Z-Test

Z-TEST OF A SINGLE MEAN

The z-test will be performed by evaluating the ratio (x¯−μ)/σx¯ —where x¯ is the sample mean, μ is the population mean under H0, and σx¯ is the standard error of the mean (standard deviation of the sampling distribution of means under H0)—when we are interested in testing the foregoing hypothesis about the mean IQ of the population. This ratio transforms a mean deviation into a z-score. Now suppose that in a sample of n = 900, randomly drawn with replacement, x¯ = 102 and s = 10. The sample is large enough, according to the central limit theorem, so the sampling distribution should be normal with mean equal to the population mean μ and variance σx¯ = σ2/n. Because σ is unknown, the variance of the sampling distribution will be estimated as ^σ2x¯ = s2/n (instead of σ2/n), and its square root gives ^σx¯. Our sample result, x¯ = 102, is subsumed under the sampling distribution with mean 100 and variance s2/n = 102/900 = 0.111. Its z-score is then z = (102 − 100)/√0.111 = 6. For this value of z or higher, we find in the standardized normal table a probability of 0.000001, which is so small (much smaller than a probability of making a type I error of α = .05 or even of α = .001) that it is highly unlikely to find this sample result under the null hypothesis. Hence, the null hypothesis is rejected, and we have found empirical support for the alternative hypothesis that μ>100. We may mention here that we do not know what the probability of type II error β is, but with a sample of large size (n = 900), we can be confident that β will be small.

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