Technology - Winona State University



Making Decisions for the Difference Between

Two Independent Population Means

The appropriate test procedure depends upon whether or not we can assume that the population variances are equal or not. In the example covered in this handout we will examine methods used to determine whether or not the population variances can be assumed to be equal for our study.

Test Statistics and Confidence Interval Formulae

Assuming Equal Population Variances/Standard Deviations (pooled t-Test)

i.e., [pic]

Test Statistic

[pic]

 

Assuming Unequal Population Variances/Standard Deviations [pic]

Test Statistic

[pic][pic]~ t-distribution df = min([pic])

Confidence Interval for the Difference Between the Pop. Means

For either case a [pic]is given by the following

(estimate) + (t-table value)(standard error of estimate)

[pic] ( (margin of error is computing using the appropriate standard error

and df based upon the equality of pop. variances assumption)

EXAMPLE 1: NORMAL HUMAN BODY TEMPERATURE:

FEMALES vs. MALES

| Data File: | Bodytemp.JMP |

|Background:  |The data for this example comes from a study of body temperature and pulse rate for adults. |

|Variables:  |Gender: gender of the individual |

| |Temperature: body temperature (degrees Farenheit) |

| |Heart.rate: heart or pulse rate (beat per minute) |

|Goal: |To be able to complete (and interpret the output from) a two-sample t-test in JMP. |

|Question of Interest:  |Do men and women have the same normal body temperature? Putting this statement into a statement involving parameters that |

| |can be tested: |

| |HO: μF = μM |

| |HA: μF ≠ μM |

| |  or |

| |HO: μF - μM= 0 |

| |HA: μF - μM  ≠ 0 |

 Intuitive Decision

In order to determine whether or not the null or alternative hypothesis is true, you could review the summary statistics for the variable you are interested in testing across the two groups.  Remember, these summary statistics and/or graphs are for the observations you sampled, and to make decisions about all observations of interest, we must apply some inferential technique (i.e. hypothesis tests or confidence intervals)

One of the best graphical displays for this situation is the side-by-side boxplots.  To get side-by-side boxplots, select Analyze > Fit Y by X.  Place Gender in the X box and Temperature in the Y box.  Place the mean diamonds on the boxplots and jitter the points.  The more separation there is in the mean diamonds, the more likely we are to reject the null hypothesis (i.e data tends to support the alternative hypothesis).

[pic]

[pic]

Assumptions

1. The two groups must be independent of each other.

2. The observation from each group should be normally distributed.

3. Decide whether or not we wish to assume the population variances are equal.

Assessing Normality of the Two Sampled Populations

To assess normality we select Normal Quantile Plot from the Oneway Analysis pull-down menu as shown below.

[pic]

Checking the Equality of the Population Variances

To test the equality of the population variances select Unequal Variances from the Oneway Analysis pull-down menu.

[pic]

The test is:

[pic]

JMP gives four different tests for examining the equality of population variances.  To use the results of these tests simply examine the resulting p-values. If any/all are less than .10 or .05 then worry about the assumption of equal variances and use the unequal variance t-Test instead of the pooled t-Test.

[pic]

Here we can see that all of the p-values exceed the 0.05 (i.e. 5%). What does this mean?   What is your conclusion about the validity of the equality of the population variances assumption? 

Performing the Test

To perform the two-sample t-test for independent samples:

• assuming equal population variances select the Means/Anova/Pooled t option from Oneway-Analysis pull-down menu.

• assuming unequal population variances select t-Test from the Oneway-Analysis pull-down menu.

[pic]

Several new boxes of output will appear below the graph once the appropriate option has been selected, some of which we will not concern ourselves with. The relevant box for us will be labeled t Test as shown below for the mean body temperature comparison.

[pic]

• What is the test statistic for this test?

 

• What is the p-value?

 

• What is your decision for the test? 

 

• Write a conclusion for your findings. 

 

Construct and Interpret a 95% CI for the Difference in the

Mean Body Temperatures [pic]

For body temperature and gender example we have:

Interpretation of the CI for [pic]

Nonparametric Alternative to the t-Test (not on an exam)

If we find that the populations we are sampling from are not normally distributed or if our samples are too small to reasonably assess normality we could use a nonparametric test instead. Nonparametric tests typically use the ranks of the observations rather than the observed values themselves to compare the “size” of the values from the two populations of interest. All the observations from both samples are ranked from smallest to largest with the smallest observation receiving a rank of 1. The general idea of the test is to compare the ranks of assigned to the observations from each population. If one population generally has larger values than the other, the observations sampled from that population should have significantly higher ranks than the observations sampled from the population with smaller values. If the discrepancy in the ranks is extreme enough we will reject the null that says the population distributions are the same in terms of “typical” value in favor of the alternative which says that one population has larger values than the other.

To perform a nonparametric test of this hypotheses in JMP select Nonparametric > Wilcoxon from the Oneway Analysis pull-down menu.  The normal approximation p-value is virtually identical to the normal approximation to the Mann -Whitney test.  Here the conclusion is the same as the parametric test, namely males and females have significantly different body temps.

[pic]

Example 2: Gender Comparisons of Drinks Per Episode

for WSU Students

Is there evidence to suggest that the average number of drinks per episode for male drinkers is greater than that for female drinkers? Using the WSU student survey data in the file STAT 110 Survey we will examine this question.

[pic]

Analysis in JMP

Using Analyze > Fit Y by X with Y = Howmuch, which is the number of drinks per episode, and X = Gender we obtain the following. Select Oneway Analysis... > Normal Quantile Plot to assess normality of the response for both groups.

[pic]

Can we assume the population variances are equal? Select Oneway Analysis > UnEqual Variances to check this assumption. The results are shown on the following page.

[pic]

What do we conclude?

Using the appropriate t-Test given the variance test results we select Oneway Analysis... > t Test.

[pic]

Results

[pic]

Additional Notes:

-----------------------

Because we have concluded that the equality of variance assumption is reasonable for these data we can refer to the output for the t-Test assuming equal variances.

Normality appears to be satisfied here.

Summary Statistics

[pic]

[pic]

Because we have no evidence against the equality of the population variances assumption we will use a pooled t-Test to compare the population means.

[pic] mean body temperature for females.

[pic] mean body temperature for males.

Summary Statistics

[pic]

HYPOTHESES

[pic]

[pic]= Pooled-estimate of the common variance

This is the conservative approach. As an alternative we could use Welch’s df formula, but for Welch’s we will let the computer do the dirty work!

[pic]

CI for [pic]

[pic]

[pic]

[pic] mean number of drinks per episode for WSU females

[pic] mean number of drinks per episode for WSU males

Both distributions are skewed right, however our sample sizes here are quite large so normality is less critical.

Conclusion:

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