Hypothesis Testing is a procedure, based on sampling data and probability, used to test statements regarding a characteristic of one or more populations.

General Procedure involves: Making a statement regarding populations Collecting Sample Data Analyzing the data to access the

probability of the statement’s validity

HYPOTHESIS TESTING IS A STEP-BY-STEP PROCEDURE: 1. From a statement of the problem (the claim),

determine the null and alternative hypothesis and the type (direction) of test.

2. Determine the significance at which the test should be run.

3. Test the null hypothesis from the experimental data using either the Classical Approach (Critical Value Method) or the P-Value Method.

4. Determine the conclusion concerning the null hypothesis.

5. Determine the conclusion concerning the original claim

The Null Hypothesis ( ) is a statement that nothing has changed, that populations are the same regarding some characteristic.

The Alternative Hypothesis ( ) is a that which we are trying to find evidence to support, that the characteristic of a population has changed.

There are four possible claims and the resulting Null and Alternative Hypothesis to go with them.

And from the Alternative Hypothesis, the type of test (direction of test) can be determined.

oH

1H

CLAIM NULL ALTERNATIVE DIRECTION OF TESTS

A = B A = B A ≠ B TWO SIDED

A ≠ B A = B A ≠ B TWO TAILED

A < B A = B A < B LEFT TAILED

A > B A = B A > B RIGHT TAILED

1H0H

EXAMPLE:The claim is made that the mean of a populationhas changed; in other words the current mean is differentthan the former mean.Null Hypothesis: Alternative Hypothesis: Type of Test: Two Tailed

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There are four possible outcomes to an Hypothesis Test:

1. The claim was true and the test found it to be true.

2. The claim was false and the test found it was false.

3. The claim was true but the test found it false.

4. The claim was false but the test found it to be true.

Types of Errors:

Type I Error = P(rejecting the Claim when it is true) =

Type II Error = P(not rejecting the Claim when it is False) =

Which is most important not to make?

REALITY ABOUT CLAIM

TRUE FALSE

CONCLUSION ABOUT

CLAIM

TRUE CORRECT CONCLUSIO

N

TYPE II ERROR

FALSE TYPE I ERROR

CORRECT CONCLUSIO

N

Significance is , the Type I error. It is the Probability of being wrong about the Claim when it is true and is typically low: 0.20, 0.10, 0.05, 0.02, 0.01.

As The significance (type I error) decreases, the type II error increases.

Significance is the area of the tail(s) of a normal distribution. If the direction of the test is right sided, the area in the right tail is the significance. If the direction of the test is left sided, the area in the left tail is the significance. If the direction of the test is two sided, the significance is divided evenly between the two tails.

Types of Tests: Classical Approach (Critical Value Method)

– From the experimental data, find a test statistic. From the significance, find the Critical Value. If the Test Statistic is in the area of the tail of the significance, then the Null Hypothesis is REJECTED.

P-Value Method – From the experimental data find the test statistic and from that the p-value (area to the right or left of the test statistic in the tail) and if the p-value is less than the significance, then the Null Hypothesis is REJECTED.

Conclusions: If the Null Hypothesis is REJECTED, the Alternative Hypothesis is ACCEPTED.

THE NULL HYPOTHESIS CAN NEVER BE ACCEPTED SO THE ALTERNATIVE HYPOTHESIS CAN NEVER BE REJECTED.

The only thing we can do is FAIL TO REJECT the Null and FAIL TO SUPPORT the Alternative.

Why?

Conclusion about the claim: If the Null is Rejected and the Null is the

Claim, the claim is Rejected. If the Null is Rejected (then the Alternative

is Accepted) and the Alternative is the Claim, the Claim is accepted.

If the Null is Not Rejected and the Null is the Claim, the Claim is Not Rejected.

If the Null is Not Rejected (then the Alternative is not Supported) and the Alternative is the Claim, the Claim is Not Supported.

Examples.

Example: The mean score on all SAT tests for Math

reasoning is 516. A certain company states the mean score of students who take their SAT prep course is higher than 516.

Find both hypothesis. If the conclusion about the Null is not rejected,

what is the conclusion about the claim? If the conclusion about the Null is wrong, what

type of error has been made?

The drug Lipator is given to reduce Chloresterol. In trials 19 out of 863 patients taking Lipator complained of flu like systems. It is known that 1.9% of patients taking a competing drug had flu like symptoms. Is there reason to believe that Lipator users experienced flu like symptoms more than 1.9% of the time to a significance level of 0.01.

Find the two hypothesis, direction of test, significance and experimental data.

First use the Classical (Critical Value) Method to test the Null: Find the Critical Value – the value of

Z for the area of significance (i.e. Use InvNorm(α) for left tailed, InvNorm(1-α) for right tailed or InvNorm(α/2) for 2 tailed).

Find the Test Statistic -

0

0 0

p p

p q

n

Second use the P-value method to test the Null: Find the Test Statistic

Use the Test Statistic to find the P-value (i.e. Use Normalcdf(-10^99, Test Statistic) for right tailed or Normalcdf(Test Statistic, 10^99) for left tailed or 2*Normalcdf(Test Statistic) for two tailed).

Use 1-PropZTest.

0

0 0

p p

p q

n

From the Classical (Critical Value) Method, if the Test statistic is in the Critical Region, then REJECT the Null, otherwise FAIL TO REJECT the Null.

If the P-value is less than the significance, then REJECT the Null, otherwise FAIL TO REJECT the Null.

Determine the conclusion about the Claim.

A poll of 676 adults aged 18 and older found that 352 believed they would not have enough money to live comfortably in retirement. Does this suggest that half of the population of adults aged 18 and older believe they will not live comfortably in retirement? Use a significance of 0.05.

58% of females 15 years old or more lived alone in 2000. Recently a survey of 500 females aged 15 and older found that 285 lived alone. Has the proportion of females living alone changed to a level of significance of 0.1.

Do other examples just giving values.

The process is the same but will use the t-distribution.

Test Statistic: or TTest Critical Values: InvT or t-table as was

done in Chapter 9. P-Value: t-table or TTest

00

xt

sn

Do students who learned English and another language score lower on SAT Critical Reading exam. A sample of 100 such persons were given the test and scored a mean of 485 with a std. dev. of 116. Test the hypothesis that such students do score lower than the mean of all test takers (501) to a significance of 0.10.

The mean household energy expenditure was $1493in 2001. The administrating organization believes this has changed. A random sample of 35 households found a mean (adjusted to 2001 dollars) household energy expenditure of $1621 and a std. dev. Of $321. Test their belief to a significance of 0.05.

Do problems just from numbers

To Test a Hypothesis about a Standard Deviation, there is not a calculator function and there is no p-value test.

Test Statistic: .

Critical Value: Use Chi-Squared Table

220 2

( 1)n s

A machine fills bottle with 64 oz. of liquid. The quality control manager has found the volumes in the bottle to be normally distributed with a std. dev of 0.42. The process engineer makes some changes to the machine and believes the standard deviation will be reduced. The manager picks a sample of 19 bottles and has a sample deviation of 0.38. Test his hypothesis to a 0.01 significance.

The NCAA as requirements for the circumference of a softball for competition. One is that a manufacturer must have a standard deviation of the circumference less than 0.05 inches. A representative of the NCAA believes a manufacture does not meet the requirement. A sample of 20 softballs are measured and the sample standard deviation is 0.09. Is there sufficient evidence to support the representative’s claim to a significance of 0.05?

Do problems with numbers only.

Proportion Mean (w/s) Std. Dev.

Must be given:

Distribution Z t

Test Statistic

Critical Value InvNorm() InvT() or t-table

p-value Normalcdf() Tcdf() N/A

Calculator Function

1-PropZTest Ttest N/A

Reject Null if p-value < significance or if test statistic is in the Critical Region (Tails)

0, , ( ), ,x n p p 0, , , ,x n s 0, , ,n s

20

0

xt

sn

00

0 0

ˆ

1

p pZ

p pn

2

20 2

( 1)n s

2 table