1=μ2 or μ1-μ2 = 0 H1 : μ1=μ2 Given x1(mean) = 22400, x2 (mean) =21800 Σ1=σ2=1000, n1=n2=50 Z= [(x1-x2) – (μ1-μ2)]/ σx1 –x2 = [(x1-x2) – (μ1-μ2)]/ √σ₁²/n1 + σ₂²/n2
65EXAMPLE 10= – /√(1000)²/50 + (1000)²/50 = 600/√ = 600/200=3 Since the calculated value z(cal )=3 is more than its critical value z (α/2) = at α =0.5 level of significance therefore Ho is rejected Hence we conclude that the tyre produced by process A last longer than that of B
66HYPOTHESIS TESTING FOR SINGLE POPULATION PROPORTION The three forms of null hypothesis and alternative hypothesis pertaining to the hypothesized population proportion Po are as followsNull HypothesisAlternative HypothesisHo: P = PoH1: P = Po ( two tailed test)Ho :P ≥ PoH1: P < Po (left tailed test)Ho : P ≤ PoH1 : P >Po (right tailed test
67HYPOTHESIS TESTING FOR SINGLE POPULATION PROPORTION Test statisticz = p-po/σp= P- Po /√Po(1-Po)/nOne tailed testTwo tailed testZ(cal) > z(α) or Z(cal) < - z(α)When H1 ; p< poZ(cal) > z(α/2) orZ(cal) < - z(α/2)P value <α