continution hypothesis testing

8/14/2019 Continution Hypothesis Testing

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Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc. 12.3

Inference With Variance Unknown Previously, we looked at estimating and testing the

population mean when the population standard deviation ( )was known or given:

But how often do we know the actual population variance?

Instead, we use the Student t-statistic , given by:


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Inference With Variance Unknown When is unknown, we use its point estimator s

and the z-statistic is replaced by the the t-statistic, where thenumber of degrees of freedom , is n 1.


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Testing when is unknown When the population standard deviation is unknown and the

population is normal, the test statistic for testing hypothesesabout is:

which is Student t distributed with = n 1 degrees offreedom. The confidence interval estimator of is given

by:


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Example 12.1 Will new workers achieve 90% of the level of experienced

workers within one week of being hired and trained?

Experienced workers can process 500 packages/hour, thus if

our conjecture is correct, we expect new workers to be ableto process .90(500) = 450 packages per hour.

Given the data , is this the case?
http://localhost/var/www/apps/conversion/tmp/scratch_6/References/Xm12-01.xlshttp://localhost/var/www/apps/conversion/tmp/scratch_6/References/Xm12-01.xls


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Example 12.1 Our objective is to describe the population of the numbers of

packages processed in 1 hour by new workers, that is wewant to know whether the new workers productivity is morethan 90% of that of experienced workers. Thus we have:

H1: > 450

Therefore we set our usual null hypothesis to:

H0: = 450

IDENTIFY


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Example 12.1 Our test statistic is:

With n=50 data points, we have n 1=49 degrees of freedom.

Our hypothesis under question is:H1: > 450

Our rejection region becomes:

Thus we will reject the null hypothesis in favor of the alternative if ourcalculated test static falls in this region.

COMPUTE


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8/38Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc. 12.10

Example 12.1 Alternatively, we can use t-test:Mean from

Tools > Data Analysis Plus in Excel

COMPUTE

:: r e

j e c

t i o n

r e g

i o n


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Example 12.2 Can we estimate the return on investment for companies that

won quality awards?

We have are given a random sample of n = 83 such

companies. We want to construct a 95% confidence intervalfor the mean return, i.e. what is: ??

IDENTIFY



Example 12.2 From the data, we calculate:

For this term

and so:

COMPUTE



Example 12.2

We are 95% confident that the population mean, , i.e.the mean return of all publicly traded companies that win

quality awards, lies between 13.20% and 16.84%

Tools > Data Analysis Plus > t-Estimate: Mean is an alternative to themanual calculation

INTERPRET



Check Requisite Conditions The Student t distribution is robust , which means that if the

population is nonnormal, the results of the t-test andconfidence interval estimate are still valid provided that the

population is not extremely nonnormal.

To check this requirement, draw a histogram of the data andsee how bell shaped the resulting figure is. If a histogramis extremely skewed (say in the case of an exponential

distribution), that could be considered extremelynonnormal and hence t -statistics would be not be valid inthis case.



Estimating Totals of Finite Populations Large populations are defined as populations that are at

least 20 times the sample size

We can use the confidence interval estimator of a mean to

produce a confidence interval estimator of the populationtotal :

Where N is the size of the finite population.



Estimating Totals of Finite Populations For example, a sample of 500 households (in a city of 1

million households) reveals a 95% confidence intervalestimate that the household mean spent on Halloween candylies between $20 & $30.

We can estimate the total amount spent in the city bymultiplying these lower and upper confidence limits by thetotal population:

Thus we estimate that the total amount spent on Halloweenin the city lies between $20 million and $30 million.



Identifying Factors Factors that identify the t-test and estimator of :



Inference About Population Variance If we are interested in drawing inferences about a

populations variability , the parameter we need toinvestigate is the population variance:

The sample variance ( s 2 ) is an unbiased, consistent andefficient point estimator for . Moreover,

the statistic, , has a chi-squared distribution,with n 1 degrees of freedom.



Testing & Estimating Population VarianceThe test statistic used to test hypotheses about is:

(which is chi-squared with = n 1 degrees of freedom).


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Testing & Estimating Population VarianceCombining this statistic:

With the probability statement:

Yields the confidence interval estimator for :

lower confidence limit upper confidence limit


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Example 12.3 Consider a container filling machine. Management wants a

machine to fill 1 liter (1,000 ccs) so that that variance of thefills is less than 1 cc 2. A random sample of n=25 1 liter fillswere taken. Does the machine perform as it should at the 5%significance level?

We want to show that:H1: < 1

(so our null hypothesis becomes: H 0: = 1). We will usethis test statistic:

Variance is less than 1 cc 2

IDENTIFY


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Example 12.3 Since our alternative hypothesis is phrased as:

H1: < 1

We will reject H 0 in favor of H 1 if our test statistic falls into

this rejection region:

We computer the sample variance to be: s 2=.8088

And thus our test statistic takes on this value

COMPUTE


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Example 12.3 Since:

There is not enough evidence to infer that the claim is true.

Excel output can also be used for this test

INTERPRET

compare


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Example 12.4 As we saw, we cannot reject the null hypothesis in favor of

the alternative. That is, there is not enough evidence to inferthat the claim is true.Note: the result does not say that the variance is greater than1, rather it merely states that we are unable to show that the variance is less than 1 .

We could estimate (at 99% confidence say) the variance of

the fills


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Example 12.4 In order to create a confidence interval estimate of the

variance, we need these formulae:

we know (n 1)s2 = 19.41 from our previous calculation, andwe have from Table 5 in Appendix B:

COMPUTE

lower confidence limit upper confidence limit


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Example 12.4 Thus the 99% confidence interval estimate is:

That is, the variance of fills lies between .426 and 1.963 cc 2.

COMPUTE


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Identifying Factors Factors that identify the chi-squared test and estimator of :


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Inference: Population Proportion When data are nominal, we count the number of occurrences

of each value and calculate proportions. Thus, the parameterof interest in describing a population of nominal data is the

population proportion p .

This parameter was based on the binomial experiment.

Recall the use of this statistic:

where p-hat ( ) is the sample proportion: x successes in asample size of n items.


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Inference: Population Proportion When np and n(1 p) are both greater than 5, the sampling

distribution of is approximately normal with

mean:

standard deviation:

Hence:


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Inference: Population Proportion Test statistic for p :

The confidence interval estimator for p is given by:

(both of which require that np >5 and n(1 p)>5)


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Example 12.5 At an exit poll , voters are asked by a certain network if they

voted Democrat (code=1) or Republican (code=2). Based ontheir small sample, can the network conclude that theRepublican candidate will win the vote?

That is:H1: p > .50

And hence our null hypothesis becomes:H0: p = .50

IDENTIFY


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Example 12.5 Since our research hypothesis is:

H1: p > .50

our rejection region becomes:

Looking at the data, we count 407 (of 765) votes for code=2.Hence, we calculate our test statistic as follows

COMPUTE


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Example 12.5 Since:

we reject H 0 in favor of H 1, that is, there is enoughevidence to believe that the Republicans win the vote.

Likewise from Excel:

INTERPRET

compare these

or look at p -value


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Selecting the Sample Size The confidence interval estimator for a population

proportion is:

Thus the (half) width of the interval is:

Solving for n , we have:


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Selecting the Sample Size For example, we want to know how many customers to

survey in order to estimate the proportion of customers who prefer our brand to within .03 (with 95% confidence).

I.e. our confidence interval after surveying will be .03,that means W=.03

Substituting into the equation

Uh Oh. Since we havent takena sample yet, we dont have

this sample proportion


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Selecting the Sample Size Two methods in each case we choose a value for then

solve the equation for n.

Method 1 : no knowledge of even a rough value of . This

is a worst case scenario so we substitute = .50

Method 2 : we have some idea about the value of . This isa better scenario and we substitute in our estimated value.


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Selecting the Sample Size Method 1 : no knowledge of value of , use 50%:

Method 2 : some idea about a possible value, say 20%:

Thus, we can sample fewer people if we already have areasonable estimate of the population proportion beforestarting.


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Estimating Totals for Large Populations In much the same way as we saw earlier, when a population

is large and finite we can estimate the total number ofsuccesses in the population by taking the product of the sizeof the population ( N ) and the confidence interval estimator:

The Nielsen Ratings (used to measure TV audiences) usesthis technique. Results from a small sample audience (say2,000 viewers) is extrapolated to the total number of TV sets(say 100 million)


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Nielsen Ratings Example Problem: describe the population of television shows

watched by viewers across the country (population), byexamining the results from 2,000 viewers (sample).

We take these values and multiply them by N =100 million toestimate that between 9.9 million and 12.7 million viewersare watching the Tonight Show.

COMPUTE
http://localhost/var/www/apps/conversion/tmp/scratch_6/References/Nielson.xlshttp://localhost/var/www/apps/conversion/tmp/scratch_6/References/Nielson.xls

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