Ho µ1 = µ2 µ1 µ2 0 - NDSU

COMPARISONS INVOLVING TWO SAMPLE MEANS

Testing Hypotheses

Two types of hypotheses: 1. Ho: Null Hypothesis - hypothesis of no difference.

? ?1 = ?2 or ?1 - ?2 = 0

2. HA: Alternate Hypothesis ? hypothesis of difference.

? ?1 ?2 or ?1 - ?2 0

Two-tail vs. One-tail Tests

? Two-tail tests have these types of hypotheses:

Ho: ?1 = ?2 HA: ?1 ?2

? Note the presence of the equal signs.

? If you reject Ho: ?1 = ?2, you don't care which mean is greater. ? ?1 can be either greater than or less than ?2.

? One-tail tests my have one of the following types of hypotheses:

Ho: ?1 ?2

Ho: ?1 ?2

HA: ?1 < ?2

HA: ?1 > ?2

? Note the presence of greater than or less than signs. ? If you reject Ho:, you are being specific that ?1 can only be on one side

of ?2.

t-Distribution

/2

/2

Two-tail test

Size of the rejection region = / 2

One-tail test Size of the rejection region =

1

Types of Errors 1. Type I Error: To reject the null hypothesis when it is actually true.

? The probability of committing a Type I Error is . ? The probability of committing a Type I Error can be reduced by the investigator

choosing a smaller . ? typical sizes of are 0.05 and 0.01. ? The probability of committing a Type I Error also can be expressed as a

percentage (i.e. *100 %) ? If =0.05, The probability of committing a Type I Error is 5%. ? If =0.05, we are testing the hypothesis at the 95% level of confidence.

2. Type II Error: The failure to reject the null hypothesis when it is false.

? The probability of committing a Type II Error is . ? can be decreased by:

a. Increasing n (i.e. the number of observations per treatment). b. Decreasing s2.

? Increase n ? Choose a more appropriate experimental design ? Improve experimental technique.

Power of the Test ? Power of the test is equal to: 1- . ? Defined as the probability of accepting the alternate hypothesis when it is true. ? We want the Power of the Test to be as large as possible.

Summary of Type I and Type II Errors

True Situation

Decision

Null hypothesis is true Null hypothesis if false

Reject the null hypothesis

Type I Error

No error

Fail to reject the null hypothesis No error

Type II Error

2

Graphical Representation of and Assume you have the hypothesis

Ho:?=?0 HA:?=?1 and we have two normal distributions, with mean equal to ?0 and ?1, respectively, and a variance for both of 2. The following distributions can be drawn,

? Let the decision to reject H0 (i.e., accept HA) or fail to reject H0 be based on one random observation.

? If the random observation is > C then we reject H0. ? Paying attention only to the distribution assuming H0 is true, we can see that a

Type I Error was committed because the rejection region is "truly" part of the distribution. ? Thus, the probability of committing a Type I Error = , which is shown graphically, is solely dependent on the null hypothesis. ? If the random observation is < C, then we fail to reject H0. ? Paying attention to the distribution assuming HA is true, we can see that a Type II Error was committed because we wrongly failed to reject H0 when we should have. ? Thus, the probability of committing a Type II Error = , which is shown graphically, is dependent on the null and alternate hypotheses.

3

Hypothesis Testing

*Summary of Testing a Hypothesis. 1. Formulate a meaningful null and alternate hypothesis. 2. Choose a level of . 3. Compute the value for the test statistic (i.e. t-statistic of F-statistic). 4. Look up the appropriate table value for the test statistic. 5. Formulate conclusions. a. If the tabular statistic > calculated statistic, then you fail to reject Ho. b. If the tabular statistic < calculated statistic, then you reject Ho.

Testing the Hypothesis that a Population Mean is a Specified Number.

Ho: = ! HA: !

Where: = Population mean and ! = specified value

Method 1: t-test = !!!!

!!

Example Suppose we know the average grade point average (GPA) of incoming students is 3.25. We wish to determine if a 3.05 GPA is significantly different than 3.25 at the 95% level of confidence given s2=64 and n=25.

Question 1: Is this a two-tail or one-tail test?

Step 1. Write the hypothesis to be tested: Ho: ? = 3.05 HA: ? 3.05

Step 2. Calculate !

! =

!! =

!

!" = 1.6

!"

Step 3. Calculate t

= !!!!

!!

(3.25 - 3.05)

=

1.6

= 0.125

4

Step 4. Look up the Table t-value

? Df = (n-1) = 25-1 = 24

? / 2 =0.05 (Note the two as the divisor of is indicating that this is a two-tail test. Thus, using the Appendix Table II of your text (Page 606), you need to use the column for =0.025.

? Table t-value=2.064.

Step 5. Make conclusions.

Since t-calc (0.125) is < t-table (2.064) we

fail to reject Ho:

at the 95% level of

confidence.

Thus we can conclude that a GPA of 3.05 is

not significantly different from 3.25 at the

95% level of confidence.

-2.064

0.125 2.064

How would this problem have been different if this was a one-tail test?

Step 1. Write the hypothesis to be tested: Ho: ? 3.05 HA: ? > 3.05

Step 2. Calculate sY

s2 64

sY =

= = 1.6 n 25

Step 3. Calculate t

= !!!!

!!

(3.25 - 3.05)

=

1.6

= 0.125

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