Before we continue, we should take a moment to review a few simple statistical transformations that are necessary for SDT calculations.
In the model we are using here, SDT is based upon two normal distributions whose variances are equal. To calculate SDT measures, we need to convert p-values to z-scores, and vice versa. In SDT, a z-score measures performance in terms of the number of standard deviations that the signal distribution is above the noise distribution, and a p-value represents the probability of observing a score greater than the observed score if we were sampling from the noise distribution.
Later in this tutorial, you will be using a computer program that performs these calculations for you. However, you will have a better understanding of how SDT measures are calculated after you have performed some of these computations yourself.
To perform the conversions between p-values and z-scores, you can use a z table which can be found in most basic statistics textbooks or you can use the WISE p-z converter applet.
To use the WISE p-z converter, input a p-value in the right text box, or input a z-score into the left text box and press enter on your keyboard. The applet shows either the ‘right-tail’ p-value for a one-tailed application, or the p-value of both tails for a two-tailed application. Select the one-tail or two-tail option depending on what you wish to calculate.
Exercise 1. Use the p/z converter applet to convert the following p-values to z-scores.
| A. z when right-tail p = .025 | |
| B. z when right-tail p = .50 | |
| C. z when right-tail p = .95 |
Check Your Answers
Question | right-tail p | z |
| A | .025 | 1.960 |
| B | .50 | 0.000 |
| C | .95 | -1.645 |
Exercise 2. Now use the applet to perform these conversions.
| A. Right-tail p when z = -1.00 | |
| B. Right-tail p when z = -0.50 | |
| C. Right-tail p when z = 0.50 | |
| D. Right-tail p when z = 2.00 |
Check Your Answers
| A | -1.00 | 0.841 |
| B | -0.50 | 0.691 |
| C | +0.50 | 0.309 |
| D | 2.00 | 0.023 |
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