Ординатура / Офтальмология / Английские материалы / Using and Understanding Medical Statistics_Matthews, Farewell_2007

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Analyzing Normally Distributed Data

9.1. Introduction

The methods described in this chapter feature prominently in most introductory textbooks on statistics. In areas of research such as the physical sciences and engineering, measurements are frequently made on a continuous scale. The assumption that such data have a normal distribution has often proved to be a satisfactory description of the observed variation. Methodology for analyzing normally distributed data is therefore important, and deserves its widespread use.

The nature of the data collected in clinical research often precludes the use of these specialized techniques. Consequently, we have emphasized simple methods of analysis which are appropriate for the type of data frequently seen in medical research, especially that concerned with chronic disease. While it is true that none of the material in this chapter is critical to an understanding of the rest of the book, statistical techniques for normally distributed data can be quite useful, and we would be negligent if we failed to mention them altogether. Nonetheless, the discussion involves more formulae than we usually include in one chapter, and readers who find it too taxing should proceed to chapter 10.

Table 9.1 presents the results of an immunological assay carried out on 14 hemophiliacs and 33 normal controls; the test was performed at two concentrations, low and high. The primary purpose of the study was to ascertain if immunological differences between hemophiliacs and normal individuals could be detected. This is one type of data arising in medical research for which the use of statistical techniques for data from a normal distribution may be appropriate. Throughout this chapter, these data will be used to illustrate the methods of analysis which will be discussed.

proach involves comparing a histogram for the data with the shape of the normal probability curve. In §1.2, we briefly described how to construct a histogram for a set of data. More detailed instructions can always be found in any elementary text on statistics. Despite the degree of subjectivity which is involved in constructing a histogram, e.g., in the choice of intervals, etc., it is wise to conclude that if the histogram does not appear roughly normal, then the use of methods which are described in this chapter may not be appropriate.

Figure 9.1a presents histograms of the high and low concentration assay results for the control samples, cf. table 9.1. We can see that the shapes of these two histograms are not radically different from the characteristic shape of a normal probability curve. However, a few large assay values give the histograms a certain asymmetry which is not characteristic of the normal distribution. To correct this problem, one might consider transforming the data. For example, figure 9.1b shows histograms for the logarithms of the high and low concentration assay results. On a log scale, the large values are perhaps less extreme, but the shapes of the corresponding histograms are not much closer to the characteristic shape of the normal distribution than those of the original data. Although other transformations could be tried, the histograms in figure 9.1a are not sufficiently non-normal to preclude at least a tentative assumption that the data are normally distributed. Therefore, we will proceed from this assumption and regard the measured values as normal observations. In §9.4.1, the need to use a transformation will be more apparent.

9.2.2. Estimating the Mean and Variance

Except in unusual circumstances where relevant prior information is available concerning the population, the mean, , and the variance, 2, will need to be estimated from the data. A natural estimate of the population mean is the sample mean. Symbolically, if xi represents the ith observation in a sample of n values, so that x1, x2, ..., xn represent the entire sample, then the formula for ˆ , the estimator of the population mean , is

In §1.3, we described the variance of a distribution as the expected value of the constructed random variable {X – E(X)}2 = (X – )2. If we knew the value of , the mean of the distribution, a natural estimate of the population variance, 2, would be

immunological test procedure; however, this information does not explicitly address the question of immunological differences between the hemophiliac and control populations. To investigate this question, we require the methods which are described in §9.4.

9.3. Analyzing a Single Sample

In most practical circumstances involving a single sample from a particular population, the principal question which is of interest concerns the value of, the population mean. Either we may wish to test the hypothesis that is equal to a specific value 0, say, or we may wish to derive a confidence interval for . We know that the sample average is a natural estimate of . Therefore, it should not be surprising that the sample average is used to test the hypothesis H: = 0, and also to derive a suitable confidence interval for .

If we let the random variables X1, X2, …, Xn represent a single sample of n observations, then the assumption that these data are normally distributed can be specified, symbolically, by writing Xi N( , 2) for i = 1, 2, …, n. It can

with mean and variance 2/n, i.e., X N( , 2/n). Therefore, the standard-

izing transformation of chapter 8 – which represents the distance between X and in units of the standard error of X – guarantees that

Z = X ~ N(0, 1),/ n

and a suitable test statistic for evaluating the significance level of the data with respect to the hypothesis H: = 0 is

T =| X 0 | ./ n

If is known and to is the observed value of T, then the significance level of the test is Pr(T 6 to). As we discovered in §8.4, the calculations which generate a confidence interval for are based on the fact that T = Z , where Z N(0, 1). The formula specifying a 95% confidence interval for is the interval

(x – 1.96 / n, x + 1.96 / n),

where x is the observed sample mean.

In most cases, will not be known. An obvious solution to this problem involves replacing by its estimate, s. In this case, the significance level is only approximate, since Z = Xs–/ n0 no longer has a normal distribution; we have estimated by s. However, if the number of observations is large, i.e., n 6 50,

mate and, therefore, how accurate an estimate s is likely to be. A Student’s t distribution with large degrees of freedom, say 50, is virtually indistinguishable from a standardized normal distribution.

Statistical tables for the Student’s t distribution are usually drawn up on the same principle as a table of critical values for the distribution of Z (cf., table 8.1). Each row of the table corresponds to an integer value of the degrees of freedom, and each column of the table represents a specified probability level; thus, the values in the body of the table are critical values for the distribution of T = t(k) . For an example of Student’s t distribution tables which correspond to this format, see table 9.4. Critical values for the distribution of Z may be found in the last row of table 9.4, since a t(G) distribution is identical with the distribution of Z. Notice that these values are all smaller than the corresponding critical values found in other rows of the table; this difference reflects the fact that the spread of the Student’s t distribution is greater than that of the standardized normal distribution.

A suitable test statistic for evaluating the significance level of the data with respect to the hypothesis H: = 0 is

T | X 0 | ;

Therefore, if to is the observed value of T, the significance level is equal to

Pr(T 6 to) = Pr( t(n–1) 6 to);

this is the reason that table 9.4 presents critical values of the distribution of T = t(k) . Similar calculations lead to the formula

(x – t* s/ n, x + t* s/ n),

which specifies a 95% confidence interval for , where t* is the appropriate 5%

critical value from table 9.4, i.e., Pr(T 6 t*) = Pr( t(n–1) 6 t*) = 0.05. Notice that this version of the 95% confidence interval for differs from the approximate

confidence interval which we obtained at the beginning of this section only in the replacement of 1.96, the 5% critical value for Z , by t*, the corresponding

critical value for the distribution of T = t(n–1) . Replacing 1.96 by t* always increases the width of the confidence interval, since t* exceeds 1.96. The increased

width reflects our increased uncertainty concerning plausible values of since we have used s, an estimate of , to derive the 95% confidence interval.

Exact 95% confidence intervals for the mean assay results at low and high concentrations in the hemophiliac and control populations may be found in table 9.5; the table also shows selected details of the calculations. If we compare corresponding intervals in tables 9.3 and 9.5, we see that the exact 95% confi-