stat-60_lect-08_April_14

stat-60_lect-08_April_14 - STATS 60, Spring 2008 April 14,...

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STATS 60, Spring 2008 April 14, Lecture 08 1 A note on interpolation When discussing percentiles, we said, for example, that the 0 . 25( n + 1) th element in an ordered list corre- sponds to the 1 st quartile. What if the elements were [2,12,34,90]? Then the 1 st quartile would be found at the 1.25th position. We did not discuss how one uses interpolation to obtain the desired result, mostly because I wanted to emphasize the big picture—that the answer was somewhere between the first and second element in the list. But sometimes you need to find an actual number, and so you need to interpolate. The formula for interpolation is low + fraction ( high - low ) , where in this example low = 12 , high = 34 , and fraction = 1 . 25 - 1 = 0 . 25 . If we were interested in the 3.56 th element, then fraction = 3 . 56 - 3 = 0 . 56 , and so on. More on z -scores It seems as if quite a few students are not clear on the basics of z -scores and the standard normal distribution. To help clarify, first note that we sometimes speak of a theoretical (perfect) normal distribution (standard or not). We also speak of the sample distribution, which can be more or less like a normal distribution but never perfectly equal to one. Samples are messy by nature. What happens when we have a sample, from any distribution, and convert it to z -scores? One thing we are not doing is transforming that sample into one that has a perfect standard normal distribution. What we are doing is converting that sample to one that has a mean of zero and SD of one. We then can compare the z -scores against a perfect standard normal distribution. In so doing, the resulting z -values can be used for a variety of purposes. What is it that we do when we calculate z -scores? It is a two step process: for any given sample value, we first subtract the mean and then divide by the sample distribution. But what are these steps doing to the sample as a whole? A non-normal sample distribution with mean of 80 and SD of 24 is shown in Figure 1.A. The first step in calculating z -values is to subtract the sample mean. But that does not change the general shape of the sample distribution, it just shifts it over until it is centered above zero (panel B in the figure). In the second step we divide each centered value by the sample SD. But what does that do? As we have noted in class, multiplying a sample times a constant, such as 4, increases the SD by a factor of 4. Likewise, dividing by 4 reduces the SD by a factor of 4. Now, we have a sample with SD = s x and we divide it by s x , so the new SD is s x s x = 1 . The sample of z -values now has a standard deviation of one. To summarize, subtracting the mean from a sample and dividing by the sample SD merely shifts the entire sample so that it is centered over zero, and then spreads out or squeezes the x -axis so that the new SD is one. Lastly, when we plot a density curve, the area under the curve is always equal to one by design. So
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stat-60_lect-08_April_14 - STATS 60, Spring 2008 April 14,...

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