Inferences About Means Values Based on a Single Sample
Let’s try to clarify this basic idea of statistics with a specific illustration. Consider an experiment that may have a direct bearing on how well you understand what’s to come. We have written the book on the assumption that the average reader has an intelligence quotient (IQ) greater than 100. If we were wrong, then readers would not be able to decipher these ramblings, and our royalties wouldn’t buy us a Big Mac. So we’d like some reassurance that we’ve targeted the book about right. How will we test this hypothesis?
To begin with, we’ll do as all good researchers are supposed to do and rephrase it as a null hypothesis; that is to say, we’ll start off by assuming that the readers are no different in IQ than the population at large. We will then do our darnedest to reject this hypothesis. So we phrase the null hypothesis as follows:
H0: mean IQ of readers = mean IQ of general population
We have deliberately chosen an experiment involving IQ tests because they are carefully standardized on very large samples to have a normal distribution with a mean of 100 and an SD of 15. So, when we state the null hypothesis, we are assuming momentarily that the readers are a random sample of this population and have a true average IQ of 100. This sounds weird because that’s what we don’t want to be the case, but bear with us.
Continuing along this line of reasoning, we then assume that if we sample 25 readers repeatedly, give them an IQ test, and calculate their mean score, these calculated means would be distributed symmetrically around the population mean of 100. The question remains, “What is the expected random variation of these mean values?” Well, from the discussion in the previous section, the SE of these means is the SD of the original distribution divided by the square root of the sample size. In our case, this is 15/√25 = 3.0.
So, suppose we went ahead with the experiment involving 25 readers (we may have to give away a few complimentary copies to pull it off) and found their mean IQ to be 107.5. We want to determine the likelihood of obtaining a sample mean IQ of 107.5 or greater from a random sample of the population with a true mean of 100 ( Fig.1129\). What we are seeking is the area in the tail to the right of 107.5. The way we approach it is to calculate the ratio of (107.5 – 100), or 7.5, to the SE of 3.0.
The ratio, 7.5/3.0 = 2.5, tells us how far we are out on the standard normal distribution; we then consult a table of values of the normal distribution and find out that the area in the tail is 0.006. Thus, the likelihood of obtaining a sample IQ of 107.5 or greater by chance, under the null hypothesis that the two population means are equal, is 0.006, or approximately 1 in 160. Because there is a low probability of obtaining a value this large or larger by chance, we reject the null hypothesis and conclude that the readers, with their average IQ of 107.5, are drawn from a different population (that is, they really do have an IQ greater than 100).
This approach of comparing a sample mean with a known population mean by calculating the ratio of the difference between means to the SE is known as the z test.
Figure 1129 – Figure 3-4: Distribution of mean IQ of sample size = 25.
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Content on this page was last changed on March 19, 2009.
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| 5343. | Norman GR, Streiner DL. PDQ Statistics . 3rd ed. Hamilton, Ontario: BC Decker Inc.; 2003. |