marketing research notes chapter 20

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So far we have talked about estimating a confidence interval alongwith the probability (the confidence level) that the true populationstatistic lies within this interval under repeated sampling. We nowexamine the principles of statistical inference to hypotheses testing.By the end of this chapter you should be able to• Understand what is hypothesis testing• Examine issues relating to the determination of level of

significance• Apply tests of hypotheses to large to management

Situations• Use of SPSS package to carry out hypotheses test and

interpretation of computer output including p- values

What is Hypothesis Testing?

What is a Hypothesis?A hypothesis is the assumption that we make about thepopulation parameter. This can be any assumption about apopulation parameter not necessarily based on statistical data. Forexample it can also be based on the gut feel of a manager. Managerialhypotheses are based on intuition; the market place decides whetherthe manager’s intuitions were in fact correct.In fact managers propose and test hypotheses all the time. Forexample:• If a manager says ‘if we drop the price of this car model by

Rs15000 , we’ll increase sales by 25000 units’ is a hypothesis.To test it in reality we have to wait to the end of the year toand count sales.

• A manager estimates that sales per territory will grow onaverage by 30% in the next quarter is also an assumption orhypotheses.

How would the manager go about testing this assumption?Suppose he has 70 territories under him.• One option for him is to audit the results of all 70 territories

and determine whether the average is growth is greater thanor less than 30%. This is a time consuming and expensiveprocedure.

• Another way is to take a sample of territories and audit salesresults for them. Once we have our sales growth figure, it islikely that it will differ somewhat from our assumed rate. Forexample we may get a sample rate of 27%. The manager isthen faced with the problem of determining whether hisassumption or hypothesized rate of growth of sales iscorrect or the sample rate of growth is more representative.

To test the validity of our assumption about the population wecollect sample data and determine the sample value of the statistic.We then determine whether the sample data supports ourhypotheses assumption regarding the average sales growth.

How is this Done?If the difference between our hypothesized value and the samplevalue is small, then it is more likely that our hypothesized value ofthe mean is correct. The larger the difference the smaller theprobability that the hypothesized value is correct.In practice however very rarely is the difference between the samplemean and the hypothesized population value larger enough orsmall enough for us to be able to accept or reject the hypothesisprima-facie. We cannot accept or reject a hypothesis about aparameter simply on intuition; instead we need to use objectivecriteria based on sampling theory to accept or reject the hypothesis.Hypotheses testing is the process of making inferences about apopulation based on a sample. The key question therefore inhypotheses testing is: how likely is it that a population such asone we have hypothesized to produce a sample such as the onewe are looking at.

Hypotheses Testing-The theoryNull Hypothesis

In testing our hypotheses we must state the assumed orhypothesized value of the population parameter before we beginsampling. The assumption we wish to test is called the NullHypotheses and is symbolized by Ho.For example if we want to test the hypotheses that the populationmean is 500. We would write it as:Ho: µ=500If we use the hypothesized value of a population mean in aproblem we represent it symbolically as: µHo.

The term null hypotheses has its origins in pharmaceutical testingwhere the null hypotheses is that the drug has no effect, i.e., thereis no difference between a sample treated with the drug anduntreated samples.

Alternative HypothesisIf our sample results fail to support the hypotheses we mustconclude that something else must be true. Whenever we rejectthe null hypothesis the alternative hypothesis is the one we haveto accept. This symbolized by Ha .There are three possible alternative hypotheses for any Ho., i.e.:Ha: µ≠500(the alternative hypothesis is not equal to 500)Ha: µ>500(the alternative hypothesis is greater than 500)Ha: µ<500( the alternative hypothesis is less than 500)

Understanding Level of SignificanceThe purpose of testing a hypothesis is not to question thecomputed value of the sample statistics but to make a judgmentabout the difference between the sample statistic and thehypothesized population parameter. Therefore the next step, afterstating our null and alternative hypotheses, is to decide what

LESSON 20:PRINCIPLE OF HYPOTHESIS TESTING

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criterion do we use for deciding whether to accept or reject the nullhypothesis.

How do We use Sampling to Accept or Reject Hypothesis?Again we go back to the normal sampling distribution. We usethe result that there is a certain fixed probability associated withintervals from the mean defined in terms of number of standarddeviations from the mean. Therefore our problem of testing ahypothesis reduces to determining the probability that a samplestatistic such as the one we have obtained could have arisen froma population with a hypothesized mean m.In the hypothesis tests we need two numbers to make our decisionwhether to accept or reject the null hypothesis:• an observed value or computed from the sample• a critical value defining the boundary between the acceptance

and rejection region .Instead of measuring the variables in original units we calculate astandardized z variable for a standard normal distribution withmean µ=0.The z statistic tells us how many how many standarddeviations above or below the mean standardized mean (z,<0,z>0) our observation falls.We can convert our observed data into the standardized scaleusing the transformation

x

xz

σµ−

=

The z statistic measures the number of standard deviations awayfrom the hypothesized mean the sample mean lies. From thestandard normal tables we can calculate the probability of thesample mean differing from the true population mean by aspecified number of standard deviations. For example:• we can find the probability that the sample mean differs

from the population mean by two or more standarddeviations.It is this probability value that will tell us how likely it is thata given sample mean can be obtained from a populationwith a hypothesized mean m. .

• If the probability is low for example less than 5% , perhapsit can be reasonably concluded that the difference betweenthe sample mean and hypothesized population mean is toolarge and the chance that the population would produce sucha random sample is too low.

What probability constitutes too low or acceptable level is ajudgment for decision makers to make. Certain situations demandthat decision makers be very sure about the characteristics of theitems being tested and even a 2% probability that the populationproduces such a sample is too high. In other situations there isgreater latitude and a decision maker may be wiling to accept ahypothesis with a 5% probability of chance variation.In each situation what needs to be determined are the costsresulting from an incorrect decision and the exact level of risk weare willing to assume. Our minimum standard for an acceptableprobability, say, 5%, is also the risk we run of rejecting a hypothesisthat is true.

The Process of Hypothesis TestingWe now look at the process of hypothesis testing. An examplewill help clarify the issues involved:Aluminum sheets have to have an average thickness of .04inchesor they are useless. A contractor takes a sample of 100 sheets anddetermines mean sample thickness as .0408 inches. On the basisof past experience he knows that the population standard deviationfor these sheets is .04 inches. The issue the contractor faces iswhether he should , on the basis of sample evidence, accept orreject a batch of 10,000 aluminum sheets.In terms of hypotheses testing the issue is :• If the true mean is .04inches and the standard deviation is

.004 inches, what are the chances of getting a sample meanthat differs from the population mean (.04 inches) by.0008inches or more?

To find this out we need to calculate the probability that a randomsample with mean .408 will be selected from a population withµ� =.04 and a standard deviation= .004 If this probability is toolow we must conclude that the aluminum company’s statementis false and the mean thickness of the consignment supplied isnot .04inches.Once we have stated out hypothesis we have to decide on a criterionto be used to accept or reject Ho. The level of significance representsthe criterion used by the decision maker to accept or reject ahypothesis. For example if the manager wishes to allow for a 5%level of significance. This means that we reject the null hypothesiswhen the observed difference between the sample mean andpopulation mean is such that it or a larger difference would onlyoccur 5 or less times in every 100 samples when the hypothesizedvalue of the population parameter is correct.It therefore indicatesthe permissible extent of sampling variation we are willing toallow whilst accepting the null hypothesis. In statistical terms 5%is called the level of significance and is denoted by a=.05We now write our data systematically :ΗΟ:µ=.04, Ha:µ≠.04 a=.045 s =.004Our sample data is as follows:

n=100, =.0408

To test any hypothesis we need to calculate the standard error ofthe mean from the population standard deviation

Next we calculate the z statistic to determine how many standarderrors away from the true mean our sample mean is. This gives usour observed value of z which can then be compared with the zcritical from the normal tables.

x

xz

σµ−

=

= .0408-.04/.0004=2This is demonstrated in the figure 1 below.


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Figure11. We have now determined that our calculated value of z

indicates that the sample mean lies two standard errors(SE)to the right of the hypothesized population mean on thestandard normal scale.

2. Our level of significance is 5%. We now determine the criticalvalue of z at 0.05 level of significance. This value is 1.96.

3. A comparison between the observed z and the z permissibleby our given level of significance :observed z: 2 Critical z:1.96

4. Since the observed value of z is greater than the critical valueof z we can infer that the difference between the value of thesample mean and the hypothesized population mean is toolarge at the 5% level of significance to be attributed tosampling variation. Hence we reject the null hypothesis. Themanager would reject the consignment of aluminum sheetsas not meeting the required specification level.

Example 2How many standard errors around the hypothesized value shouldwe use to be 99.44 certain that we accept the hypothesis when it istrue?This problem requires that we leave a probability 1-.994 =.0056 inthe tail. Since it is two tailed test we have to halve this probabilityto determine z such that there is .0056/2 =.0028 area in each tail.Area under one half of the normal curve =.5-.0028=.4972Looking up the normal tables we find for positive values of z , aprobability of .4972 is associated with a z =2.77 this is illustratedin the figure 2 below.

Activities

1. What do we mean when we reject a hypothesis on the basisof a sample?

Interpreting the Level of SignificanceThe level of significance is demonstrated diagrammatically belowin figure 3. Here .95 of the area under the curve is where we wouldaccept the null hypotheses. The two coloured parts under thecurve representing a total of 5% of the area under the curve arethe regions where we would reject the null hypotheses.

Figure 3A word of caution regarding areas of acceptance and rejection.Even if our sample statistic does not fall in the non shaded regionthis does not prove that our Ho is true. The sample results merelydo not provide statistical evidence to reject the hypothesis. This isbecause the only way a hypothesis can be accepted or rejected withcertainty is for us to know the true population parameter.Therefore we say the sample data is such as to cause us to not rejectnull hypotheses.

Selecting a Level of SignificanceThere is no standard or externally given level of significance fortesting hypotheses. The level of significance at which we want totest a hypothesis is set externally by the manager based on hisevaluation of the costs and benefits associated with acceptance orrejection of a null hypothesis. We can however not a few pointsregarding this issue:1. The higher the level of significance the greater the probability

of rejecting Ho when it is true.This is illustrated in the figure 4 below, which shows three levelsof significance: .01, .1, .50. The location of the sample statistic isalso shown in each of the three distributions. It obviously remainsthe same.• In fig 4a &b we would accept Ho that the sample mean does

not differ significantly from the population mean.• In fig4c we would reject Ho.Why? Because out level of significance of .5 is so high that wewould rarely accept Ho when it is true and frequently reject Howhen it is true.


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figure 4a, 4b, 4c

Type 1 and Type2 Errors

Type 1 Error: defined as the probability of rejecting Ho when itis true. This is also the probability of the level of significance. It issymbolized by a .Type2 error: This is the probability of accepting Ho when it isfalse. It is symbolized by ß .There exists a tradeoff between these two errors: To get a lowwe have to have a higher ß.We can reduce the probability of making a type 1 error if we arewilling to increase the probability of making a type 2 error. Withreference to the earlier figure, in figure 3 we illustrate a 50% levelof significance. Here we have reduced the acceptance region . Thuswe will rarely accept a Ho when it is not true. The price that we haveto pay for this higher level of certainty is that we will frequentlyreject Ho when it is true.To deal with this tradeoff, a manager has to weigh the costs andbenefits involved with each type of error before setting the levelof significance. This is best seen with a example.

Example of a TradeoffSuppose a type 1 error involves the time and trouble of workinga batch of chemicals that should have been accepted. (rejecting Howhen it is true.)Type2 error means that an entire group of users may be poisoned!!( accepting Ho when it is false).In this situation the companywould prefer to minimize type 2 error and will set a high level ofsignificance. That is it will set very high levels of significance (possibly >50%) to get low ßs.

Consider another ScenarioType 1 error( rejecting Ho when it is true):The costs associated with a type 1 error involves disassembling anentire engine at the factory-i.e., it is associated with high productioncosts. This happens, for example, if a random test sample ofproducts throws up product specifications, which are on theborderlines of acceptability.

Type2 (Accepting ho When it is False)This involves accepting products with defective specifications i.e,that some customers may get a somewhat defective product. Thecosts associated with this maybe recalling the defective parts oroffering a involves giving a special warranty to repair the defect.Offering a repair warranty is a relatively less costly option.In this case the manufacturer will set lower levels of significance tominimize type 1 errors. For example he may set 95% 05 99% levelof significance.

Activities

1. Explain why there is no single level of probability used toaccept or reject a hypothesis.

2. If we reject a hypothesis because it differs from a samplestatistic by more than 1.75 SE, what is the probability thatwe have rejected a hypothesis that is in fact true?

3. Define Type1 and 2 errors.4. In a criminal trial, the null hypothesis is that an individual is

innocent of a certain crime. Would the legal system prefer tocommit a type1 or type2 error with this hypothesis.

5. If out goal is to accept a null hypothesis that µ =36.5 with96% certainty when it is true. Our sample size is 50 .Diagram the acceptance and rejection regions for thefollowing hypothesis:µ≠36.5µ>36.5µ<36.5

6. Your null hypothesis is that the battery for a heart pacemakerhas an average life of 300 days, with the alternativehypothesis being that the battery life is more than 300 days.You are the quality control engineer for the batterymanufacturer.a. Would you rather make a type 1 or type 2 errors?b. Based on your answer to a. would you choose a high

or low level of significance.

One Tailed TestsSo far we have discussed two tailed tests where the sample statisticcan differ from the hypothesized mean i.e., it can either be morethan the hypothesized mean or less than the hypothesized mean.Thus a 5 % level overall level of significance implies that we cantest whether 2.5% probability whether it lies above thehypothesized mean and 2.5% probability whether it lies below it.However there may be many cases where we have some priorinformation which enables us to test whether the sample statisticis significantly more than or less than the true population statistic.In this case we go in for a one-tail test.The null hypothesis continues to be that of no difference. Thealternative hypothesis can be one of two:a. Ho: µ=500b. Ha: µ>500 – Upper tailed testorc. Ha:µ<500 – lower tailed test


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In this case if we wish to accept or reject a hypotheses at 5% levelwe then determine z critical such that the entire 5% lies on eitherthe right side (upper tailed test) or on the left side ( lower tailedtest).This is illustrated by the coloured regions in figures 5a&5b.

Figure 5a

Figure5bThe procedure for testing the hypothesis remains the same as inthe two tailed case. The only difference will be in the value of thez critical, which is determined by the entire level of significance ononly one side of the normal distribution.

An Example will Clarify the SituationA hospital uses large quantities of packaged doses of a particulardrug. Excessive doses will pass harmlessly out of the system.Insufficient doses do not produce the desired medical treatment.The hospital has purchased the drug from the same manufacturerfor many years. The hospital inspects 50 doses randomly andfinds the mean dose to be 99.75cc. The population standarddeviation of doses is 2cc.Our problem suggests that the hospital faces a problem if dosesare significantly less than 100cc as patient’s treatment will be affected.However if the dose is more than 100cc there appears to be nomajor problem. Therefore the null hypotheses remains unchanged,we are only interested in testing as the alternative hypotheseswhether the sample mean strength is significantly below 100cc.The hypotheses can be stated as follows:Ho: µ=100ccHa: µ<100ccThis is a left tailed test and the coloured region corresponds to .10level of significance. The acceptance region consists of 40% onthe left side of the distribution plus the entire 50% on the rightside for a total area of 90%. This is shown in figure5a&5b.

We can calcu late the SE of sample mean dose:

28.0502

===n

SEσ

We can now calculate the standardized z statistic:

88.00289.0

10075.99=

−=

−=

SEx

zµ

z. critical= 1.28Therefore since -.88<1.28, the sample mean lies within theacceptance region and the hospital can accept the null hypothesis:the observed mean of the sample is not significantly differentfrom the hypothesized mean dose.

Applications of One Tailed TestsMany managerial situations call for a one tailed test. Typically if aproblem requires you to test whether the sample statistic is:• More than a given population statistic• Less than a given population statistic a one tailed test is

appropriate.If the problem requires us to assess whether the sample statistic isnot equal to a population statistic then we use a two tailed test.

Example

1. A highway safety engineer decides to test the load bearingcapacity of a bridge that is 20 years old. Considerable data areavailable from similar tests on the same type of bridges.Which type of hypothesis is appropriate one or two tail testIf the minimum load bearing capacity of this bridge mustbe 10 tons , what are the null and alternative hypotheses?The engineer would be interested in whether a bridge of thisage could withstand minimum load bearing capacitiesnecessary for safety purposes . She therefore wants its capacityto be above a certain minimum level; so a one tailed test, i.e.,a right tailed test would be used.

The hypotheses are:Ho: µ=10 tons Ha: µ>10 tons2. Hinton press hypothesizes that the average life of its press is

14500 hours. The standard deviation of a press life is 2100hours. From a sample of 25 presses the company findssample mean life to be 13000 hours. At a .001 significancelevel should the company conclude that the average life ofthe press life is less than the hypothesized 14500 hours.

Our problem requires us to assess whether average press life issignificantly less than the hypothesized press life. Therefore it isa one tail test.Ho: µ=14500 Ha: µ<14500 n=25, s =2100 µ=.01SE= 2100/√25=420Z = 13000-14500/420= - 3.57z. critical for a one tail test= -2.33-3.57<-2.33 implies we should reject Ho. The average life issignificantly less than the hypothesized life.


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Activities

1. Aaj Ka films is a film distribution company. They know thata hit movie runs for an average of 84 days in a city with astandard deviation of 10 days. The manager of a SouthEastern district is interested in comparing the moviespopularity in his region as compared to the all India average.He randomly selects 75 theatres in his region and finds theyran the movie for an average of 81.5 days. The managerwants to know whether mean running time in the SouthEast is below that of the national average. Test theappropriate hypothesis at .01 level of significance.

2. Atlas Sporting goods has implemented a special tradepromotion policy for its stoves. They think the promotionshould result in a significant price change for customers.Before the promotion began the average retail price of astove was $44.95 with a standard deviation of $5.75. Afterthe promotion they sample 25 retailers and finds mean priceto be $42.95. At a .02 level of significance does Atlas havereason to believe the average retail price to consumers ahsdecreased?

3. Under what conditions is it appropriate to use a one tailedtest? A two tailed test?

4. The statistics department installed energy efficient lights,heaters, ACs. Now they want to determine whether theaverage monthly energy usage has decreased. Should theyperform a one or two tail test? If their previous averagemonthly usage was 3124 KW hours, what are the null andalternative hypotheses?

In the last lecture we learnt about the general principles ofhypothesis testing and how to carryout a test a sample statisticwith a hypothesized population value. We now extend the principlesof hypothesis testing to include testing of hypothesis to sampleproportions vis a vis a hypothesized population proportion. Wewill also learn to test for the differences between two samples.That is do two separate samples differ significantly from eachother. Similarly we try to determine whether the two differentsample proportions differ significantly from each other.

Hypothesis Testing of ProportionsSo far we have talked about the principles of hypothesis testingwhere we have compared a sample mean with a hypothesized orpopulation mean.We can also apply the principles of hypothesis testing alreadydetermined to testing for differences between proportions ofoccurrence in a sample with a hypothesized level of occurrence.Theoretically the binomial distribution is the correct distributionto use when dealing with proportions. As sample size increasesthe binomial distribution approaches a normal distribution interms of characteristics. Therefore we can use the normaldistribution to approximate the sampling distribution. To dothis we need to satisfy the following conditions:np>5, nq>5where p is the proportion of successesq: proportion of failures.

Testing of Hypothesis for ProportionsWhen testing significance of hypotheses for proportions we beginwith a similar procedure to the earlier case:

First we define various proportions or percentages of occurrenceof a event:pHo: the population or hypothesized population of successqHo Hypothesized value of the population proportion of failuresp: sample proportion of successesq: sample proportion of failuresAgain we calculate a z statistic

p

Hppz

σ0

−=

where � p is the standard error of the proportion which iscalculated using the hypothesized population proportion .

n

qp HHp

00=σ

We can then check the calculated z with z critical for the appropriatelevel of significance to determine whether to accept or reject Ho.An example shall make the process clearer..A ketchup manufactureris in the process of deciding whether to produce a new extra spicyketchup. The company’s marketing research department used anational telephone survey of 6000 households and found that335 would purchase extra spicy ketchup. A much more extensivestudy made two years ago found showed that 5% of householdswould purchase would purchase the brand then. At a two percentlevel of significance should the company conclude that there is anincreased interest in the extra spicy flavour?

z critical for a one tailed test is 2.05 Since the observed z > z criticalwe should reject Ho and current levels of interest are significantlygreater than interest two years ago.

Activities

1. Steve Cutter sells lawn mowers . He is interested incomparing the reliability of his lawn mowers with aninternational brand. He knows only 15% of the internationalbrands require repairs. A sample of 120 of Steve’s customersshows that 22 of them required repairs. At .02 level ofsignificance is there evidence that Steve’s mowers differ fromthe international brand.

Ans. Observed z value=1.03< 2.33(z critical) Therefore we acceptHo.2. Grant Inc., a manufacturer of women’s dress blouses knows

that its brand is carried in 19% of the women’s clothingstores in Delhi. Grant recently samples 85 women’s clothingstores in Chandigarh and found that 14.12% of the storescarried the brand. At .04 level of significance is there evidence


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that Grant’s has a poorer distribution in Chandigarh than inDelhi?

3. Macroswift estimated last year that 35% of its potentialsoftware buyer were planning to wait to purchase the newoperating system Window Panes, until an upgrade has beenreleased. After an advertising campaign to reassure the public, Macroswift surveyed 3000 people and found 950 who werestill skeptical. At the 5% level of significance can the companyconclude the proportion of skeptical people has decreased?

Hypothesis Tests of Differences between MeansSo far we have examined the case where we are testing the resultsof a sample against a hypothesized value of a population statistic.We now turn to case where we wish to compare the parameters fortwo different populations and determine whether these differfrom each other. In this case we are not really interested in theactual value of the two parameters but the relation between thetwo parameter, i.e. is there a significant difference between them.Example of hypothesis of this type are:• Whether female employees earn less than males for the same

work.• A drug manufacturer may need to compare reactions of one

group of animals administered the drug and the controlgroup.

• A company may want to see if the proportion ofpromotable employees in one installation is different fromanother.

In each case we are not interested in the specific values of theindividual parameters as the relation between the two parameters.The core problem reduces to one of determining whether themeans from two samples taken from two different populationsis significantly different from each other. If they are not, we canhypothesize that the two samples are not significantly differentfrom each other.

Theoretical BasisShown below are three different but related distributions.Figure 1a shows the population distribution for two differentpopulations 1 and 2. They have respectively the followingcharacteristics:Mean s 1 and s 2 and standard deviations s 1 and s 2.

Figure 1b shows the respective sampling distribution of sample.This distribution is defined by the following statistics:Mean of sampling distribution of sample means : µ1

and µ2

Standard deviation of sampling distribution of mean or thestandard error of sampling mean: s ¯x1 and s ¯x2.

We have two populations 1, 2 with mean µ1 and µ2 with standarddeviation s 1 and s 2. The associated sampling distribution forsampling means –x1 and –x2. However what we are now interestedin is the difference between the two values of the sampling means.i.e., the sampling distribution of the difference between thesampling means –x1 - –x2. How do we derive this distribution ?Suppose we take a random sample from the distribution ofPopulation 1 and another random sample from the distributionof Population 2. If we then subtract the two sample means, we

get –x1 - –x2 . This difference will be positive if –x1 is larger than –x2 and vice versa. By constructing a distribution of all possiblesample differences –x1 - –x2 we end up with a distribution of thedifference between sample means shown below in figure 6c.

Figure 1a&1b

Figure 1c

The mean of this distribution is

The standard deviation of the distribution of the differencebetween sample means is called the standard error of the differencebetween two means.

The testing procedure for a hypothesis is similar to the earliercases.(µ1 - µ2) Ho: µ1 = µ2

Ha: µ1≠ µ2 a=.05

The z statistic = (–x1 - –x2 ) – (µ1 - µ2) Ho / s –x1 - –x2

Since we will usually we will be testing for equality between thetwo population means hence:(µ1 - µ2) Ho =0 since µ1 = µ2

An example will make the process clearer:A manpower statistician is asked to determine whether hourlywages of semi skilled labour are the same in two cities. The resultof the survey is given in the table below. The company wants totest the hypothesis at .05 level of significance that there is nosignificant difference between the hourly wage rate across the twocities.


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City Mean hourly wage

Standard deviation of sample

Size of sample

apex $8.95 $.4 200 Eden $9.10 $.6 175 Since standard deviation of the two populations are not known

we estimate s ˆ1 and s ˆ2 by using the sample standard deviation s1

and Ho: µ1 = µ2

Ha: µ1 ≠ µ2 a=.05The estimated standard error of the difference between the twomeans is –x1 - –x2 =√( s1

2/n1+ s22 /n2) =√( (.4)2/200+ (.6)2 /175)=$.053

We then calculate the z statistic:z. =(–x1 - –x2 ) – (� 1 - � 2) Ho / –x1 - –x2

8.95-9.10-0/.053=-2.83We can mark the standardized difference on a sketch of thesampling distribution and compare with the critical value ofz=±1.96 in figure 7 As we can see the calculated z lies outside theacceptance region. Therefore we reject the null hypothesis.

Figure 2

Example 2

1. Two independent samples of observations were collected.For the first sample of 60 elements, the mean was 86 andthe standard deviation 6. The second sample of 75 elementshad a mean of 82 and a standard deviation of 9.a. Compute the estimated standard error of the

difference between the two means.b. Using a = 0.01, test whether the two samples can

reasonably be considered to have come frompopulations with the same mean.

= (86-82)-0/1.296=3.09Since 3.09>2.58, we reject Ho and it is reasonable to conclude thatthe two samples come from different populations.

Activities

1. In 1993, the Financial Accounting Standards Board (FASB)was considering a proposal to require companies to reportthe potential effect of employees’ stock options on earningsper share (EPS. A random sample of 41 high-technologyfirms revealed that the new proposal would reduce EPS byan average of 13.8 percent with a standard deviation of 18.9%. A random sample of 35 producers of consumer goodsshowed that the proposal would reduce EPS by 9.hpercenton average, with a standard deviation of 8:7 percent. On thebasis of these samples, is it reasonable for the FASB toconclude (at a=.10) that the FASB proposal will cause agreater reduction in EPS for high technology firms than forproducers of consumer goods?

2. A sample of 32 money market funds was chosen on Jan 1.1996, and the average annual rate of return over the past 30days was found to be 3.23% and the sample standarddeviation was.51% . A year earlier , a sample of 38 moneymarket funds showed an average rate of return of 4.36% andthe sample standard deviation was .84%. Is it reasonable toconclude (at �=.05) that money m arket interest ratesdeclined during 1995?

3. Despite the Equal pay act it still appeared that in 1993 menearned more than women in similar jobs. A random sampleof 38 male machine tool operators found a mean hourlywage of $11.38, and a sample standard deviation was $1.84.A random sample of 45 female operators found their meanwage to be $8.42 and the sample standard deviation was$1.31. On the basis of these samples is it reasonable toconclude (at a=.01) that the male operators are earning over$2.00 more per hour than the female operators?

Tests for differences between proportions for large samplesThe analysis for testing differences in proportions of two samplesis broadly similar to earlier case. As long as samples are greaterthan 30 we can use the normal approximation to binomial.Suppose we have two sample proportions p 1 and p1 which measurethe probability of occurrence of an event or characteristic in twosamples. We wish to test whether the probability of occurrence issignificantly different across the two samples .

p1: sample proportion of success in sample 1

p2: sample proportion of success in sample 2 n1 : sample size 1 n2 : sample size 1Standard error of the difference between two proportionsSince we do not know the population proportions, we need to

estimate them from sample statistics:

We hypothesize that there is no difference between the twoproportions. In which case our best estimate of the overallpopulation proportion successes is the combined proportion ofsuccesses in both samples. This can be calculated from thefollowing formula:


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Once we have the estimated proportion of successes in twopopulations. The estimated standard error of the differencebetween two proportions is as follows:

The standard z statistic in this case is calculated as :

This can be best illustrated with the help of an example:A pharmaceutical company tests two new compounds to reduceblood pressure. They are administered to two different groups oflab animals. In group1 71 of 100 animals respond to drug1 withlower BPS. In group2 58 of 90 animals showed lower BP levels.The company wants to test at .05 level of significance whetherthere is a difference between the two drugs.

p1: .71 (sample proportion of success in sample 1)

q1: .29 n1 : 100 p2: .644 (sample proportion of success in sample 2)

q2: .356 n2 : 90 Ho: p1=p2 Ha: : p1≠ p2 a=.05

The overall population or hypothesized percentage of occurrenceassumes that there is no difference between the two populationproportions. We therefore estimate it as the combined proportionof successes in both samples. ˆp=100(.71)+90(.644)/190=.6789The estimated standard error of the difference between two sampleproportions is:

The observed z statistic is:z= (.71-.644)-0/.0678=.973z. critical for .o5 level of significance is 1.96. Since observed z isless than z critical we accept Ho. This is shown in the figure 3below:

Figure 3

Example 2For tax purposes a city government requires two methods oflisting property. One requires property owner to appear in personbefore a tax lister. The second one allows the form to be mailed.The manager thinks the personal appearance method leads tolower fewer mistakes. She authorizes an examination of 50personal appearances, and 75 mail forms. The results show that

10% of personal appearances have errors whereas 13.3 % of mailsforms had errors. The manager wants to test at the .15 level ofsignificance, the hypothesis that that personal appearance methodproduces lower errors. The hypothesis is a one tailed test. Theprocedure for this as the same as for carrying out a one tailed testfor comparing sample means. The data is as follows:

Since it is a one tail test we do not divide the level of significanceon both sides of the normal curve. A .15 level of significanceimplies we determine z critical for area under one side of thenormal curve i.e., .35 (.5-.15). Specifically it is a left tailed test as themanager wishes to test that method 1 , i.e., personal appearances,result in significantly lower errors. This is shown by the markedof region in the figure4

Figure 4

therefore calculated z <critical z we accept the null hypotheses.Hint: If a test is concerned with whether one proportion issignificantly different from another, use a two-tailed test. If thetest asks whether one proportion is significantly higher or lowerthan the other, a one tailed test is appropriate.

Activities

1. A large hotel chain is trying to decide whether to convert moreof its roe=.: to nonsmoking rooms. In a random sample of400 guests last year, 166 have requested nonsmoking rooms.This year, 205 guests in a sample of 38C preferred thenonsmoking rooms. Would you recommend that the hotelchain convert more rooms to nonsmoking? Support yourrecommendation: testing the appropriate hypotheses at a 0.01level of significance.


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Ans:z calculated = -3.48< -z critical=-2.33 Therefore reject Ho.2. Two different areas of a large Eastern city are being considered

as sites for day-care centers. of 200 households surveyed inone section, the proportion in which the mother’s worked full-time was 0.52. In another section of the city 40 percent of the150 households surveyed had mothers working at fulltimejobs. At the 0.04 level of significance is there a significantdifference in the proportions of working mothers in the twoareas of the city?

Ans:z calculated =2.23>z critical=2.05. Therefore reject Ho.4. On Friday 11 stocks in a sample of 40 of the 2500 stocks

traded on the BBSE advanced, i.e., their price of their sharesincreases. In a sample of 60 BSE stocks on Thursday , 24 hadadvanced At α� =.10 , can you conclude that a smaller proportionof BSE stocks advanced on Friday than did on Thursday?

5. A coal fired power plant is considering two different systemsfor reducing pollution. The first system has reduced theemission of pollutants to acceptable levels68% asdetermined from 200 air samples. The second moreexpensive system has reduced the emission of pollutants toacceptable levels 76% of the time as determined on the basisof 250 air samples. If the expensive system is significantlymore effective than the inexpensive system, the managementof the power plant will install the inexpensive system. Whichsystem will be installed if the management uses a significancelevel of .02 in making its decisions?

In this chapter we will wrap up our analysis of hypothesis testingfor large samples. By now you should have a good idea how toapply the principles of hypothesis testing to different types ofmanagerial problems. However these days any managementproblem that we may wish to analyze generates such a largevolume of data that it is virtually impossible to analyze and testhypotheses manually. Given the widespread availability ofcomputers and statistical packages we can easily run such tests onthe computer. Therefore it becomes important to understandand interpret how these tests are run on the computer.Obviously the basic theory and principles of statistical analysis donot change when a test is carried out on computer. However thereare some differences in the way the level of significance is presented.Computer outputs usually present the prob value or p-value. Weshall look at what prob values mean and compare them withconventional tests of significance.We shall also look at what is considered a good hypothesis test.This is done by measuring the power of a test. This concept willalso be presented in detail.

Probability ValuesSo far we have tested a hypothesis at a given level of significance.In other words before we take the sample we specify how unlikelythe observed result will have to be in order for us to reject Ho. Forexample we test the hypothesis that observing a sample meanthis far away from the true population mean is less than 5%,where %% is our externally given level of significance.There is another way to approach the decision whether to accept orreject Ho which does not require us to prespecify the level ofsignificance before taking a sample. In this case we take a sample,

compute mean and ask: suppose Ho were true what is theprobability of getting a sample mean this far away from µHo.

This is called a probability value or p- value of the sample mean.The two methods are equivalent and essentially represent twosides of a coin.• In the earlier case we prespecify a level of probability and

compare the observed probability of getting a samplestatistic with the prespecified level of probability(a).

• We now ask what is the probability value of getting such aresult. This is termed the p- value of a statistic.

Once the p- value is determined the decision maker can thenweigh all relevant factors and decide whether to accept/reject Howithout being bound by a prespecified level of significance.The p- value can also be more informative. For example if wereject a Ho at a=.05, we only know that the observed value wasatleast 1.96 SE away from the mean. A p- value tells us the exactprobability of the getting a sample mean 1.96SE away from themean.The concept will be made clearer with the help of an example.

ExampleA machine is used to cut Swiss Cheese into blocks of specifiedweight. On the basis of experience the weight of a block has astandard deviation of .3gm. The machine is currently set to cutblocks of 12 gm. A sample of 25 blocks is found to have anaverage weight of 12.25g. Should we conclude the machine needsto be recalibrated?Since this is a two tailed test we need to determine the probabilityof observing a value of >x atleast as far away from 12 as 12.25 or11.75gm if Ho is true.We therefore need to calculate the probabilityP(x>12.25 or x<d 11.75) if Ho is true.Our hypothesis can be stated as :

Ho: µ=12 Ha: µ≠12s=.3 n=25 >x=12.15s >x = s/√n=.3/5=.06

We can then convert >x to a standard z score.

From the normal tables we can find the probability that a z greaterthan 2.5 is .5-.4938=.0062Since this is a two tailed test the p- value is 2*.0062=.0124 thisinformation is shown in the figure 1 below;


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Figure 1Given the above information the cheese packer can now decidewhether to recalibrate the machine or not. As we can see the p –value is very low and he probably will not go in for recalibration. If he had he carried out a conventional hypotheses test at .05 levelof significance is also illustrated in the figure 1.On the basis of thez test he would have rejected the Ho 5 % level. However at asignificance level of .01 we would have accepted the hypotheses, asthe critical z value would have been 2.58.The p- value tells us the largest significance level at which wewould have accepted Ho, i.e, .0124 and the associated z value(±2.5). Thus at any level of significance above .0124 we wouldreject Ho.

Uses of p- valuesUse of p values saves the tedium of looking up tables. The smallerthe probability value, the greater the significance of the finding.The simple rule of thumb is: As long as µ>p reject Ho.For example if we have a p-value=.01 and µ=.05 . Then thismeans that the probability of getting our sample result is .01. Wecompare this with our standard of accepting/ rejecting Ho whichis .05. since .05>.01 we reject Ho as the probability of getting sucha result is much lower than our level of significance.

Example 2The Coffee Institute has claimed that more than 40% of Americanadults regularly have a cup of coffee with breakfast. A randomsample of 450 individuals showed that 200 of them were regularcoffee drinkers at breakfast. What is the prob value of a test ofhypotheses seeking to show that the Coffee Institute’s claim wascorrect.

Activities

1. A car retailer Hunks that a 40000 mile claim for tyre life by themanufacturer is too high. She carefully records the mileageobtained from a sample of 64 such tyres. The mean turnsout to be 38,500 miles. The standard deviation of the life ofall tyres of this type has previously been calculated by themanufacturer to be 7,600 miles. Assuming that the mileageis normally distributed, determine the largest significancelevel at which we would accept the manufacturer’s mileageclaim, that is, at which we would not conclude the mileage issignificantly less than 40,000 miles.

2. The North Carolina Department of Transportation hasclaimed that at most, 18 percent of passenger cars exceed 70mph on Interstate 40 between Raleigh and Durham. Arandom sample of 300 cars found 48 cars exceeding 70 mph.What is the prob value for a test of hypothesis seeking toshow the NCDOT’s claim is correct?

3. Kelly’s machine shop uses a machine-controlled metal saw tocut sections of tubing used in pressure-measuring devices.The length of the sections is normally distributed with astandard deviation of 0.06". Twenty-five pieces have been cutwith the machine set to cut sections 5.00" long. When thesepieces were measured, their mean length was found to be4.97". Use prob values to determine whether the machineshould be recalibrated because the mean length issignificantly different from 5.00".

3. SAT Services advertises that 80 percent of the time, itspreparatory course will increase an individual’s score on theCollege Board exams by at least 50 points on the combinedverbal and quantitative total score. Lisle Johns, SAT’smarketing director, wants to see whether this is a reasonableclaim. Lisle has reviewed the records of 125 students whotook the course and found that 94 of them did, indeed,increase their scores by at least 50 points. Use prob values todetermine whether SAT’s ads should be changed because thepercentage of students whose scores increase by 50 or morepoints is significantly different from 80 percent.

Using the computer to Test the HypothesesThese days in actual managerial situations hypotheses tests arerarely done manually. Therefore it is important that students caninterpret computer output generated for hypotheses tests byvarious standard statistical analysis packages. The most popularof these packages are SPSS and Minitab. Broadly all programmesfollow the same principles. Instead of a comparing the calculatedz value with a predetermined level of significance, most packagesdisplay the prob values or p- values.To accept or reject an hypotheses we compare the level of significance(a) and the p- value. If a>p- value we reject Ho at the relevant levelof significance and vice versa.An example will help show how computer outputs results forhypotheses testing.

ExampleWhile designing a test it was expected that the average grade wouldbe 75%, i.e., 56.25 out of 75. This hypotheses was tested againstactual test results for a sample of 199 students.Ho: µ=56.25 Ha: µ≠56.25This hypotheses was tested at a=.05The computer output for this test is shown below using theMinitab package. The observed t value for this test was -15.45,with an associated (two-tailed) prob value of 0.0000. Because thisprob value is less than our significance level of a= 0.05, we mustreject Ho and conclude that the test did not achieve the desiredlevel of difficulty.This is shown in Table 1T test for a meanTest for µ=56.25 vs µ≠56.25Varaible N Mean Stdev SEMean T P-valueResult 199 45.281 10.014 .710 -15.45 0.00

Table1T test for difference between two sample meansHere we test hypotheses of equality of two means.


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The university had been receiving many complaints about thecaliber of teaching being done by the graduate-student teachingassistants. As a result, they decided to test whether students insections taught by the graduate TAs really did worse in the examthan those students in sections taught by the faculty.If we let the TAs’ sections be sample 1 and the faculty’s sectionsbe sample 2, then the appropriate hypotheses for testing thisconcern areHo: µ1 = µ2

Ha: µ1< µ2

The underlying population is assumed to be equal for bothsamples.The Minitab output for doing this test is given below. The testresults are reported assuming that the two population variancesare equal. If we can assume that the two variances are equal, thenthe test reported by Minitab is the test using a pooled estimate fors2. This is shown in table 2

Table2

Two sample T- Test

Instrnum N Mean stdev SE mean

1 89 44.93 9.76 1.0

2 110 45.6 10.2 .98

T test µ1 = µ2 vs µ1< µ2 : T=-.44 P=.33 Df =197

Both use pooled stdev=10.0

What does the data tell us regarding the efficacy of TA ? The probvalue is quite high, i.e., .33. Therefore if we compare this with alevel of significance of .05 (a = 0.05) we would accept the nullhypotheses that there is no difference in results between TAs andfaculty. In this case we would accept the null hypotheses at anylevel of significance up to .33.

Measuring the Power of a TestWhat should a good hypothesis test do ?Ideally and (the probabilities of Type I and Type 2 errors) shouldboth be small. A Type I error occurs when we reject a nullhypothesis that is true. a (the significance level of the test) is theprobability of making a Type I error. Once we decide on thesignificance level, there is nothing else we can do about a.A Type 2 error occurs when we accept a null hypothesis that isfalse; the probability of a Type 2 error is ß. Ideally a managerwould want a hypothesis to reject a null hypothesis when it isfalse. Suppose the null hypothesis is false. Then managers wouldlike the hypothesis test to reject it all the time. Unfortunately,hypothesis tests cannot be foolproof; sometimes when the nullhypothesis is false, a test does not reject it, and a Type 2 error ismade.When the null hypothesis is false, µ (the true population mean)does not equal µHo (the hypothesized population mean);instead,equals some other value. For each possible value of for

which the alternative hypothesis is true, there is a differentprobability of incorrectly accepting the null hypothesis.Of course, we would like this ß (the probability of accepting a nullhypothesis when it is false) to be as small as possible or we wouldlike 1-ß (the probability of rejecting a null hypothesis when it isfalse) to be as large as possible.Therefore a high value of 1-ß(something near 1.0) means the testis working quite well ( i.e it is rejecting the null hypothesis when itis false); a low value of 1-ß (something near 0.0) means that thetest is working very poorly( i.e. not rejecting the null hypotheseswhen it is false).The value of 1-ß is a measure of how well the test is working andis known as the power of the test. If we plot the values of 1-ß foreach value of for which the alternative hypothesis is true, theresulting curve is known as a power curve. This can be explainedbetter with the help of an example.

ExampleWe were deciding whether to accept a drug shipment. Our testindicates that we should reject the null hypothesis if thestandardized sample mean is less than - 1.28, that is, if samplemean dosage is less than 100.00 - 1.28 (0.2829), or 99.64 cc.In Figure 9a, we show a left-tailed test. In Figure 9b, we showthe power curve which is plotted by computing the values of 1

Figure 2a


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Figure 2bPoint C on the power curve in Figure 2 b shows populationmean dosage is 99.42 cc. Given that the population mean is 99.42cc, we must compute the probability that the mean of a randomsample of 50 doses from this population will be less than 99.64 cc(the point below which we decided to reject the null hypothesisi.e., the value of the dose at which we rejected the null hypothesis.This shown in Figure 2c .

Figure 2 cWe had computed the standard error of the mean to be 0.2829 cc.So 99.64 cc is (99.64- 99.42)/0.2829 = 0.78Thus 99.64 is .78 SE above the true population mean when ittakes a value µ= 99.42 cc.The probability of observing a sample mean less than 99.64 ccand thus rejecting the null hypothesis is 0.7823, when we take thetrue population mean to be µ= 99.42 cc. This is given by thecolored area in Figure 9c. Thus, the power of the test 1 - ßat = 99.42 is 0.7823. This simply means that if = 99.42, theprobability that this test will reject the null hypothesis when it isfalse is 0.7823.Point D in Figure 9b shows that if the population mean dosageis 99.61 cc. We then ask what is the probability that the mean ofa random sample of 50 doses from this population will be lessthan 99.64cc and thus cause the test to reject the null hypothesis?This is illustrated in Figure 2d. Here we see that 99.64 is (99.64 -99.61)/0.2829, or 0.11 standard error above 99.61 cc. Theprobability of observing a sample mean less than 99.64cc andthus rejecting the null hypothesis is 0.5438, the colored area inFigure 9d. Thus, the power of the test (1 - µ ) at µ= 99.61 cc is0.5438.

Figure 2dUsing the same procedure at point E, we find the power of thetest at = 99.80 cc is 0.2843; this is illustrated as the colored area inFigure 2e.

Figure 2eAs we can see the values of 1 - ß continue to decrease to the rightof point E. This is because as the population mean gets closerand closer to 100.00 cc, the power of the test (1 - ß ) gets closer andcloser to the probability of rejecting the null hypothesis when thepopulation mean is exactly 100.00 cc. This probability is nothingbut the significance level of the test which in this case is 0.10. Thecurve terminates at point F, which lies at a height of 0.10 directlyover the population mean.

What does the Power Curve in figure 2b Tell Us?As the shipment becomes less satisfactory (as the doses in theshipment become smaller), our test is more powerful (it has agreater probability of recognizing that the shipment isunsatisfactory). It also shows us, however, that because ofsampling error, when the dosage is only slightly less than 100.00cc, the power of the test to recognize this situation is quite low.Thus, if having any dosage below 100.00 cc is completelyunsatisfactory, the test we have been discussing would not beappropriate.

Example:Before the 1973 oil embargo and subsequent increase in oil price,petrol usage in the US had grown at a seasonally adjusted rate of.57% per month, with a standard deviation of of .10% per month.In 15 randomly chosen months between 1975 and 1985, petrolusage grew at an average rate of .33% per month. At a .01 level ofsignificance can you conclude that the growth in the use of gasolinehad decreased as a result of the embargo?Compute the power ofthe test for �=.50, .45 and .4% per month. =.10 n=15,Ho: µ=.57, Ha: µ<.57 At a=.01, the lower limit of the

acceptance region is :µHo – 2.33 µ/√n=.57-2.33(.10) /√15=.510

a. At µ=.50, the power of the test is : P(>x<.510)=P(z<.510-.50/.10/√15=

P(z<.39)=.5+.1517=.6517b. At �=.45, the power of the test is P(>x<.510)=P(z<.510-.45/.10/”15=

P(z<2.32)=.5+.4898=.9898c. At �=.40, the power of the testis P(>x<.510)=P(z<.510-.40/.10/”15=

P(z<.4.26)=.5+.1517=1.00


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Problems

1. A manufacturer of petite women’s sportswear hashypothesized that the average weight of the women itsbuying1.ts_clothing is 110 pounds. The company takes twosamples of its customers and finds one sample’s estimate ofthe population mean is 98 pounds, and the other sampleproduces a mean weight of 122,pqunds. In the test of thecompany’s hypothesis that the population mean is 110pounds versus the hypothesis that the mean does equal 110pounds, is one of these sample values more likely to leadaccept the null hypothesis? Why or why not?

2. On an average day; about 5 percent of the stocks on the NewYork Stock set a new high for the year. On Friday Sept 18th ,1992 the Dow Jones closed at closed at 3,282,on a robustvolume of over 136 million shares traded. A randomsample of 120, stocks showed that 16 had set new annualhighs that day. Using a significance level of 0.01, should weconclude that more stocks than usual set new highs on thatday?

3. A finance developed a theory that predicted that closed endequity funds should sell at a premium of about 5% onaverage. Assuming that the discount /premium populationis approximately normally distributed does the sampleinformation support his theory? Test at .05 level ofsignificance.

4. A company recently criticized for not paying women as muchas men claims that its average salary paid to all employees is$23500. From a random sample of 29 women, the averagesalary was calculated to be $23,000. If the populationstandard deviation is known to be $1250 for these jobsdetermine whether we could reasonably, within 2 standarderrors expect to find $23000 as the sample mean if, in fact ,the company’s claim is true.

5. A manufacturer of vitamins for infants inserts a coupon fora free sample of its production a package that is distributedat hospitals to new parents. Historically about 185 of thecoupons have been redeemed. Given current trends forhaving fewer children and starting families later, the firmsuspects that today’s parents are better educated on averageand as a result more likely to use vitamin supplements fortheir infants. A sample of 1500 new parents redeemed 295coupons. Does this support at a significance level of 2percentthe firm’s beliefs about today’s new parents.

6. From a sample of 10,200 loans made by a state employeescredit union in the most recent five year period, 350 weresampled to determine what proportion was made towomen. This sample showed that 39% of the loans weremade to women employees. A complete census of loans 5years ago showed that 41% were women borrowers. At asignificance level of .02 can you conclude that the proportionof loans made to women has changed significantly in the lastfive years?

Points to Ponder

• Hypothesis testing cab be viewed as a six-step procedure• Establish a Null Hypothesis as well as alternative

Hypothesis. It is a one tail test of significance if the

alternative hypothesis states the direction of differences. Ifno direction of difference is given, it is two tailed test.

• Choose the statistical test on the basis of the assumptionabout the population distribution and measurement level.The form of the data can also be a factor. In light of theseconsiderations, one typically chooses the test that has thegeneral power efficiency or ability to reduce decision error.

• Select the desired level of confidence. While 05.0=α isthe most frequently used level, many others are also used.The α is the significant level that we desire and is set inadvance of the study.

• Compute the actual test value of the data.• Obtain the critical test value, usually by referring to a table for

the appropriate type of distribution.• Interpret the result by comparing the actual test value with

the critical test value.

Notes

marketing research notes chapter 20

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