Introduction ~ Collecting data for the variables to be ...



The Basics ~ The focus for this discussion will be the two sample case for the mean, or the Independent T test. The Independent T test is employed when information regarding the differences between two population means are desired, but the parameter information is unknown. Two randomly selected samples are used to represent the populations and are analyzed. The test is a determination of whether two samples are mathematically different from each other. In this procedure, an inference is being drawn about two unknown populations as represented by the selected samples (See Figure 1).

Figure 1. The Two-Sample Independent T Test – A Test for Mean Differences

The null hypothesis for this test is written as H0: μ1 = μ2 or H0: μ1 - μ2 = 0. This states that the means of the populations (μ1 and μ2) from which we selected the samples (X1 and X2) are essentially equivalent. In other words, there is no difference in the mean values. The nondirectional alternative hypothesis is written as Ha: μ1 ≠ μ2 or Ha: μ1 - μ2 = 0. It is evaluated with a two-tailed test. A directional alternative hypothesis may be used if the researcher suspects an increased or decreased difference in the mean values. A directional alternative hypothesis could be written as Ha: μ1 - μ2 > 0 and is evaluated with a one-tailed test. In either case, if there is a sufficiently large difference in the mean values, then we have a statistically significant difference and the null hypothesis is rejected.

The Independent T Test Statistic ~ The test statistic for the two-sample (Independent) T test should look very similar to the t test score calculations previously discussed. The two relevant formulas used to calculate the Independent T test statistic are:

Naturally X1 is the sample mean for group 1 and X2 is the sample mean for group 2. The standard error of the mean differences (sx1 – x2) is the new element. This value is calculated by using the pooled estimate of the population variance, or (s2) and the sample sizes (n1 and n2). This value is an estimate of σx1 – x2. Once the test statistic has been calculated, it is evaluated against the critical values (from a table of the t distribution) based on the selected alpha level, the degrees of freedom, and the directionality of the alternative hypothesis.

The T calculated value must exceed the T critical value for the null hypothesis to be rejected. Because the T distributions are a family of distributions, the shape of the bell-shaped curve will change as the sample size changes. As sample size increases, the differences between the normal distribution and the corresponding T distribution decrease. From a practical standpoint, the normal distribution is an adequate approximation of the T distribution when the degrees of freedom exceeds a sample size of 120, however, the value of n=30 is often touted as sufficiently large for the approximation of the normal curve.

The Confidence Interval ~ In addition to the hypothesis test, we can also construct a confidence interval to make a decision regarding the null hypothesis. The confidence interval is a range of values that are tenable for the unknown population mean. We can establish a specific level of certainty that we are confident contains the population mean. The formula needed to create this range of mean values is CI(1-α) = X1- X2 ± (Tcv)(sx1 – x2). The most regularly constructed confidence interval is the 95% confidence interval, so the corresponding alpha level would be .05.

The sample means and the standard error of the mean in the formula are the same values used for the T test statistic. The critical value of the confidence interval will be different from the critical value of the hypothesis test when a one-tailed test is conducted. The confidence interval is based on a two-tailed test, so be careful to select the appropriate critical value. To confirm your decision from the hypothesis test, the value of the difference between the sample means (X1-X2) should NOT be contained within the range of values for the confidence interval. If this mean difference value falls within the range of values for the confidence interval, an incorrect decision regarding the null hypothesis was made. This

confidence interval should never be considered to specify the chances (or probability) that the population mean is “caught” by the interval.

Example ~ As a statistician, you are interested in the following research question: Do group 1 and group 2 differ in math achievement (as measured by the variable MATHACH) as the result of two teaching methods? Two Random SAMPLES are taken from two populations of children. The mean for the first sample (the control group - X1) was 12.16. The mean for the second sample (the experimental group - X2) was 14.30. The standard deviation for the first sample (s1) was 6.38 and the standard deviation for the second sample (s2) was 6.70. The sample size for group 1 (n1) is 280 and the sample size for group 2 (n2) is 220. To answer our research question, we will work with a five-step process.

STEP 1: STATE THE HYPOTHESIS

The Null Hypothesis H0: μ1 - μ2 = 0

The (Non-Directional) Ha: μ1 - μ2 ≠ 0

Alternative Hypothesis

STEP 2: SET-UP THE REGION OF REJECTION

We have a non-directional alternative hypothesis, so the region of rejection will be split between the two tails of our standard curve (we are running a two-tailed test). I have chosen to use an alpha level of .05 (α = 05), so the Tcv is ±1.96 with 498 degrees of freedom.

The critical values (Tcv) for the region of rejection.

STEP 3: COMPUTING THE TEST STATISTIC

Here are the two formulas you’ll need to answer this question.

The T test statistic: The standard error of the mean:

STEP 4: DECIDE ABOUT H0

Since Tcalc (-3.64) does fall within the region of rejection, we conclude the following: The math achievement score means for the two populations as represented by the two samples are statistically significant from each other at an alpha level of .05 and 498 degrees of freedom. We reject the null hypothesis of H0: μ1 - μ2 = 0.

STEP 5: CALCULATE THE CI95

Use the following formula to calculate the 95% confidence interval: CI95 = X - X ± (Tcv)(sx1-x2).

CI95 = (-2.14) ± (1.96)(.58765) => (-3.29,-0.98)

The mean difference of 0 is not captured by the 95% confidence interval, so we can confirm our decision to reject the null hypothesis.

The Assumptions/Requirements for the Independent T Test ~ There are two important assumptions for using the Independent T test procedure. The first assumption is the Assumption of Independent Samples. This refers to the scores of one data set being unrelated to the scores of the second dataset. If the samples are selescted at random, or if subjects are assigned randomly to experimental treatments, the two samples will initially vary in terms of random fluctuation. The second assumption is the Assumption of Homogeneity of Variance. For the calculation of the estimated standard error for the difference of the

means, the pooled estimate of the sample variances is used. In order to use the pooled variances, we must assume that the variances from the two representative populations are essentially equivalent, or that σ21 = σ22. The effect of violating this assumption depends on the sample sizes. If the two sample sizes are equal, then the effect of the violation of the homogeneity assumption is not serious. But if the samples sizes are not equal and the variances are not equal, then this could be problematic. To test whether the variances are equivalent, the F ratio must be used. The null hypothesis for this test is, H0: σ21 = σ22 or written in a different form, H0: σ21/σ22 = 1. The statistic used to test this hypothesis is as follows:

If the F calculated value exceeds the F critical value at a specified alpha level with n-1 and n-2 degrees of freedom, then the assumption of the homogeneity of variances has been violated. Alterations to the formulas for the standard error of difference between the means and the related degrees of freedom for the test statistic must be made. The new formulas are:

Effect Size Calculations ~ Computing the effect size for the Independent T test can be completed in one of two ways. One of the most common effect size formulas is the Cohen’s d. It can be calculated in the following manner:

From the previous data presentation, d = 6.20.

A second formula is a standard score format detailed by the formula:

This calculation gives you the difference of the two means in the standard deviation of the control group (See Figure 2). From the previous data presentation, Δ = .335

Figure 2. The Δ Effect Size Measure

Related Information ~

• Sampling Distributions

• One-Sample Case for the Mean (z Test)

• The Two-Sample Case for the Mean (Paired T Test)

• ANOVA Designs: One-way (Single Factor) ANOVA with Two Levels

References ~

Glass, G. V., & Hopkins, K. D. (1996). Statistical methods in education and psychology (3rded.). Boston: Allyn and Bacon.

Hinkle, D. E., Wiersma, W., & Jurs, S. G. (1998).

Applied statistics for the behavioral sciences (4thed.). New York: Houghton Mifflin.

Additional Notes ~

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

+

T =

Fcalc =

2.5%%

-3.29 -0.98 0

95%

Where: s2 =

Population1

Sample 2 (X2) represents one population (¼2) to be tested against the X1 statistic

¼2 is unknown

Ã2 is unknown

¼1 is unknown

Ã1 is unknown

Population2

(X1 - X2)

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