Critical Value

The critical value (CV) is used in significance testing to establish the critical and noncritical regions of a distribution. If the test value or statistic falls within the range of the critical region (the region of rejection), then there is a significant difference or association; thus, the null hypothesis should be rejected. Conversely, if the test value or statistic falls within the range of the noncritical region (also known as the nonrejection region), then the difference or association is possibly due to chance; thus, the null hypothesis should be accepted.

[Page 308]

When using a one-tailed test (either left-tailed or right-tailed), the CV will be on either the left or the right side of the mean. Whether the CV is on the left or right side of the mean is dependent on the conditions of an alternative hypothesis. For example, a scientist might be interested in increasing the average life span of a fruit fly; therefore, the alternative hypothesis might be H1: μ > 40 days. Subsequently, the CV is on the right side of the mean. Likewise, the null hypothesis would be rejected only if the sample mean is greater than 40 days. This example would be referred to as a one-tailed right test.

To use the CV to determine the significance of a statistic, the researcher must state the null and alternative hypotheses; set the level of significance, or alpha level, at which the null hypothesis will be rejected; and compute the test value (and the corresponding degrees of freedom, or df, if necessary). The investigator can then use that information to select the CV from a table (or calculation) for the appropriate test and compare it to the statistic. The statistical test the researcher chooses to use (e.g., z-score test, z test, single sample t test, independent samples t test, dependent samples t test, one-way analysis of variance, Pearson product-moment correlation coefficient, chi-square) determines which table he or she will reference to obtain the appropriate CV (e.g., z-distribution table, t-distribution table, F-distribution table, Pearson's table, chi-square distribution table). These tables are often included in the appendixes of introductory statistics textbooks.

For the following examples, the Pearson's table, which gives the CVs for determining whether a Pearson product-moment correlation (r) is statistically significant, is used. Using an alpha level of .05 for a two-tailed test, with a sample size of 12 (df = 10), the CV is .576. In other words, for a correlation to be statistically significant at the .05 significance level using a two-tailed test for a sample size of 12, then the absolute value of Pearson's r must be greater than or equal to .576. Using a significance level of .05 for a one-tailed test, with a sample size of 12 (df = 10), the CV is .497. Thus, for a correlation to be statistically significant at the .05 level using a one-tailed test for a sample size of 12, then the absolute value of Pearson's r must be greater than or equal to .497.

When using a statistical table to reference a CV, it is sometimes necessary to interpolate, or estimate values, between CVs in a table because such tables are not exhaustive lists of CVs. For the following example, the t-distribution table is used. Assume that we want to find the critical t value that corresponds to 42 df using a significance level or alpha of .05 for a two-tailed test. The table has CVs only for 40 df (CV = 2.021) and 50 df (CV = 2.009). In order to calculate the desired CV, we must first find the distance between the two known dfs (50 − 40 = 10). Then we find the distance between the desired df and the lower known df (42 − 40 = 2). Next, we calculate the proportion of the distance that the desired df falls from the lower known

. Then we find the distance between the CVs for 40 df and 50 df (2:021 − 2:009 =: 012). The desired CV is .20 of the distance between 2.021 and 2.009 (.20 ×: 012 =: 0024). Since the CVs decrease as the dfs increase, we subtract .0024 from CV for 40 df (2:021 – .0024 = 2:0186); therefore, the CV for 42 df with an alpha of .05 for a two-tailed test is t = 2.0186.

...