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Test of Hypotheses

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Tests of hypotheses and Significanc e Statistical decision : Decision about populations on the basis of sample information. Ex.) A new vac cine is reall y effective in curing a disease. One educational procedure is better than another.
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Statistical Hypothesis : (Null Hypothesis, Alternative hypothesis) Assumptions or guesses about the populations involved.

Ex.) If we want to decide whether a given coin is loaded, we formulate thehypothesis that the coin is fair.

Ex.) If we want to decide whether one procedure is better than another,

we formulate the hypothesis that there is no difference between theprocedures.

Ex.) Null Hypothesis (Hypothesis) : the coin is fair, there is no difference between the procedures.

 Alternative Hypothesis : alternative to the null hypothesis

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Tests of hypotheses and Significance :

If on the supposition that a particular hypothesis is true we find that results observed in a random sample differ markedly from thoseexpected under the hypothesis on the basis of pure chance usingsampling theory, we would say that the observed differences aresignificant and we would be inclined to reject the hypothesis.

Type I and Type II Errors :

If we reject a hypothesis when it happens to be true,

we say that a TYPE I error has been made.If we accept a hypothesis when it should be rejected,we say that a TYPE II error has been made.

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Level of significance :In testing a given hypothesis, the maximum probabilitywith which we would be willing to risk a Type I error is calledthe level of significance of the test.

Ex.) If a 0.05 or 5% level of significance is chosen, in designing a test of ahypothesis, then there are about 5 chances in 100 that we would reject 

the hypothesis when it should be accept, i.e. whenever the null hypothesisis true, we are about 95% confident that we would make the right decision.The hypothesis has been rejected at a 0.5 level of significance, which

means that we would be wrong with probability 0.05.

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Tests involving the Normal Distribution

Suppose that under a given hypothesis the sampling distributionof a statistic is a normal distribution with mean and standarddeviation .

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We can be 95% confident that, if the hypothesis is true, the Z scoreof an actual sampling statistic will lie between -1.96 and 1.96.

If on choosing a single sample at random we find that the Z score of its statistic lies outside the range -1.96 to 1.96, we would conclude that such an event could happen with the probability of 0.05 if the givenhypothesis were true.

We would say that this Z score differed significantly from what wouldbe expected under the hypothesis, and we would be inclined to reject the hypothesis.

The level of significance represents the probability of our being wrongin rejecting the hypothesis, i.e., the probability of making a Type I

error.

Hypothesis is rejected at 0.05 level of significance or that the Z scoreof the given statistic is significant at a 0.05 level of significance.

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Two-tailed tests, one-tailed tests :

Decision Rule :

(a) Reject the hypothesis at a 0.05 level of significance if the Z score of thestatistic lies outside the range -1.96 to 1.96. Equivalently, theobserved sample statistic is significant at the 0.05 level.

(b) Otherwise, accept the hypothesis.

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p-value : the probability that a value of in the direction(s) of if were true.

: Assertion that a population parameter has a specific value.

: The parameter is greater than the state value.

The parameter is less than the state value.

The parameter is either greater or less than the state value.

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W Ex.) Suppose that the standard deviation, , of a normal distribution is knownto be 3 and asserts that the mean is 12. A random sample of size 36 yields

.

 A)

P value is the probability that a random sample of size 36 would yield a samplemean of 12.95 or more, if true mean were 12.

B)

p-value is Chances are about 97 in 100 that if .

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C) 12 : H0 ! 

12 : H1 { 

p-value is the probability that a random sample of size 36

would yield a sample mean 12.95 or more away from 12,if the true mean were 12.

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Small p-value : Rejecting the null hypothesis in favor of the alternativehypothesis.

Large p-value : Not rejecting the null hypothesis in favor of the alternativehypothesis.

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1. Means X S !

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: Infinite population ORfor sampling with replacement 

: Sampling without replacement from finite population

Special tests of Significance for the Large samples- Many statistics have nearly normal distribution with and .S 

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Ex.) To test that the population mean is ,

(level of significance = 0.05)

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Decision Rule : (two-tailed test)we would accept at the 0.05 level of if for a particularsample of size having mean

and would reject it otherwise.

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Ex.)(level of significance = 0.05)

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Decision Rule : (One-tailed test)we would accept at the 0.05 level

if and reject it otherwise.

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Ex.) a  ! : H0

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Decision Ruleaccept at the 0.05 level

if . Otherwise reject it.

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2. Proportion

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In case ( : actual number of successes in a sample)

: the proportion of successes in a sample.

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21 Q Q !

3. Difference of Means

: sample mean obtained in large samples of size drawn

from population having , .

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: sample mean obtained in large samples of size drawnfrom population having , .

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There is no difference between the population mean ( ).Sampling distribution of difference in means is

approximately normal with , and

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 p

4. Difference of Proportions

: sample proportion obtained in large samples of size

drawn from population having proportion .

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1 p

: sample proportion obtained in large samples of sizedrawn from population having proportion .

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Sampling distribution of differences in proportion

is approximately normal with,

where is used as an estimate of population

proportion .

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Special tests of Significance for the Small samples( )30n

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1. Means X 

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: mean of a sample of size

T  ~ Students distributiont 

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2. Difference of Means

where

( & are sample standard deviation.)

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degree of freedom 221 ! nnv

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Ex1) Find the probability of getting between 40 and 60 heads inclusive

in 100 tosses of a fair coin.

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Ex2) To test the hypothesis that a coin is fair, the following decision rules are

adopted 1) Accept the hypothesis if the number of heads in a single sample of 

100 tosses is between 40 and 60 inclusive, (2) reject the hypothesis otherwise.

(a) Find the probability of rejecting the hypothesis when it is actually correct.(b) Interpret graphically the decision rule and the result of part(a).

(c) What conclusions would you draw if the sample of 100 tosses yields 53

heads? 60 heads?

(d) Could you be wrong in your conclusion to (c)? Explain.

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Ex3) Design a decision rule to test the hypothesis that a coin is fair if a

sample of 64 tosses of the coin is taken and if a level of significance of 

(a) 0.05, (b) 0.01 is used.

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Ex4) In an experiment on extrasensory perception(ESP) a subject in one room is

asked to state the color(red or blue) of a card chosen from a deck of 50 well-

shuffled cards by an individual in another room. It is unknown to the subject how

many red or blue cards are in the deck. If the subject identifies 32 cards correctly,

determine whether the results are significant at the (a) 0.05, (b) 0.01 level of significance. (c) Find and interpret the P value of the

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Ex5) The manufacturer of a patent medicine claimed that it was 90% effective in

relieving an allergy for a period of 8 hours. In a sample of 200 people who had

the allergy, the medicine provided relief for 160 people.

(a) Determine whether the manufacturer¶s claim is legitimate by using 0.01 as

the level of significance.(b) Find the P value of the test.

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Ex6) The mean lifetime of a sample of 100 fluorescent light bulbs produced

by a company is computed to be 1570 hours and the standard deviation of all

the bulbs produced by a company is 120 hours. If is the mean lifetime of all the

bulbs produced by the company, test the hypothesis hours against the

alternative hypothesis hours, using a level of significance of (a) 0.05

and (b) 0.01. (c) Find the P value of the test.

 Q1600! Q

1600{ Q

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Ex7) In Ex.6) test the hypothesis hours against the alternative

hypothesis hours, using a level of significance of (a) 0.05, (b)

0.01. (c) Find the P value of the test.

1600! Q1600£ Q

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Ex8) The breaking strengths of cables produced by a manufacturer have mean

1800 lb and standard deviation 100lb. By a new technique in the manufacturing

process it is claimed that the breaking strength can be increased. To test this

claim, a sample of 50 cables is tested, and it is found that the mean breaking

strength is 18500 lb.(a) Can we support the claim at a 0.01 level of significance?

(b) What is the P value of the test.


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