nuisance parameters are plugged in as discussed above, it is important to use estimates appropriate for the way the data were sampled. In the special case of Z-tests for the one or two sample location problem, the usual sample standard deviation is only appropriate if the data were collected as an independent sample. In some situations, it is possible to devise a test that properly accounts for the variation in plug-in estimates of nuisance parameters. In the case of one and two sample location problems, a t-test does this. Example Suppose that in a particular geographic region, the mean and standard deviation of scores on a reading test are 100 points, and 12 points, respectively. Our interest is in the scores of 55 students in a particular school who received a mean score of 96. We can ask whether this mean score is significantly lower than the regional mean—that is, are the students in this school comparable to a simple random sample of 55 students from the region as a whole, or are their scores surprisingly low? First calculate the standard error of the mean: where is the population standard deviation. Next calculate the z-score, which is the distance from the sample mean to the population mean in units of the standard error: In this example, we treat the population mean and variance as known, which would be appropriate if all students in the region were tested. When population parameters are unknown, a Student's t-test should be conducted instead. The classroom mean score is 96, which is −2.47 standard error units from the population mean of 100. Looking up the z-score in a table of the standard normal distribution cumulative probability, we find that the probability of observing a standard normal value below −2.47 is approximately 0.5 − 0.4932 = 0.0068. This is the one-sided p-value for the null hypothesis that the 55 students are comparable to a simple random sample from the population of all test-takers. The two-sided p-value is approximately 0.014 (twice the one-sided p-value). Another way of stating things is that with probability 1 − 0.014 = 0.986, a simple random sample of 55 students would have a mean test score within 4 units of the population mean. We could also say that with 98.6% confidence we reject the null hypothesis that the 55 test takers are comparable to a simple random sample from the population of