What conditions must be met for a confidence interval?

Data Mining to Understand the Average Donation Amount

Consider the donations database with 20,000 entries on the companion site. The total amount given by these 20,000 people in response to the current mailing was $15,592.07, with 989 making a current donation and 19,011 not donating at this time. Thus, the average donation is $0.7796035, or about 78 cents per person. Certainly, the amount donated will vary according to the circumstances of a particular mailing. One source of variation is pure statistical variation, leading to the following question: If we were to send a mailing to a similar (but much larger) group of people that these 20,000 people represent (viewing these 20,000 as a random sample from the larger group), how much, on average, should we expect to receive from each person in the new mailing? An answer may be found using the confidence interval.

The standard deviation of the 20,000 donations is $4.2916438 and the standard error is $0.0303465, leading to a 95% confidence interval extending from $0.720122 to $0.839085. If we plan a new mailing to 500,000 people, then we would expect to receive donations totaling between $360,061 and $419,543 (obtained by multiplying the ends of the confidence interval by 500,000 people).

What about the assumptions for validity of this confidence interval from about 72 to 84 cents for the population mean donation amount? The first assumption requires that the data be a random sample from the population of interest, and this would be true, for example, if the 20,000 were initially selected randomly from the 500,000 as part of a pilot study to see if it would be worthwhile mailing to all 500,000 at this time.12 The second assumption requires that the quantity being measured be normally distributed; this assumption does not appear to be satisfied, as is seen from the very nonnormal histogram for the 20,000 donation amounts in Fig. 9.2.3. However, the confidence interval is OK in this case, even though the distribution of individual donations is very skewed, because the sample size is large enough to make the distribution of “averages of 20,000 donations” approximately normal. To show that the distribution of “averages of 20,000 donations” is normally distributed, Fig. 9.2.4 shows a histogram of 500 “bootstrap samples,” with each bootstrap sample of size 20,000 chosen by sampling with replacement from the database of 20,000 donation amounts. Even though the individual donation amounts are highly skewed, the averages of 20,000 donations are actually very close to a normal distribution because of the central limit theorem.

For the population slope, β:

Fromb−tSbtob+tSb

For the population intercept, α:

Froma−tSatoa+tSa

Variable Costs of Production

For the production cost data, the estimated slope is b = 51.66, the standard error is Sb = 7.35, and the two-sided critical t value for n − 2 = 16 degrees of freedom is 2.119905 for 95% confidence interval. The 95% confidence interval for β extends from 51.66 − (7.35)(2.119905) = 36.08 to 51.66 + (7.35)(2.119905) = 67.24. Your confidence interval statement is as follows:

We are 95% confident that the long-run (population) variable costs are somewhere between $36.08 and $67.24 per item produced.

As often happens, the confidence interval reminds you that the estimate ($51.66) is not nearly as exact as it appears. Viewing your data as a random sample from the population of production and cost experiences that might have happened under similar circumstances, you find that with just 18 weeks’ worth of data, there is substantial uncertainty in the variable cost.

A one-sided confidence interval provides a reasonable upper bound that you might use for budgeting purposes. This reflects the fact that you do not know what the variable costs really are; you just have an estimate of them. In this example, the one-sided critical t value is 1.745884, so your upper bound is 51.66 + (7.35)(1.746) = 64.49. The one-sided confidence interval statement is as follows:

We are 95% confident that the long-run (population) variable costs are no greater than $64.49 per item produced.

Note that this bound ($64.49) is smaller than the upper bound of the two-sided interval ($67.24) because you are interested only in the upper side. That is, because you are not interested at all in the lower side (and will not take any error on that side into account), you can obtain an upper bound that is closer to the estimated value of $51.66.

PREDICTION

A frequent use of a straight-line fit is to predict a dependent variable on the y-axis by an independent variable on the x-axis. In the leukocyte example, WBC count can be predicted by the infection culture count. Of course, because the fit is estimated from variable statistical data rather than exact mathematical data, the prediction also will be an estimate, subject to probability and susceptible to confidence interval statements.

How TO CALCULATE A FIT

It seems intuitively obvious that a fit should use all the data and satisfy the most important mathematical criteria of a good fit. The simplest and most commonly used fitting technique of this sort is named least squares. The name comes from minimizing the squared vertical distances from the data points to the proposed line. A primitive mechanical way of thinking about it would be to imagine the points as tacks, each holding a rubber band. A thin steel bar representing the line segment is threaded through all the rubber bands. The tension on each rubber band is the square of the distance stretched. The bar is shifted about until it reaches the position at which the total tension (sum of tensions of each rubber band) is a minimum. In the form of Eq. (8.1), the point in the point–slope fit is given by the sample means of x and y, and the slope can be shown mathematically to be estimated by the sample estimates of the x,y covariance divided by the x variance, or Sxy/S2x. Let us denote by b values the estimates of the β values, to conform with the convention of Greek letters for population (or theoretical) values and Roman letters for sample values. Then, b1 = Sxy/S2x, which yields

(8.2)y−my=b1 (x−mx)=Sxysx2 (x−mx).

For the data plotted in Fig. 8.2, mx= 40.96, my = 16.09, S2x. = 163.07 (sx == 12.77), and Sxy= 25.86. Substituting these quantities in Eq. (8.2), we obtain Eq. (8.3)

as the fit:

(8.3)y−16.09=0.1586(x−40.96)

Figure 8.3 shows the data with the fit superposed.

THE TERM REGRESSION

The term regression has a historical origin that is unnecessary to remember. It may well be thought of as just a name for fitting a model to data by least squares methods. Interestingly, however, the name arose from a genetic observation. Evidence that certain characteristics of outstanding people were genetic anomalies rather than evolutionary changes was given by fitting the characteristics of the children of outstanding people to those of their parents and grandparents. Because the children's characteristics could be better predicted by their grandparents’ characteristics than their parents’, the children were said to have “regressed” to the more normal grandparent state.

CORRELATION AS RELATED TO REGRESSION

Correlation was introduced in Section 3.2. Clearly, both correlation and the straight-line regression slope express information about the relationship between x and y. However, they clearly are not exactly the same because the sample correlation r, by Eq. (3.10), is

(8.4)r=SxySxSy=cov(x,y)sd(x)sd(y)

and the sample slope b1, from Eq. (8.2), is

(8.5)b1=Sxy Sx2=cov(x,y)sd(x)sd(x).

Simple algebra will verify that

(8.6)r=bSx Sy.

What does this difference imply? In correlation, the fit arises from simultaneously minimizing the distances from each point perpendicular to the proposed line. In regression, the fit arises from simultaneously minimizing the vertical (y) distances from each point to the proposed line. Thus, regression makes the assumption that the x measurements are made without random error. Another difference is that regression may be generalized to curved models, whereas correlation is restricted to straight lines only. A difference in interpretation is that correlation primarily is used to express how closely two variables agree (the width of an envelope enclosing all the points), whereas regression can indicate this (the amount of slope in the fit) and also provide a prediction of the most likely y-value, and a confidence interval on it, for a given x-value. Both can be subjected to hypothesis tests, but tests on regression are more incisive, can be generalized to curved models, and can often be related to other statistical methods (such as analysis of variance).

ASSUMPTIONS

As in most of statistics, certain assumptions underlie the development of the method. If these assumptions are violated, we can no longer be sure of our conclusions arising from the results. In the case of least squares regression fits, we assume not only that x-values are measured without error but also that, for each x, the distribution of data vertically about the regression line is approximately normal and that the variability of data vertically about the regression line is the same from one end of the line segment to the other.