This means that the difference in group means is 12.79 standard deviations away from the mean of the. Example of a p-value The two-tailed t-test of the difference in test scores generates a t-value of 12.79. That value is then called a critical value, and is sometimes denoted as z 90, indicating that 90% of the area is to the left of that value. The t-score which generates a p-value below your threshold for statistical significance is known as the critical value of t, or t. The area under the curve, to the left of z is 0.90. In other words we are trying to find the height that corresponds to being taller than 90 percent of the women (in this case). If instead of looking for the probability, we look for the value of x for a given percentile (= probability or area), the problem can be called finding the x-score.Įxample: Suppose that for the example above we try to find the height corresponding to the 90th percentile. For a population with unknown mean and known standard deviation, a confidence interval for the population mean, based on a simple random sample (SRS) of size n, is + z, where z is the upper (1-C)/2 critical value for the standard normal distribution. Notice that the first value is negative, which means that it is below the mean. For the second problem we have to values of x to standarize, x 1=60.3 and x 2=65. This is the best answer based on feedback and ratings. The standard normal distribution table is a compilation of areas from the standard normal distribution, more commonly known as a bell curve, which provides the area of the region located under the bell curve and to the left of a given z- score to represent probabilities of occurrence in a given population. Round to two decimal places, and enter the answers separated by a comma if needed. P(X>70.4) = P(Z>2.56) =0.5 - 0.4948 = 0.0012.Ģ. Using the z table (The Standard Normal Distribution Table), find the critical value (or values) for the left-tailed test with 0.09. Therefore, the probability P(X>70.4) is equal to P(Z>2.56), where X is the normally distributed height with mean $\mu$ = 64 inches and the standard deviation $\sigma$ = 2.5 inches (X~N(64,2.5) for short), and Z is a standard normal distribution (Z~N(0,1)).