20.10 – Growth equations and dose response calculations

Introduction

In biology, growth may refer to increase in cell number, change in size of an individual across development, or increase of number of individuals in a population over time. Nonlinear, time-series, several models proposed to fit growth data, including the Gompertz, logistic, and the von Bertalanffy. These models fit many S-shaped growth curves. These models are special cases of generalized linear models, also called Richard curves.

Growth example

This page describes how to use R to analyze growth curve data sets.

Hours Abs
0.000 0.002207
0.274 0.010443
0.384 0.033688
0.658 0.063257
0.986 0.111848
1.260 0.249240
1.479 0.416236
1.699 0.515578
1.973 0.572632
2.137 0.589528
2.466 0.619091
2.795 0.608486
3.123 0.621136
3.671 0.616850
4.110 0.614689
4.548 0.614643
5.151 0.612465
5.534 0.606082
5.863 0.603933
6.521 0.595407
7.068 0.589006
7.671 0.578372
8.164 0.567749
8.877 0.559217
9.644 0.546451
10.466 0.537907
11.233 0.537826
11.890 0.529300
12.493 0.516551
13.205 0.505905
14.082 0.491013

Key to parameter estimates: y0 is the lag, mumax is the growth rate, and K is the asymptotic stationary growth phase. The spline function does not return an estimate for K.

R code

require(growthrates)
#Enter the data. Replace these example values with your own 
#time variable (Hours)
Hours <- c(0.000, 0.274, 0.384, 0.658, 0.986, 1.260, 1.479, 1.699, 1.973, 2.137, 2.466, 2.795, 3.123, 3.671, 4.110, 4.548, 5.151, 5.534, 5.863, 6.521, 7.068, 7.671, 8.164, 8.877, 9.644, 10.466, 11.233, 11.890, 12.493, 13.205, 14.082)
#absorbance or concentration variable (Abs)
Abs <- c(0.002207, 0.010443, 0.033688, 0.063257, 0.111848, 0.249240, 0.416236, 0.515578, 0.572632, 0.589528, 0.619091, 0.608486, 0.621136, 0.616850, 0.614689, 0.614643, 0.612465, 0.606082, 0.603933, 0.595407, 0.589006, 0.578372, 0.567749, 0.559217, 0.546451, 0.537907, 0.537826, 0.529300, 0.516551, 0.505905, 0.491013)
#Make a dataframe and check the data; If error, then check that variables have equal numbers of entries
Yeast <- data.frame(Hours,Abs); Yeast
#Obtain growth parameters from fit of a parametric growth model
#First, try some reasonable starting values
p <- c(y0 = 0.001, mumax = 0.5, K = 0.6)
model.par <- fit_growthmodel(FUN = grow_logistic, p = p, Hours, Abs, method=c("L-BFGS-B"))
summary(model.par)
coef(model.par)
#Obtain growth parameters from fit of a nonparametric smoothing spline
model.npar <- fit_spline(Hours,Abs)
summary(model.npar)
coef(spline.md)
#Make plots
par(mfrow = c(2, 1))
plot(Yeast, ylim=c(0,1), cex=1.5,pch=16, main="Parametric Nonlinear Growth Model", xlab="Hours", ylab="Abs")
lines(model.par, col="blue", lwd=2)
plot(model.npar, ylim=c(0,1), lwd=2, main="Nonparametric Spline Fit", xlab="Hours", ylab="Abs")

Results from example code

Top: Parametric Nonlinear Growth Model Bottom: Nonparametric Spline Fit

Figure 1. Top: Parametric Nonlinear Growth Model; Bottom: Nonparametric Spline Fit

LD50

In toxicology, the dose of a pathogen, radiation, or toxin required to kill half the members of a tested population of animals or cells is called the lethal dose, 50%, or LD50. This measure is also known as the lethal concentration, LC50, or properly after a specified test duration, the LCt50 indicating the lethal concentration and time of exposure. LD50 figures are frequently used as a general indicator of a substance’s acute toxicity. A lower LD50 is indicative of increased toxicity.

The point at which 50% response of studied organisms to range of doses of a substance (e.g., agonist, antagonist, inhibitor, etc.) to any response, from change in behavior or life history characteristics up to and including death can be described by the methods described in this chapter. The procedures outlined below assume that there is but one inflection point, i.e., an “s-shaped” curve, either up or down; if there are more than one inflection points, then the logistic equations described will not fit the data well and other choices need to be made (see Di Veroli et al 2015). We will use the drc package (Ritz et al 2015).

Example

First we’ll work through use of R. We’ll follow up with how to use Solver in Microsoft Excel.

After starting R, load the drc library.

library(drc)

Consider some hypothetical 24-hour survival data for yeast exposed to salt solutions. Let resp equal the variable for frequency of survival (e.g., calculated from OD660 readings) and NaCl equal the millimolar (mm) salt concentrations or doses.

At the R prompt type

resp <- c(1,1,1,.9,.7,.3,.4,.2,0,0,0)
NaCl=seq(0,1000,100) 
#Confirm sequence was correctly created; alternatively, enter the values.
NaCl 
[1] 0 100 200 300 400 500 600 700 800 900 1000 
#Make a plot
plot(NaCl,resp,pch=19,cex=1.2,col="blue",xlab="NaCl [mm]",ylab="Survival frequency")

And here is the plot (Fig. 2).

 

plot of simulated yeast survival by salt concentration salt

Figure 2. Hypothetical data set, survival of yeast in different salt concentrations.

Note the sigmoidal “S” shape — we’ll need an logistic equation to describe the relationship between survival of yeast and NaCl doses.

The equation for the four parameter logistic curve, also called the Hill-Slope model, is

    \begin{align*} f\left ( b,c,d,e \right ) = c + \frac{d-c}{1+exp^{b\left ( log\left ( x \right )-log\left ( e \right ) \right )}} \end{align*}

where c is the parameter for the lower limit of the response, d is the parameter for the upper limit of the response, e is the relative EC50, the dose fitted half-way between the limits c and d, and b is the relative slope around the EC50. The slope, b, is also known as the Hill slope. Because this experiment included a dose of zero, a three parameter logistic curve would be appropriate. The equation simplifies to

    \begin{align*} f\left ( b,d,e \right ) = \frac{d}{1+exp^{b\left ( log\left ( x \right )-log\left ( e \right ) \right )}} \end{align*}

EC50 from 4 parameter model

Let’s first make a data frame

dose <- data.frame(NaCl, resp)

Then call up a function, drm, from the drc library and specify the model as the four parameter logistic equation, specified as LL.4(). We follow with a call to the summary command to retrieve output from the drm function. Note that the four-parameter logistic equation

model.dose1 = drm(dose,fct=LL.4())
summary(model.dose1)

And here is the R output.

Model fitted: Log-logistic (ED50 as parameter) (4 parms) 
Parameter estimates: 
                   Estimate  Std. Error    t-value  p-value 
b:(Intercept)      3.753415    1.074050   3.494636   0.0101 
c:(Intercept)     -0.084487    0.127962  -0.660251   0.5302 
d:(Intercept)      1.017592    0.052460  19.397441   0.0000 
e:(Intercept)    492.645128   47.679765  10.332373   0.0000 

Residual standard error: 
0.0845254 (7 degrees of freedom)

The EC50, technically the LD50 because the data were for survival, is the value of e: 492.65 mM NaCl.

You should always plot the predicted line from your model against the real data and inspect the fit.

At the R prompt type

plot(model.dose1, log="",pch=19,cex=1.2,col="blue",xlab="NaCl [mm]",ylab="Survival frequency")

As long as the plot you made in earlier steps is still available, R will add the line specified in the lines command. Here is the plot with the predicted logistic line displayed (Fig. 3).

Hypothetical survival plot with logistic curve of yeast exposed to sodium chloride salt

Figure 3. Logistic curve added to Figure 1 plot.

While there are additional steps we can take to decide is the fit of the logistic curve was good to the data, visual inspection suggests that indeed the curve fits the data reasonably well.

More work to do

Because the EC50 calculations are estimates, we should also obtain confidence intervals. The drc library provides a function called ED which will accomplish this. We can also ask what the survival was at 10% and 90% in addition to 50%, along with the confidence intervals for each.

At the R prompt type

ED(model.dose1,c(10,50,90), interval="delta")

And the output is shown below.

Estimated effective doses
(Delta method-based confidence interval(s))
     Estimate Std. Error   Lower   Upper
1:10  274.348     38.291 183.803  364.89
1:50  492.645     47.680 379.900  605.39
1:90  884.642    208.171 392.395 1376.89

Thus, the 95% confidence interval for the EC50 calculated from the four parameter logistic curve was between the lower limit of 379.9 and an upper limit of 605.39 mm NaCl.

EC50 from 3 parameter model

Looking at the summary output from the four parameter logistic function we see that the value for c was -0.085 and p-value was 0.53, which suggests that the lower limit was not statistically different from zero. We would expect this given that the experiment had included a control of zero mm added salt. Thus, we can explore by how much the EC50 estimate changes when the additional parameter c is no longer estimated by calculating a three parameter model with LL.3().

model.dose2 = drm(dose,fct=LL.3())
summary(model.dose2)

R output follows.

Model fitted: Log-logistic (ED50 as parameter) with lower limit at 0 (3 parms)
Parameter estimates:
               Estimate Std. Error   t-value p-value
b:(Intercept)   4.46194    0.76880   5.80378   4e-04
d:(Intercept)   1.00982    0.04866  20.75272   0e+00
e:(Intercept) 467.87842   25.24633  18.53253   0e+00
Residual standard error:
 0.08267671 (8 degrees of freedom)

The EC50 is the value of e: 467.88 mM NaCl.

How do the four and three parameter models compare? We can rephrase this as as statistical test of fit; which model fits the data better, a three parameter or a four parameter model?

At the R prompt type

anova(model.dose1, model.dose2)

The R output follows.

1st mode
fct:      LL.3()
2nd model
fct:      LL.4()
ANOVA table
          ModelDf      RSS Df F value p value
2nd model       8 0.054684                   
1st model       7 0.050012  1  0.6539  0.4453

Because the p-value is much greater than 5% we may conclude that the fit of the four parameter model was not significantly better than the fit of the three parameter model. Thus, based on our criteria we established in discussions of model fit in Chapter  16 – 18, we would conclude that the three parameter model is the preferred model.

The plot below now includes the fit of the four parameter model (red line) and the three parameter model (green line) to the data (Fig. 4).

Hypothetical survival plot with 3 (green) & 4 (red) parameter logistic curves of yeast exposed to sodium chloride salt

Figure 4. Four parameter (red) and three parameter (green) logistic models fitted to data.

The R command to make this addition to our active plot was

lines(dose,predict(model.dose2, data.frame(x=dose)),col="green")

We continue with our analysis of the three parameter model and produce the confidence intervals for the EC50 (modify the ED() statement above for model.dose2 in place of model.dose1).

Estimated effective doses
(Delta method-based confidence interval(s))
     Estimate Std. Error   Lower  Upper
1:10  285.937     33.154 209.483 362.39
1:50  467.878     25.246 409.660 526.10
1:90  765.589     63.026 620.251 910.93

Thus, the 95% confidence interval for the EC50 calculated from the three parameter logistic curve was between the lower limit of 409.7 and an upper limit of 526.1 mm NaCl. The difference between upper and lower limits was 116.4 mm NaCl, a smaller difference than the interval calculated for the 95% confidence intervals from the four parameter model (225.5 mm NaCl). This demonstrates the estimation trade-off: more parameters to estimate reduces the confidence in any one parameter estimate.

Additional notes of EC50 calculations

Care must be taken that the model fits the data well. What if we did not have observations throughout the range of the sigmoidal shape? We can explore this by taking a subset of the data.

dd = dose[1:6,]

Here, all values after dose 500 were dropped

dd
  resp dose
1  1.0    0
2  1.0  100
3  1.0  200
4  0.9  300
5  0.7  400
6  0.3  500

and the plot does not show an obvious sigmoidal shape (Fig. 5).

Subset of hypothetical survival plot of yeast exposed to sodium chloride salt

Figure 5. Plot of reduced data set.

We run the three parameter model again, this time on the subset of the data.

model.dosedd = drm(dd,fct=LL.3())
summary(model.dosedd)

Output from the results are

Model fitted: Log-logistic (ED50 as parameter) with lower limit at 0 (3 parms)
Parameter estimates:
                Estimate Std. Error    t-value p-value
b:(Intercept)   6.989842   0.760112   9.195801  0.0027
d:(Intercept)   0.993391   0.014793  67.153883  0.0000
e:(Intercept) 446.882542   5.905728  75.669344  0.0000
Residual standard error:
 0.02574154 (3 degrees of freedom)

Conclusion? The estimate is different, but only just so, 447 vs. 468 mm NaCl.

Thus, within reason, the drc function performs well for the calculation of EC50. Not all tools available to the student will do as well.

NLopt and nloptr

draft

Free open source library for nonlinear optimization.

Steven G. Johnson, The NLopt nonlinear-optimization package, http://github.com/stevengj/nlopt

https://cran.r-project.org/web/packages/nloptr/vignettes/nloptr.html

 

Alternatives to R

What about online tools? There are several online sites that will allow students to perform these kinds of calculations. Students familiar with MatLab know that it can be used to solve nonlinear equation problems. An open source alternative to MatLab is called GNU Octave, which can be installed on a personal computer or run online at http://octave-online.net. Students also may be aware of other sites, e.g., mycurvefit.com and IC50.tk.

Both of these free services performed well on the full dataset (results not shown), but fared poorly on the reduced subset: mycurvefit returned a value of 747.64 and IC50.tk returned an EC50 estimate of 103.6 (Michael Dohm, pers. obs.).

Simple inspection of the plotted values shows that these values are unreasonable.

EC50 calculations with Microsoft Excel

Most versions of Microsoft Excel include an add-in called Solver, which will permit mathematical modeling. The add-in is not installed as part of the default installation, but can be installed via the Options tab in the File menu for a local installation or via Insert for the Microsoft Office online version of Excel (you need a free Microsoft account). The following instructions are for a local installation of Microsoft Office 365 and were modified from information provided by sharpstatistics.co.uk.

After opening Excel, set up worksheet as follows. Note that in order to use my formulas your spreadsheet values need to be set up exactly as I describe.

  • Dose values in column A, beginning with row 2
  • Response values in column B, beginning with row 2
  • In column F type b in row 2, c in row 3, d in row 4, and e in row 5.
  • In cells G2 – G5, enter the starting values. For this example, I used b = 1, c = 0, d = 1, e = 400.
    • For your own data you will have to explore use of different starting values.
  • Enter headers in row 1.
    • Column A Dose
    • Column B Response
    • Column C Predicted
    • Column D Squared difference
    • Column F Constants
  • In cell C14 enter sum squares
  • In cell D14 enter =sum(D2:D12)

Here is an image of the worksheet (Fig. 6), with equations entered and values updated, but before Solver run.

Screenshot Microsoft Excel worksheet containing our data set

Figure 6. Screenshot Microsoft Excel worksheet containing our data set (col A & B), with formulas added and calculated. Starting values for constants in column G, rows 2 – 4.

Next, enter the functions.

  • In cell C2 enter the four parameter logistic formula — type everything between the double quotes: “=$G$3+(($G$4-$G$3)/(1+(A2/$G$5)^$G$2))“. Next, copy or drag the formula to cell C12 to complete the predictions.
    • Note: for a three parameter model, replace the above formula with “=(($G$4)/(1+(A2/$G$5)^$G$2))“.
  • In cell D2 type everything between the double quotes: “=(B2-C2)^2“.
  • Next, copy or drag the formula to cell D12 to complete the predictions.
  • Now you are ready to run solver to estimate the values for b, c, d, and e.
    • Reminder: starting values must be in cells G2:G5
  • Select cell D14 and start solver by clicking on Data and looking to the right in the ribbon.

Solver is not normally installed as part of the default installation of Microsoft Excel 365. If the add-in has been installed you will see Solver in the Analyze box (Fig. 7).

Screenshot Microsoft Excel, Solver add-in available

Figure 7. Screenshot Microsoft Excel, Solver add-in available.

If Solver is not installed, go to File → Options → Add-ins (Fig. 8).

Figure 7. Screenshot Microsoft Excel, Solver add-in available.

Figure 8. Screenshot Microsoft Excel, Solver add-in available and ready for use.

Go to Microsoft support for assistance with add-ins.

With the spreadsheet completed and Solver available, click on Solver in the ribbon (Fig. 7) to begin. The screen shot from the first screen of Solver is shown below (Fig. 9).

Screenshot Microsoft Excel solver menu.

Figure 9. Screenshot Microsoft Excel solver menu.

Setup Solver

  • Set Objective, enter the absolute cell reference to the sum squares value, \$D$14
  • Set To: Min.
  • For By Changing Variable Cells, enter the range of cells for the four parameters, $G$2:$G$5
  • Uncheck the box by “Make Unconstrained Variables Non-Negative.”
    • Where constraints refers to any system of equalities or inequalities equations imposed on the algorithm.
  • Select a Solving method, choose “GRG Nonlinear,” the nonlinear programming solver option (not shown in Fig. 7, select by clicking on the down arrow).
  • Click Solve button to proceed.

GRG Nonlinear is one of many optimization methods. In this case we calculate the minimum of the sum of squares — the difference between observed and predicted values from the logistic equation — given the range of observed values. GRG stands for Generalized Reduced Gradient and finds the local optima — in this case the minimum or valley — without any imposed constraints. See solver.com for additional discussion of this algorithm.

If all goes well, this next screen (Fig. 10) will appear which shows the message, “Solver has converged to the current solution. All constraints are satisfied.”

Screenshot solver completed run

Figure 10. Screenshot solver completed run.

Click OK and note that the values of the parameters have been updated (Table 1).

Table 1. Four parameter logistic model, results from Solver 

Constant Starting values Values after solver
b 1 3.723008382
c 0 -0.088865487
d 1 1.017964505
e 400 493.9703594

Note the values obtained by Solver are virtually identical to the values obtained in the drc R package. The differences are probably because of the solver algorithm.

More interestingly, how did Solver do on the subset data set? Here are the results from a three parameter logistic model (Table 2).

Table 2. Three parameter logistic model, results from Solver

Constant Starting values Full dataset,
Solver results
Subset,
Solver results
b 1 3.723008382 6.989855948
d 1 1.017964505 0.993391264
e 400 493.9703594 446.8819282

The results are again very close to results from the drc R package.

Thus, we would conclude that Solver and Microsoft Excel would be a reasonable choice for EC50 calculations, and much better than IC50.tk and mycurvefit.com. The advantages of R over Microsoft Excel is that the model building is more straightforward than entering formulas in the cell reference format required by Excel.

Questions

[pending]

References and suggested readings

Beck B, Chen YF, Dere W, et al. Assay Operations for SAR Support. 2012 May 1 [Updated 2012 Oct 1]. In: Sittampalam GS, Coussens NP, Nelson H, et al., editors. Assay Guidance Manual [Internet]. Bethesda (MD): Eli Lilly & Company and the National Center for Advancing Translational Sciences; 2004-. Available from: www.ncbi.nlm.nih.gov/books/NBK91994/

Di Veroli G. Y., Fornari C., Goldlust I., Mills G., Koh S. B., Bramhall J. L., Richards, F. M., Jodrell D. I. (2015) An automated fitting procedure and software for dose-response curves with multiphasic features. Scientific Reports 5: 14701.

Ritz, C., Baty, F., Streibig, J. C., Gerhard, D. (2015) Dose-Response Analysis Using R PLOS ONE, 10(12), e0146021

Tjørve, K. M., & Tjørve, E. (2017). The use of Gompertz models in growth analyses, and new Gompertz-model approach: An addition to the Unified-Richards family. PloS one, 12(6), e0178691.

Chapter 20 contents

/MD

8.6 – Confidence limits for the estimate of population mean

Introduction

In Chapter 3.4 and Chapter 8.3, we introduced the concept of providing a confidence interval for estimates. We gave a calculation for an approximate confidence interval for proportions and for the Number Needed to Treat (Chapter 7.3). Even an approximate confidence interval gives the reader a range of possible values of a population parameter from a sample of observations.

In this chapter we review and expand how to calculate the confidence interval for a sample mean, \bar{X}. Because \bar{X} is derived from a sample of observations, we use the t-distribution to calculate the confidence interval. Note that if the population was known (population standard deviation), then you would use normal distribution. This was the basis for our recommendation to adjust your very approximate estimate of a confidence interval for an estimate by replacing the “2” with “1.96” when you multiply the standard error of the estimate (SE) in the equation estimate \bar{X} \pm 2SE. As you can imagine, the approximation works for large sample size, but is less useful as sample size decreases.

Consider \bar{X}; it is a point estimate of \mu, the population mean (a parameter). But our estimate of \bar{X} is but one of an infinite number of possible estimates. The confidence interval, however, gives us a way to communicate how reliable our estimate is for the population parameter. A 95% confidence interval, for example, tells the reader that we are willing to say (95% confident) the true value of the parameter is between these two numbers (a lower limit and an upper limit). The point estimate (the sample mean) will of course be included between the two limits.

Instead of 95% confidence, we could calculate intervals for 99%. Since 99% is greater than 95%, we would communicate our certainty of our estimate.

Note 1: Again, the caveats about p-value extend to confidence intervals. See Chapter 8.2.

Question 1: For 99% confidence interval, the lower limit would be smaller than the lower limit for a 95% confidence interval.

  1. True \color{red} \leftarrow Answer
  2. False

When we set the Type I error rate, \alpha (alpha) = 0.05 (5%), that means that 5% of all possible sample means from a population with mean, \mu, will result in t values that are larger than + t_{0.05(2),df} OR smaller than - t_{0.05(2),df}.

Why the t-distribution?

We use the t-test because, technically, we have a limited sample size and the t-distribution is more accurate than the normal distribution for small samples. Note that as sample size increases, the t-distribution is not distinguishable from the normal distribution and we could use \pm 1.96  (Fig 1).

Figure 1. Critical values at Type I rate of 5% of t-distribution \pm t_{0.05(2),df}. Blue dashed line is z = 1.96

Figure 1. Critical values at Type I rate of 5% of t-distribution \pm t_{0.05(2),df}. Blue dashed line is z = 1.96.

Here’s the equation for calculating the confidence interval based on the t-distribution. These set the limits around our estimate of the sample mean. Together, they’re called the 95% confidence interval, 1-0.05 = 0.95.

    \begin{align*} \left [ -t_{0.05\left ( 2 \right ),df} \leq \frac{\bar{X}-\mu}{s_{\bar{X}}}\leq +t_{0.05\left ( 2 \right ),df}\right ]=0.95 \end{align*}

Here’s a simplified version of the same thing, but generalized to any Type I level…

    \begin{align*} \bar{X} \pm t_{\alpha/2,v} \cdot s_{\bar{X}} \end{align*}

This statistic allows us to say that we are 95% confident that the interval \pm t_{0.05(2),df} includes the true value for \mu. For this confidence interval you need to identify the critical t value at 5%. Thus, you need to know the degrees of freedom for this problem, which is simply n - 1, the sample size minus one.

It is straightforward to calculate these by hand, but…

Set the Type I error rate, calculate the degrees of freedom (df):

  • n- 1 samples for one sample test
  • n - 1 pairs of samples for paired test
  • n - 2 samples for two independent sample test

and lookup the critical value from the t table (or from the t distribution in R). Of course, it is easier to use R.

In R, for the one tail critical value with seven degrees of freedom, type at the R prompt

qt(c(0.05), df=7, lower.tail=FALSE)
[1] 1.894579

For the two-tail critical value

qt(c(0.025), df=7, lower.tail=FALSE)
[1] 2.364624

Or, if you prefer to use R Commander, then follow the menu prompts to bring up the t quantiles function (Fig 2 and Fig 3).

Figure 2. Drop down menu to get t-distribution

Figure 2. Drop down menu to get t-distribution.

Note 2: Quantiles divide probability distribution into equal parts or intervals. Quartiles have four groups, deciles have ten groups, and percentiles have 100 groups.

Figure 3. Menu for t quantiles, with values entered for the one-tail example. Note: for 2-tailed probability 5%, divide 0.05/2 = 0.025

Figure 3. Menu for t quantiles, with values entered for the two-tail example.

You should confirm that what R calculates agrees with the critical values tabulated in the Table of Critical values for the t distribution provided in the Appendix.

A worked example

Let’s revisit our lizard example from last time (see Chapter 8.5). Prior to conducting any inference test, we decide acceptable Type I error rates (cf justify alpha discussion in Ch8.1); For this example, we set Type I error rate to be 1% for a 99% confidence interval.

The Rcmdr output was

t.test(lizz$bm, alternative='two.sided', mu=5, conf.level=.99)
data: lizz$bm
t = -1.5079, df = 7, p-value = 0.1753
alternative hypothesis: true mean is not equal to 5 
99 percent confidence interval:
1.984737 6.199263
sample estimates:
mean of x
4.092

Sort through the output and identify what you need to know.

Question 1: What was the sample mean?

  1. 5
  2. -1.5079
  3. 7
  4. 0.1753
  5. 1.984737
  6. 6.199263
  7. 4.092 \color{red} \leftarrow Answer

Question 2: What was the most likely population mean?

  1. 5 \color{red} \leftarrow Answer
  2.  Answer
  3. -1.5079
  4. 7
  5. 0.1753
  6. 1.984737
  7. 6.199263
  8. 4.092

Question 3: This was a “one-tailed” test of the null hypothesis?

  1. True
  2. False \color{red} \leftarrow Answer

The output states “alternative hypothesis: true mean is not equal to 5” — so it was a two-tailed test.

Question 4: What was the lower limit of the confidence interval?

  1. 5
  2. -1.5079
  3. 7
  4. 0.1753
  5. 1.984737 \color{red} \leftarrow Answer
  6. 6.199263
  7. 4.092

The 99% confidence interval, CI_{99\%}, is 1.984737, 6.199263, which means we are 99% certain that the population mean is between 1.984737(lower limit) and 6.199263 (upper limit). In Chapter 8.5 we calculated the CI_{99\%}, is 2.667, 5.517.

Confidence intervals by nonparametric bootstrap sampling.

Bootstrapping is a general approach to estimation or statistical inference that utilizes random sampling with replacement (Kulesa et al. 2015). In classic frequentist approach, a sample is drawn at random from the population and assumptions about the population distribution are made in order to conduct statistical inference. By resampling with replacement from the sample many times, the bootstrap samples can be viewed as if we drew from the population many times without invoking a theoretical distribution. A clear advantage of the bootstrap is that it allows estimation of confidence intervals without assuming a particular theoretical distribution and thus avoids the burden of repeating the experiment. Which method to prefer? For cases where assumption of a particular distribution is unwarranted (eg, what is the appropriate distribution when we compare medians among samples?), bootstrap may be preferred (and for small data sets, percentile bootstrap may be better). We cover bootstrap sampling of confidence intervals in Chapter 19.2 Bootstrap sampling.

Conclusions

The take home message is simple.

  • All estimates should — must? — be accompanied by a Confidence Interval
  • The more confident we wish to be, the wider the confidence interval will be

Note that the confidence interval concept combines DESCRIPTION (the population mean is between these limits) and INFERENCE (and we are 95% certain about the values of these limits). It is good statistical practice to include estimates of confidence intervals for any estimate you share with readers. Any statistic that can be estimated should be accompanied by a confidence interval and, as you can imagine, formulas are available to do just this. For example, earlier this semester we calculated NNT.

Questions

  1. To gain practice with calculations of confidence intervals, calculate the approximate confidence interval, the 95% and the 99% confidence intervals based on the t distribution, for each of the following.
    • \bar{X} = 13, s = 1.3, n = 10
    • \bar{X} = 13, s = 1.3, n = 30
    • \bar{X} = 13, s = 2.6, n = 10
    • \bar{X} = 13, s = 2.6, n = 30

Quiz Chapter 8.6

Confidence limits for the estimate of population mean


Chapter 8 contents

8.3 – Sampling distribution and hypothesis testing

Introduction

Understanding the relationship between sampling distributions, probability distributions, and hypothesis testing is the crucial concept in the NHST — Null Hypothesis Significance Testing — approach to inferential statistics. is crucial, and many introductory text books are excellent here. I will add some here to their discussion, perhaps with a different approach, but the important points to take from the lecture and text are as follows.

Our motivation in conducting research often culminates in the ability (or inability) to make claims like:

  1. “Total cholesterol greater than 185 mg/dl increases risk of coronary artery disease.”
  2. “Average height of US men aged 20 is 70 inches (1.78 m).”
  3. “Species of amphibians are disappearing at unprecedented rates.”

Lurking beneath these statements of “fact” for populations (just what IS the population for #1, for #2, and for #3?) is the understanding that not ALL members of the population were recorded.

How do we go from our sample to the population we are interested in? Put another way — How good is our sample? We’ve talked about how “biostatistics” can be generalized as sets of procedures you use to make inferences about what’s happening in populations. These procedures include:

  • Have an interesting question
  • Experimental design (Observational study? Experimental study?)
  • Sampling from populations (Random? Haphazard?)
  • Hypotheses: HO and HA
  • Estimate parameters (characterize the population)
  • Tests of hypotheses (inferences)

We have control of each of these — we choose what to study, we design experiments to test our hypotheses…We have already introduced these topics (Chapters 6 – 8).

We also obtain estimates of parameters, and inferential statistics applies to how we report our descriptive statistics (Chapter 3). Estimates of parameters like the sample mean and sample standard deviation can be assessed for accuracy and precision (eg, confidence intervals).

Sampling distribution

Imagine drawing a sample of 30 from a population, calculating the sample mean for a variable (eg, systolic blood pressure), then calculating a second sample mean after drawing a new sample of 30 from the same population. Repeat, accumulating one estimate of the mean, over and over again. What will be the shape of this distribution of sample means? The Central Limit Theorem states that the shape will be a normal distribution, regardless of whether or not the population distribution was normal, as long as the sample size is large (ie, Law of Large Numbers). We alluded to this concept when we introduced discrete and continuous distributions (Chapter 6).

It’s this result from theoretical statistics that allows us to calculate the probability of an event from a sample without actually carrying out repeated sampling or measuring the entire population.

A worked example

To demonstrate the CLT we want R to help us generate many samples from a particular distribution and calculate the same statistic on each sample. We could make a for loop, but the replicate() function provides a simpler framework.  We’ll sample from the chi-square distribution. You should extend this example to other distributions on your own, see Question 5 below.

Note 1: This example is much simpler to enter and run code in the script window, modifying code directly as needed. If you wish to try to run this through Rcmdr, you’ll need to take a number of steps, and likely need to modify the code and rerun anyway. Some of the steps in would be Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Sample from chi-square distribution…, then running Numerical summaries and saving the output to an object (eg, out), extracting the values from the object (eg, out$Table, confirm by running command str(out)— str() is an R utility to display the structure of an object), then testing the object for normality Rcmdr: Statistics → Test of normality, select Shapiro-Wilk, etc.. In other words, sometimes a GUI is a good idea, but in many cases, work with the script!

Generate x replicate samples (eg, x = 10, 100, 1000, one million) of 30 each from chi-square distribution with one degree of freedom, test the distribution against null hypothesis (assume normal distributed, eg, Shapiro-Wilk test, see Chapter 13.3), then make a histogram (Chapter 4.2), like Figure 1 or Figure 2.

x.10 <- replicate(10, {
my.mean <- rchisq(30, 1)
mean(my.mean)
})

normalityTest(~x.10, test="shapiro.test")
hist(x.10, col="orange")

Result from R

Shapiro-Wilk normality test

data: x.10
W = 0.87016, p-value = 0.1004

histogram of means from ten repeated samples of 30 from the chi-square distribution

Figure 1. means of ten replicate samples drawn at random from chi-square distribution, df = 1.

Modify the code to draw 100 samples, we get Fig 2.

Modify the code to draw 100 samples, we get  means of 100 replicate samples drawn at random from chi-square distribution, df = 1.

Figure 2. means of 100 replicate samples drawn at random from chi-square distribution, df = 1. Results from Shapiro-Wilks test: W = 0.97426, p-value = 0.04721.

And finally, modify the code to draw one million samples, we get Figure 3.

means of one million replicate samples drawn at random from chi-square distribution, df = 1.

Figure 3. means of one million replicate samples drawn at random from chi-square distribution, df = 1. Normality test will fail to run, sample size of 5000 limit.

How to apply sampling distribution to hypothesis testing

First, a reminder of some definitions.

Estimate = we will always (almost) concern ourselves with how good our sample mean (such values are called estimates) is relative to the population mean, the thing we really want, but can only hope to get an estimate of.

Accuracy = how close to the true value is our measure?

Precision = how repeatable is our measure?

How can we tell if we have a good estimate? We want an estimate with an evaluation for accuracy and for precision. The sampling error provides an assessment of precision, whereas the confidence interval provides a statement of accuracy. We need an estimate of the sampling error for the statistic,

Sample standard error of the mean

We introduced sample error of the mean in section 3.4 of Chapter 3. Everything we measure can have a corresponding statement about how accurate (sampling error) is our estimate! First, we begin by asking, “how accurate is the mean that we estimate from a sample of a population?” How do we answer this? We could prove it in the mathematical sense of proof (and people have and do) OR we can use the computer to help. We’ll try this approach in a minute.

What we will show relates to the standard error of the population mean (SEM) or

    \[s_{\bar{X}}\]

, whose equation is shown below.

    \begin{equation*} \label{eq:SEMvar} SEM=\frac{s^2}{n} \end{equation*}

or equivalently, from the standard deviation we have

    \begin{equation*} \label{eq:SEMsd} SEM=s_{\bar{X}}=\frac{s}{\sqrt{n}} \end{equation*}

Note that the SEM takes the variance and divides through by the sample size. In general, then, the larger the sample size, the smaller the “error” around the mean. As we work through the different statistical tests, t-tests, analysis of variance, and related, you will notice that the test statistic is calculated as a ratio between a difference or comparison divided by some form of an error measurement. This is to remind you that “everything is variable.”

A note on standard deviation (SD) and standard error of the mean (SEM): SD estimates the variability of a sample of Xi‘s whereas SEM estimates the variability of a sample of means.

Let’s return to our thought problem and see how to demonstrate a solution. First, what is the population? Second, can we get the true population mean?

One way, a direct (but impossible?) approach would be to measure it — get all of the individuals in a population and measure them, then calculate the population mean. Then, we could compare our original sample mean against the true mean and see how close it was. This can be accomplished in some limited cases. For example, the USA conducts a census of her population every ten years, a procedure which costs billions of dollars. We can then compare samples from the different states or counties to the USA mean. And these statistics are indeed available via the census.gov website. But even the census uses sampling — individuals are randomly selected to answer more questions and from this sample trends in the population are inferred.

So, sampling from populations is the way to go for most questions we will encounter. The procedures we will use to show how a sample mean relates to the population mean are general and may be used to show how any estimate of a variable (sample mean and sample standard deviation, etc.), relates to properties of a parameter. We’ll get to the other issues, but for now, think about sample size.

Sampling from populations is necessary and inevitable, and, to a certain extent, under your control. But how many individuals do we need? The quick answer is for me to direct your attention to the equation for the SEM. Can you see in that ratio the secret to obtaining more precise estimates? There are many ways to approach this question, but let’s use the tools from last time, those based on properties of a normal distribution.

If we can view the sampling as having come from a population at least approximately normally distributed for our variable, then we can now examine empirically the effect of different sample sizes on the estimate of the mean.

A hint: variability is important!

From one population we obtain two samples, A and B. Sample sizes are

    \begin{align*} &Group \ A, n = 9 \\ &Group \ B, n = 50 \end{align*}

Assume for now that we know the true mean (μ) and standard deviation (σ) for the population. Note. This is one of the points of why we use computer simulation so much to teach statistics — it allows us to specify what the truth is, then see how our statistical tools work or how our assumptions affect our statistically based conclusions.

    \begin{align*} \mu = 47.0 \ mm \\ \sigma = 12.0 \ mm \end{align*}

Confidence intervals

Reliability is another word for precision. We define a confidence interval as a statistic to report the reliability of our estimated statistic. We introduced confidence interval in Chapter 3.4. At least in principle, confidence intervals can be calculated for all statistics (mean, variance, etc.,) and for all data types. Confidence intervals define a lower limit, L, and an upper limit, U, and that you are making a statement that you are “95% certain that the true value (parameter value) is between these two limits.

We previously reported how to calculate an approximate confidence intervals for proportions and for NNT; simply multiple standard error estimate by 2. Here we introduce an improved approximate calculation of the 95% confidence interval for the sample mean

    \begin{equation*} CI95\% = \bar{X}\pm Z\cdot s_{\bar{X}} \end{equation*}

where Z is something you would look up from the table of the normal distribution. For a 95% confidence interval, 100% – 95% = 5% and divide 5% by two: the lower limit corresponds to 2.5% and the upper limit corresponds to 2.5% on our normal distribution. We look up the table and we find that Z for 0.025 is 1.96 and that is the value we would plug into our equation above. For large sample sizes, you can get a pretty decent estimate of the confidence interval by replacing 1.96 with “2.”

Questions

1. What is the probability of having a sample mean greater than 50 (mean > 50) for a sample of n = 9 ?

We’ll use a slight modification of the Z-score equation we introduced in Chapter 6.6 — the modification here is that previously we referred to the distribution of Xi‘s and how likely a particular observation would be. Instead, we can use the Z score with the standard normal distribution (aka Z-distribution), approach to solving how likely an estimated sample mean is given the population parameters μ and σ. Recall the Z score

    \begin{equation*} Z = \frac{X_{i}-\mu}{\sigma} \end{equation*}

We have everything we need except the SEM, which we can calculate by dividing the standard deviation by squared root of sample size.

For

    \[\bar{X} = 50\]

, σ = 12.0 (given above), and μ = 47, n = 9, plug in the values:

    \begin{equation*} s_{\bar{X}} = \frac{12.0}{\sqrt{9}}=4 \end{equation*}

Therefore, after applying the equation for Z score, Z = 0.75. This corresponds to how far away from the standard mean of zero.

Look up Z = 0.75 from the table of normal distribution. The answer is 0.22663, which corresponds to that Z is EQUAL to or GREATER than 0.75, which is what we wanted. Translated, this implies that, given the level of variability in the sample, 22.66% of your sample means would be greater than 50! We write: P(X > 50.0) = P (Z > 0.75) = 0.2266.

Some care needs to be taken when reading these tables — make sure you understand how the direction (less than, greater than) away from the mean is tabulated.

2. Instead of greater, how would you get the probability less than 50?

Total area under the curve is 1 (100%), so subtract 1 -0.22663 = 0.7734.

I recommend that you do these by hand first, then check your answers. You’ll need to be able to do this for exams.

Here’s how to use Rcmdr to do these kind of problems.

Rcmdr: Distributions → Continuous distributions → Normal distribution → Normal probabilities …

Screenshot Rcmdr normal probability

Figure 5. Screenshot Rcmdr menu to get normal probability.

Here’s the answer from Rcmdr

pnorm(c(50), mean=47, sd=12, lower.tail=TRUE)

[1] 0.5987063

3. Now, try a larger sample size. For N = 50, what is the probability of having a sample mean greater than 50 (bar{X} \gt 50)?

    \[\bar{X} = 50\]

, μ = 47, σ = 12, n = 50 and

    \[SEM = \frac{12.0}{\sqrt{50}} = 1.697\]

Therefore, after applying the equation for Z score, <em>Z</em> = 1.768. Look up <em>Z</em> = 1.768 (Normal table, subtract answer from 1) and we get 0.0384.

Or 3.84% of your sample means would be greater than 50! We write: <em>P(X \gt 50.0) = P (Z \gt> 1.768) = 0.0384.

Said another way: If you have a sample size of 50 (N = 50) and you obtain a mean greater than 50 then there is only a 3.84% chance that the TRUE MEAN IS 47.

4. What happens if the variability is smaller? Chance σ from 12 to 6 then repeat questions 1 and 4.

5. Repeat the demonstration of Central Limit Theorem and Law of Large Numbers for discrete  distributions

  1. binomial distribution. Replace rchisq() with rbinom(n, size, prob) in the replicate() function example. See Chapter 6.5
  2. poisson distribution. Replace rchisq() with rpois(n, lambda) in the replicate() function example. See Chapter 6.5

Quiz Chapter 8.3

Sampling distribution and hypothesis testing


Chapter 8 contents

 

 

5 – Experimental design

Beyond numbers.

During this course, you will learn about statistics, yes, but my hope and goal for your experience in this class is much more than that: statistical reasoning. To have statistical reasoning skills, you need to be comfortable with the context of how data are acquired, ie, data acquisition. It’s not just about knowing how an instrument performs, eg, under or over the limits of quantification characteristics of the instrument lead to censored missing values, missing not at random (MNAR). In a broad stoke view, data are obtained from two kinds of studies: observational studies and experimental (manipulative) studies. what a correctly designed experiment can tell you about the world, and how a poorly designed experiment works against you. At the end of the semester, you should be familiar with the issues of Randomization, Control, Independence, and Replication.

The basics explained.

Experimental design is a discipline within statistics concerned with the analysis and design of experiments. Design is intended to help research create experiments such that cause and effect can be established from tests of the hypothesis. We introduced elements of experimental design in Chapter 2.4. I recommend a quick re-read: We talk about the different types of bias that can affect research. We also point out parts of the problem you’ll need to think about when planning a project, so you can avoid or reduce bias from the start.

Note 1: Just the highlights from Chapter 2.4: use randomization, blinding, and control measures in the study design. Different strategies may be needed depending on the type of bias.

Here, we expand our discussion of experimental design. We begin with review of definitions and an outline of an important concept in statistics. First, we need to be clear about why we measure or conduct experiments. We do so to collect data (datum is the singular) about a characteristic or trait from a population. Data are observations: they include observations and measurements from instruments. At their best, data are the “facts” of science.

Measurement, Observation, Variables, Values.

Measurement is how we as scientists acquire our data (Chapter 3.4). The process of measuring involves the assignment of numbers, codes, or labels to observations according to rules established prior to any data collection (Stevens 1946, Houle et al 2011). Observations refer to the units of measurement, whereas variables are the characteristics or traits that are measured. A value refers to the particular number, score, or label assigned to a particular sample for a variable. Variables are generic, the thing being measured, values are specific to the subject or sample being measured for the variable. Each discipline in biology has its own set of variables and samples may or may not have different values for each variable measured. Variables are summarized as a statistic (eg, the sample mean), which is a number taken to estimate a parameter, which pertains to the population. Variables and parameters in statistics were discussed in Chapter 3.4. Because numbers or scores or labels can be assigned according to different rules, this means that variables may be measured on different kinds of scales or data types. The different kinds of data types were presented in Chapter 3.1.

Missing values.

A missing value refers to lack of a value for an observation or variable (Chapter 3.5).  Censored values, a type of missing value, include observations for which limited or partial information is available, eg, range, while a missing value lacks any information about the true value.

Missing values can affect analysis and many R algorithms are sensitive or may fail to run in the presence of missing values. Missing data may be of three kinds, and one of them, missing not at random (MNAR), can substantially influence the analysis. MNAR implies some observations are missing because of a systematic bias. Instrument limits of quantification are an example of systematic bias — for example, spectrophotometric absorbance readings of zero for a colorimetric assay (eg, Bradford protein assay) may not represent complete absence of the target, but rather, the lower detection limits of the instrument or the assay method (0.1 mg protein/mL in this case).

The other kinds of missing values are missing completely at random (MCAR) and missing at random (MAR). MCAR implies there is no association between any element of the experiment and the absence of a value. Analysis on MCAR data sets may support unbiased conclusions. MAR includes the random errors that occur during data acquisition: data may be lost because of operator error or power outage (shudder). Unlike MNAR, analysis of MAR, like MCAR data sets may still result in unbiased conclusions.

Note 2: The word “may” is doing some heavy lifting here; obviously, the size of the data set influences whether this claim holds.

One consequence of missing values — a decision how best to handle the lost information. The options are either pairwise deletion or listwise deletion. Listwise means the entire row, all variables, are lost for any record with one or more missing values. You recorded root length, stem length, numbers of secondary roots, time to germinate, but the SPAD meter failed and there is no chlorophyl reading for seedling #212? Sorry, the entire row is deleted. Optionally, the missing data is deleted pairwise — only chlorophyl reading for #212 is lost, all other variables remain.

Users of R will quickly learn some algorithms immediately fail to run if not instructed how to process missing values. In some cases, missing values can be replaced by imputed values.

Note 3: This being an introduction to applied biostatistics, we won’t dwell on the pros and cons of imputation. Instead, I point interested readers to two heavily cited articles: Schafer 1999; Schafer and Graham 2002.

Cause & Effect.

Observations or measurements gathered under controlled conditions (experiments) are essential if we are to answer questions about populations, to separate cause and effect, where one or more events is directly the result of another factor, as opposed to anecdote, a story about an event which, by itself, cannot be used to cause from association (eg, spurious correlations, see Ch 16.2). Recall that anecdotal evidence typically comes from personal experience, where observations may be obtained by non-systematic methods. Well designed experiments, in the classical sense, permit discrimination among competing hypotheses in large part because observations are collected according to strict rules. Well developed hypotheses tested by well-designed experiments permit ruling out alternative explanations (sensu Pratt [1964] strong inference). Observational studies, or epidemiology studies if we are talking about investigations of risk assessment, also may contribute to discussions of cause and effect (see history of smoking by Doll 1998).

Medicine is replete with stories about how a patient showed a particular set of symptoms, and how a physician applied a set of diagnostic protocols. An outcome was achieved, and the physician reports the outcome and circumstances related to the patient to her colleagues. This is an example of a case study, and the focus of investigation is the individual, the patient. The doctor’s report will sound like, “I tried the standard treatment given the diagnosis, and the patient’s symptoms diminished, but later returned. I tried a higher dose, but the patient’s symptoms persisted unabated.” No inferences are made to a wider set of individuals and the report is anecdotal. If observations are made on several patients, this may be a case series.

In ecology, a field biologist may notice a six-legged adult frog (Scott 1999, Alvarez et al 2021) — since frogs typically have four legs, the six-legged frog attracts the biologist’s eye and he jots down the circumstances in which the frog were found: relative humidity, air temperature, ground temperature, where the frogs was found (on lower leaf of a philodendron plant). A water source near where the frog was found is tested for pH with a meter the biologist carries, and a water sample is taken for later testing of herbicides. The frog is collected so that it can be checked for skin parasites. Upon further inspection, the frog did indeed have parasites known to cause deformities in other frog species. However, note that this example too, is a case study. Although the biologist makes additional observations, any conclusions about why the frog has six legs is anecdotal.

From these case studies, no conclusions can be drawn. We cannot say why the patient failed to respond to treatment, nor can we say why the frog has six legs. Why? Because these are singular events, and a variety of explanations can be given as to their causes — importantly, no controls are available, so there’s no way to distinguish among possibilities.

From such anecdotes, however, experiments can be designed. The physician may decide to recruit additional patients with the diagnosed illness and apply the standard treatment to see if her anecdote is a single, unique event, or more indicative of a problem with the treatment. The biologist may collect other frogs from the area near where he found the six-legged frog and check to see if they, too, have the parasites. If additional patients fail to respond to the treatment, then the singular even is more likely to be a phenomenon. If the normal frogs also have similar levels of the parasite then it is unlikely that these parasites caused the malformed frog. With this simple step (recruiting similar patients, finding additional frogs), we can begin to make inferences about cause and effect and in some cases, to generalize our findings.

This is the objective of most statistical procedures, the concept of sampling from a reference population and making distinctions between groups within the sample. The difference between observational and experimental studies then is how the subjects are selected with respect to the groups. In an experiment, the researcher controls and decides which subject receives the treatment, therefore, allocation to groups is manipulated by the researcher. In contrast, subjects included in an observational study have already been “assigned” to a group, but not by us. Assignment to groups, smokers or non smokers, Type II diabetes or no diabetes (etc.), is done by nature.

Now, I do not wish to imply that research that cannot be generalized back to a reference population are worthless. Far from it. In fact, there is a strong argument for specificity. For example, much basic biological research depends on work or model organisms, which in turn may be further partitioned into specific genetic lines (cf. discussion in Rothman et al 2013). And my goodness, what we have learned about the devastation to oceanic islands like Guam when the brown tree snake was introduced (Fritts and Rodda 1998).  Strictly speaking, what has happened to Guam is a case history. But no one would argue that what has happened to Guam cannot happen to Hawaii and other oceanic islands (eg, United States Federal law 384-108 “Brown Tree Snake Control and Eradication Act of 2004”). In other words, even from case histories, generalizations can sometimes be made.

There can also be real reasons to ignore the issue of generality. One benefit of specificity is experimental control. Transgenic lines may differ by single gene knockout or by gene duplication, and clearly the aim of such studies is to evaluate the function (hence purpose) of the gene product (or its absence) on some phenotype. In this sense it may not seem important that an transgenic mouse is not representative of a wild outbred mouse population. However, this argument is fundamentally one of expedience — such studies do result in specific results, results that cannot be generalized beyond the strains involved. It ignores the issue of genetic background — all of the genes that affect a trait in addition to the candidate gene under study (Sigmund 2000; Lariviere et al 2001). Transgenic mice of different inbred strains or their hybrids may have very different alleles at other genes that may influence a phenotype, hence, the results of the gene knockdown or other engineering result in different outcomes. Results of genetic manipulations on inbred strains, no matter how sophisticated, the conclusions are strain- or hybrid cross-specific. Thus, although technically and financially difficult, conclusions are better, more generalizable, when conducted with many different inbred lines and verified in outbred mouse populations precisely because genetic background often influences function of single genes (Sigmund 2000; Lariviere et al 2001).

“Causal criteria:” Logic of causation in medicine.

This section is in progress. Just a list of key points and references

Throughout this text, emphasis on the power of experimentation is emphasized. Well designed experiments …

Henle-Koch’s postulates (1877, 1882), working on tuberculosis reported a set of causal criteria to establish link between a microorganism and a disease, are the following:

  • The microorganism must be found in abundance in all organisms suffering from the disease, but should not be found in healthy organisms.
  • The microorganism must be isolated from a diseased organism and grown in pure culture.
  • The cultured microorganism should cause disease when introduced into a healthy organism.
  • The microorganism must be reisolated from the inoculated, diseased experimental host and identified as being identical to the original specific causative agent.

Robert Koch wrote these more than 100 years ago, so, clearly, understanding of infectious disease has improved. Evan’s postulates, quoted from A Dictionary of Epidemiology, 5th edition (pp. 86-87).

  1. Prevalence of the disease should be significantly higher in those exposed to the hypothesized cause than in controls not so exposed.
  2. Exposure to the hypothesized cause should be more frequent among those with the disease than in controls without the disease—when all other risk factors are held constant.
  3. Incidence of the disease should be significantly higher in those exposed to the hypothesized cause than in those not so exposed, as shown by prospective studies.
  4. The disease should follow exposure to the hypothesized causative agent with a normal or log-normal distribution of incubation periods.
  5. A spectrum of host responses should follow exposure to the hypothesized agent along a logical biological gradient from mild to severe.
  6. A measurable host response following exposure to the hypothesized cause should have a high probability of appearing in those lacking this before exposure (eg, antibody, cancer cells) or should increase in magnitude if present before exposure. This response pattern should occur infrequently in persons not so exposed.
  7. Experimental reproduction of the disease should occur more frequently in animals or humans appropriately exposed to the hypothesized cause than in those not so exposed; this exposure may be deliberate in volunteers, experimentally induced in the laboratory, or may represent a regulation of natural exposure.
  8. Elimination or modification of the hypothesized cause should decrease the incidence of the disease (eg, attenuation of a virus, removal of tar from cigarettes).
  9. Prevention or modification of the host’s response on exposure to the hypothesized cause should decrease or eliminate the disease (eg, immunization, drugs to lower cholesterol, specific lymphocyte transfer factor in cancer).
  10. All of the relationships and findings should make biological and epidemiological sense.

Fredericks, D. N., & Relman, D. A. (1996). Sequence-based identification of microbial pathogens: a reconsideration of Koch’s postulates. Clinical microbiology reviews, 9(1), 18-33.

Correlation (association) does not imply causation, a well-worn truism in any application of critical thinking skills. Conversely, causation does imply association (correlation); If a change in one variable directly changes another variable, there will be a statistical association between the two variables. Thus, a correlation between two variables is necessary, but not sufficient evidence for cause and effect between two variables.

Note 4: Association is the more general term for a possible relationship between two or more variables. A correlation in statistics generally refers to a linear association (Chapter 16); the aforementioned truism should be restated as association (or statistical relationship) does not imply causation.

However, sometimes association does point to a cause. A familiar example is association between tobacco cigarette smoking causes lung cancer.  Surgeon General Luthar Terry’s 1964 report (link to document in National Library of Medicine), presented a strong case linking smoking to elevated risk of lung cancer and coronary artery disease.

Bradford Hill’s guidelines to evaluate causal effects based on epidemiology (Hill 1965, see also Sussar 1999, Fedak et al 2015). Set of necessary and sufficient conditions, inductive reasoning.

  1. Strength of association
  2. Consistency of observed association
  3. Specificity of association
  4. Temporal relationship of the association
  5. Biological gradient, eg, a dose-response curve
  6. Biological plausibility
  7. Coherence, the cause and effect inference should not conflict with what is known about the etiology of a disease.

Follows and extends David Hume’s (1739) causation criteria: association (Hill #1), cause precedes effect (Hill #4), direction of connection.

Validity.

An obvious objective of research is to reach valid conclusions about fundamental questions. A helpful distinction between the specific and the generalizable experiment is to recognize there are two forms of validity in research: internal validity and external validity (Elwood 2013). Internal validity is the quality of a designed study that determines whether cause and effect can be determined. Random assignment of subjects to treatment groups enhances the internal validity of the study. External validity relates to how general the assessment of cause and effect can be to other populations. Thus, random sampling from a reference population has to do with whether or not the study has external validity.

Additional definitions.

We proceed now with definitions. We use the term population in a special and restrictive way in statistics. Our definition includes the one you are already familiar with, but it also means more than that.

Populations are the entire group of individuals that you want to investigate. In statistics, the entire groups is actually the entire class with the observation — so if we are referring to the average body weight of house mice, we’re actually referring to the body weight as the population — it’s a subtle distinction, not essential for our introduction to biostatistics. When we conduct experiments and apply statistical tests on collected data, we generally intend to make inferences (draw conclusions) from our results back to the population.

Population has a strict application in statistics, but the definition also includes our general understanding of the word population. For example, Examples of a population in the general sense one may refer to include: .… the entire human population existing today .… the entire collection of U.S. citizens. .… all the individuals in an entire species. .… all individuals in a population of a species (eg, house mice in a dairy barn in Hawai’i) .… all of us in this classroom (if we are ONLY interested in US) If you could measure the entire population then there would be no need to do (or learn) statistics! Populations usually are in the thousands, millions, or billions of individuals. Here, population is used in the everyday sense that we think of — a collection of individuals that share a characteristic.

A more formal definition of “population” in statistics reads as follows: A statistical population is the complete set of possible measurements on a trait or characteristic corresponding to the entire collection of sampling units for which conclusions are intended.

To conclude, in this class, when we talk about population, we will generally be using it in the everyday sense of the word. However, keep in mind that the definition is more restrictive than that and the key is to identify what sampling units are measured.

Conclusions.

This is only the beginning, the basics of experimental design. Beginning with RA Fisher‘s 1935 book, The Design of Experiments (nine editions, some freely accessible via internet search), which is still very much worth reading, entire books are written on the subject, as you can well imagine.

Note 5: Fisher’s The Design of Experiments concerned experimental design, randomization, blocking, and replication, particularly from perspective of agriculture research. Most readers of this eBook likely know that RA Fisher’s broader career absolutely included eugenics and hereditarian views about human populations and race. Those views are not in The Design of Experiments, which is largely about methods and statistical framework for inference.

We will also return to the subject of Experimental Design throughout the book. We will return to random sampling in Chapter 5.5. Next we discuss distinctions between experiments and observational studies with respect to sampling of populations.

A bit of a disclaimer here before proceeding; while I cite several papers for examples in experimental design in Chapter 5, reader’s should not read into this that I am either criticizing or endorsing the published experiments. Experimental design will always have elements of compromise — the trick of course is knowing which choices influence validity (Thompson and Panacek 2006).

Questions.

  1. Define in your own words the following terms
  • reference population
  • subjects
  • specific versus general conclusions
  • random sampling
  • convenience sampling
  • haphazard sampling
  • research validity
  1. Revisit our cell experiment, “What is the sampling unit in the following cell experiment?” How would you change this experiment so that there will be biological and not just technical replication?
  2. Describe the type of sampling for each research scenario described?
  • All African snails on a staircase at Chaminade University are collected on a Thursday evening.
  • All African snails on a staircase at Chaminade University are collected every Thursday evening for six months.
  • All African snails on all staircases at Chaminade University are collected.
  • African snails are studied in the lab, then returned to the areas from which they were collected. Days later, the researcher collects snails from the same area.
  • African snails are studied in the lab, then returned to the areas from which they were collected. Days later, the researcher collects snails from a different area.
  • The Chaminade University campus is divided into grids. Grids that include stairwells are marked. Before collecting snails, the researcher randomly selects from the list of grids and searches for snails only in those grids selected from the list.
  • Giant African Snail (Lissachatina fulica, formerly Achatina fulica). Image by M. Dohm
    Figure 1. Giant African Snail (Lissachatina fulica, formerly Achatina fulica). Image by M. Dohm.
  1. This next scenario may be evaluated by you for potential sources of bias. Review the list of bias listed above. A researcher wants to do a population count of feral cats on campus. Feral cats are active at night, so he decides to set up a feeding station near a light post. The researcher sits all night in a parked car yards away and watches the feeding station for visits by cats. The researcher repeats these observations over the course of a week, moving the feeding station to different campus locations each night, and reports the total number of cats seen during the week as an estimate of the population size. Be able to discuss this study in terms of potential and actual bias.

Quiz

Including the quiz on this page, a total of 65 questions among the several subchapters, a mix of true or false and multiple choice question format.

Quiz Chapter 5

Experimental design


Chapter 5 contents