Many analysis tools that are used for hypothesis testing in STATISTICA give a calculated test statistic and a p-value. (For more information about hypothesis testing, see the article, How to Interpret Statistical Analysis Results.) At times, it may be necessary in your hypothesis test to report the test’s critical value. This article describes how to find the critical value for a statistical test.
In statistical hypothesis testing, a critical value is the cut-off value for the computed test statistics that defines statistical significance of the test. This statistical test follows a known distribution, and the analyst will have selected alpha, the type I error rate. The critical value is the value from the distribution of the test for which P(X>X critical value) = alpha, where X is the observed test statistic and X critical value is the critical value for the test.
Statistical tests can be either one or two-sided, and this is important for finding the critical value of a test. Below are three possible null and alternative hypotheses for a single sample mean test. Given the same alpha, each of the three tests would have a different critical value (or values).
The distribution a test follows is an important piece of finding the critical value. Tests comparing population means may follow a standard Normal or Studentized T distribution. ANOVA significance tests follow the F distribution. Chi Square is another common distribution of test statistics. T, F, and Chi Square all have one or more degree of freedom parameters that are needed to find the critical value.
You can use STATISTICA's Probability Distribution Calculator to find values or areas from various distributions. This tool can be used to find the critical value for a test. Once the distribution is selected, alpha is entered as well as any other required parameters such as degrees of freedom. Then the tool computes the critical value of the test.
For these examples, reading comprehension was measured for students in three grade levels who were administered one of two teaching methods.
In this one-sample t-test, researchers hypothesized that the average reading comprehension score is significantly different from 55. This is a two-sided test. Data was collected and the one-sample t-test was calculated with the results shown below.
The calculated test statistic is -1.96166 and p=0.054526. At alpha = 0.05, this two-sided test is not statistically significant, p-value = 0.054526 > 0.05 = alpha. Let’s find the critical value of the test.
On the Statistics tab in the Base group, click Basic Statistics to display the Basic Statistics and Tables dialog box. Select Probability calculator.
Click the OK button to display the Probability Distribution Calculator.
Since the test follows the t (Student) distribution, select this in the Distribution list. Select the Inverse, Two-tailed and (1-Cumulative p) check boxes. Enter 59 as df, which comes from the t-test output above. We are using alpha = 0.05, so enter this value for p. Click the Compute button to calculate the t critical value. It is found to be 2.000995.
Since the test has a two-sided alternative, the critical region is -2.000995 < t calc < 2.000995. The computed test statistic is -1.96166, which is between -2.000995 and 2.000995, so the conclusion is to fail to reject the null hypothesis. There is insufficient evidence to conclude that the average reading comprehension score is significantly different from 55.
Now let’s assume that the test had a one-sided alternative, i.e. we hypothesize that average reading comprehension scores are significantly less than 55. (Note that this example is for illustrating the calculation of critical values only. It is not an acceptable statistical practice to change your hypothesis to suit the data. The one-sided alternative critical values are calculated slightly different and this example aims to highlight this difference.)
The critical value for this test is very similar, but the Two-tailed check box is cleared. Additionally, depending on the direction of the hypothesis, the (1-Cumulative p) check box may also be cleared. The alternative hypothesis is that µ<55. The direction of the inequality also tells the direction of the test. For this test, the (1-Cumulative p) check box should be cleared.
For the one-sided test, the critical value is -1.671093. If t calc < -1.671093, reject H0. Thus, -1.96166<-1.671093, so reject H0. The conclusion for this test is different from the two-sided test, which failed to reject the null. Here, the conclusion would be that the average reading comprehension scores are significantly less than 55. (Note that the t-test output from STATISTICA gives p-values based on the two-sided alternative. For a one-sided alternative test, the p-value should be divided by 2.)
Using an example from an ANOVA analysis, let’s compute the critical F value for the significance test for an effect. The ANOVA table below tests for a significant effect of Method, Grade, and the interaction between the two on a student’s reading comprehension. This ANOVA output table gives the calculated F statistics, the associated p-values, and degrees of freedom for each test. (The F distribution requires 2 degrees of freedom parameters, nominator and denominator. The F statistics are calculated as MEeffect/MSerror. The degrees of freedom for the numerator and denominator come from the effect and error respectively. Additionally, F tests are always one-sided tests.) Using this information and the Probability Distribution Calculator, we can find the critical values of each test for a given alpha.
In the Probability Distribution Calculator, select F (Fisher) in the Distribution list. Select the (1-Cumulative p) check box. Enter numerator and denominator degrees of freedom, df1 =1 and df2 =54. We are using alpha = 0.05, so enter this value for p.
Click Compute to compute the F statistic: 4.019541. This is the critical value for the significance test for the METHOD effect.
Given that the null hypothesis is true, P(F calc >4.019541)=0.05, so if F calc > F critical = 4.019541, reject H0. The computed test statistic is 1.884, which is less than 4.019541, so the conclusion is to fail to reject the null hypothesis. There is insufficient evidence to conclude that there is a significant difference in the teaching methods.
Critical values can be calculated for any statistical test that follows a known distribution. The Probability Distribution Calculator makes it easy to find these test critical values.
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