Germán Rodríguez

Demographic Methods
Princeton University
We will work through the example in Preston et al., sections 2.2 and 2.3.

I copied the counts of mid-year population and deaths by Age for Sweden and Kazakhstan from Table 2.1 into a text file which is available in the course website. The file is in "long" format. As usual I show the analysis in R and Stata.

. infile str10 country str5 ageg pop deaths /// > using https://data.princeton.edu/eco572/datasets/preston21long.dat (38 observations read) . list in 1/3 +-------------------------------------+ | country ageg pop deaths | |-------------------------------------| 1. | Kazakhstan 0 174078 3720 | 2. | Kazakhstan 01-4 754758 1220 | 3. | Kazakhstan 05-9 879129 396 | +-------------------------------------+

> p21 <- read.table("https://data.princeton.edu/eco572/datasets/preston21long.dat", + header=FALSE, col.names = c("country", "ageg", "pop", "deaths")) > head(p21, 3) country ageg pop deaths 1 Kazakhstan 0 174078 3720 2 Kazakhstan 01-4 754758 1220 3 Kazakhstan 05-9 879129 396

The first thing to do is to calculate the age-specific rates, dividing deaths by population and multiplying by 1000:

. gen rates = 1000 * deaths / pop

> p21 <- mutate(p21, rates = 1000 * deaths / pop)

Crude death rates are just a weighted average of age-specific rates using the population in each age group as the weight. We can easily compute them like this:

. tabstat rates [fw=pop], by(country) Summary for variables: rates by categories of: country country | mean -----------+---------- Kazakhstan | 7.423042 Sweden | 10.54756 -----------+---------- Total | 8.470285 ----------------------

> group_by(p21, country) %>% + summarize(crude = weighted.mean(rates, pop)) Source: local data frame [2 x 2] country crude (fctr) (dbl) 1 Kazakhstan 7.423042 2 Sweden 10.547561

The interesting result here is that mortality appears to be *lower* in
Kazakhstan than in Sweden.

Following Preston et al., we will standardize the rates using the weighted average of the two population compositions as the standard. To do this we calculate first the percent distribution in each country and then average. While we are at it we will also calculate the average rates.

> p21 <- group_by(p21, country) %>% + mutate(pct = pop/sum(pop)) > avg <- group_by(p21, ageg) %>% + summarize(pop = mean(pct), rates = mean(rates))

. egen pcpop = pc(pop), by(country) . egen avgcomp = mean(pcpop), by(ageg)

Now we can easily compute the standardized rates using the average
composition as the standard.
Because we assigned the grouped data to `p21`

we can skip the grouping step.

> summarize(p21, std = weighted.mean(rates, avg$pop)) Source: local data frame [2 x 2] country std (fctr) (dbl) 1 Kazakhstan 11.881997 2 Sweden 7.374094

. tabstat rates [aw=avgcomp], by(country) Summary for variables: rates by categories of: country country | mean -----------+---------- Kazakhstan | 11.882 Sweden | 7.374094 -----------+---------- Total | 9.628045 ----------------------

We see that once we adjust for age composition the mortality rate is in
fact *higher* in Kzakhstan than in Sweden, in fact 61.1% higher.

Frequently we don't have age-specific rates, but can easily obtain the age distribution. We can still do a form of standardization applying the rates of one country (or any other standard) to the two age distributions. Let us use the rates for Sweden and apply them to both countries. To do this we sort by age and country, and for each age pick the second value

. bysort age (country): gen swrates = rates[2] . tabstat swrates [fw=pop], by(country) Summary for variables: swrates by categories of: country country | mean -----------+---------- Kazakhstan | 4.200627 Sweden | 10.54756 -----------+---------- Total | 6.327926 ----------------------

> sweden <- filter(p21, country == "Sweden") > summarize(p21, istd = weighted.mean(sweden$rates, pop)) Source: local data frame [2 x 2] country istd (fctr) (dbl) 1 Kazakhstan 4.200627 2 Sweden 10.547561

So if Kazakhstan has Swedish rates the CDR would be 4.2. The ratio of
the actual CDR to the expected rate under the standard age structure
is known as the *standardized mortality ratio* and turns out to be

. quietly sum rates [fw=pop] if country == "Kazakhstan" . scalar kzcmr = r(mean) . quietly sum swrates [fw=pop] if country == "Kazakhstan" . di "SMR = ", kzcmr/r(mean) SMR = 1.7671271

> filter(p21, country == "Kazakhstan") %>% + summarize(istd = weighted.mean(sweden$rates,pop), + crude = weighted.mean(rates,pop), smr = crude/istd) Source: local data frame [1 x 4] country istd crude smr (fctr) (dbl) (dbl) (dbl) 1 Kazakhstan 4.200627 7.423042 1.767127

Mortality is in fact 77% higher than it would be at Swedish rates with the observed composition.

Preston et al. show how to decompose a difference between two rates into a part due to compositional differences and a part due to differences in rates. The first part is obtained applying the average composition to the observed rates. To assess the second component we apply the observed composition to the average rates.

. egen avgrates = mean(rates), by(ageg) . tabstat avgrates [fw=pop], by(country) Summary for variables: avgrates by categories of: country country | mean -----------+---------- Kazakhstan | 5.811834 Sweden | 13.44426 -----------+---------- Total | 8.36999 ----------------------

> dbr <- summarize(p21, + crude = weighted.mean(rates,pop), + comp = weighted.mean(rates, avg$pop), + rates = weighted.mean(avg$rates, pop)); dbr Source: local data frame [2 x 4] country crude comp rates (fctr) (dbl) (dbl) (dbl) 1 Kazakhstan 7.423042 11.881997 5.811834 2 Sweden 10.547561 7.374094 13.444256

We see that with the same rates the older composition of Sweden would result in much higher mortality than observed. We can now verify the decomposition:

. display "Difference = " 10.55-7.42 /// > "; Composition = " 13.44-5.81 /// > "; Rates = " 7.37-11.88 Difference = 3.13; Composition = 7.63; Rates = -4.51

> c(-1, 1) %*% as.matrix(data.frame(dbr[,-1])) crude comp rates [1,] 3.124519 -4.507903 7.632422

So a difference in 3.13 points in the CBR between Sweden and Kazakhstan
results from a compositional effect of +7.63 and a difference in rates of -4.51.
I wrote a Stata program that can do these calculations for data in "wide" format.
In net-aware Stata type `net from https://data.princeton.edu/eco572/stata`

and check out `ddrate`

.

So far we have focused on attempts to summarize mortality in a single number, but why not look at the complete set of rates? I would like to plot the rates using the mid-points of the ages groups (with 87.5 for 85+). To compute these I first convert age from a string to numbers (1 to 19) and then figure out the midpoint of each group.

. encode ageg, gen(agei) // "05-9" becomes 3 . gen agem = 5*agei - 7.5 // 3 becomes 7.5 . replace agem = 0.5 if ageg == "0" (2 real changes made) . replace agem = 3 if ageg == "01-4" (2 real changes made)

> p21 <- mutate(p21, + agei = as.numeric(ageg), + age = ifelse(agei > 2, 5*agei - 7.5, ifelse(agei == 1, 0, 3)))

We then plot the rates using a log-scale (otherwise we wouldn't see much)

. twoway (line rate agem if country=="Sweden") /// > (line rate agem if country=="Kazakhstan") /// > , yscale(log) ylabel(50 100) /// > title("Age-Specific Mortality Rates") /// > xtitle("Age") ytitle("log(ASMR)") /// > legend(order(2 "Kazakhstan" 1 "Sweden") ring(0) pos(5) cols(1)) . graph export swkzasmr.png, replace width(500) height(400) (file swkzasmr.png written in PNG format)

> ggplot(p21, aes(x=age, y=log(rates), color=country)) + geom_line() > ggsave("swkzasmrr.png", width=500/72, height=400/72, dpi=72)

We see clearly that mortality is higher in Kazakhstan than in Sweden at *every* age.
The lines are not exactly parallel, in fact the difference seems to become smaller
at older ages. (We will discuss later how heterogeneity can produce this type of
convergence.)

The aim of standardization is to control for a compositional variable. This, of course, is also one of the aims of regression analysis. I now show how one could analyze the data using Poisson regression, a method appropriate for count data.

This regression technique assumes that the number of deaths has mean (and variance) given by the product of a rate and exposure time. The log of the mean is then the log of the rate (which is modelled using a linear predictor, just as in linear regression) times the log of exposure (which is defined as an offset, or a fixed part of the linear predictor). That's why the code below computes the log of the mid-year population counts. We also represent country using a dummy variable for Kazakhstan.

. gen logexpo = log(pop) . gen kz = country == "Kazakhstan" . poisson deaths i.agei kz, offset(logexpo) Iteration 0: log likelihood = -49623.369 Iteration 1: log likelihood = -10315.221 Iteration 2: log likelihood = -1482.9267 Iteration 3: log likelihood = -1275.2697 Iteration 4: log likelihood = -1274.853 Iteration 5: log likelihood = -1274.853 Poisson regression Number of obs = 38 LR chi2(19) = 315966.62 Prob > chi2 = 0.0000 Log likelihood = -1274.853 Pseudo R2 = 0.9920 ------------------------------------------------------------------------------ deaths | Coef. Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- agei | 01-4 | -2.600295 .0322872 -80.54 0.000 -2.663577 -2.537013 05-9 | -3.823097 .0509118 -75.09 0.000 -3.922882 -3.723311 10-14 | -4.003534 .0571946 -70.00 0.000 -4.115633 -3.891434 15-19 | -3.29342 .043102 -76.41 0.000 -3.377899 -3.208942 20-24 | -2.99395 .0395719 -75.66 0.000 -3.071509 -2.91639 25-29 | -3.02696 .03777 -80.14 0.000 -3.100988 -2.952932 30-34 | -2.743422 .033986 -80.72 0.000 -2.810034 -2.676811 35-39 | -2.436497 .0318358 -76.53 0.000 -2.498894 -2.3741 40-44 | -1.985014 .0285733 -69.47 0.000 -2.041017 -1.929012 45-49 | -1.588591 .0280586 -56.62 0.000 -1.643584 -1.533597 50-54 | -1.231015 .0229505 -53.64 0.000 -1.275997 -1.186032 55-59 | -.6591381 .0220602 -29.88 0.000 -.7023753 -.6159008 60-64 | -.3400267 .0199076 -17.08 0.000 -.3790448 -.3010086 65-69 | .2502514 .0188483 13.28 0.000 .2133094 .2871935 70-74 | .6572624 .0186292 35.28 0.000 .6207498 .693775 75-79 | 1.094818 .0178766 61.24 0.000 1.05978 1.129855 80-84 | 1.719851 .017546 98.02 0.000 1.685462 1.754241 85+ | 2.451387 .0170575 143.71 0.000 2.417955 2.484819 | kz | .4794218 .0062738 76.42 0.000 .4671254 .4917181 _cons | -4.445508 .0166385 -267.18 0.000 -4.478119 -4.412897 logexpo | 1 (offset) ------------------------------------------------------------------------------ . di exp(_b[kz]) 1.6151402

> p21 <- mutate(p21, logexpo = log(pop), kz = as.numeric(country=="Kazakhstan")) > fit <- glm(deaths ~ ageg + kz + offset(log(pop)), data = p21, family = poisson) > coef(summary(fit)) Estimate Std. Error z value Pr(>|z|) (Intercept) -4.4455076 0.01663845 -267.18275 0.000000e+00 ageg01-4 -2.6002957 0.03228717 -80.53649 0.000000e+00 ageg05-9 -3.8230970 0.05091182 -75.09253 0.000000e+00 ageg10-14 -4.0035340 0.05719463 -69.99842 0.000000e+00 ageg15-19 -3.2934201 0.04310201 -76.40989 0.000000e+00 ageg20-24 -2.9939502 0.03957193 -75.65843 0.000000e+00 ageg25-29 -3.0269600 0.03777003 -80.14185 0.000000e+00 ageg30-34 -2.7434226 0.03398604 -80.72205 0.000000e+00 ageg35-39 -2.4364968 0.03183579 -76.53326 0.000000e+00 ageg40-44 -1.9850146 0.02857332 -69.47090 0.000000e+00 ageg45-49 -1.5885907 0.02805860 -56.61690 0.000000e+00 ageg50-54 -1.2310144 0.02295053 -53.63773 0.000000e+00 ageg55-59 -0.6591382 0.02206023 -29.87903 3.685603e-196 ageg60-64 -0.3400269 0.01990755 -17.08030 2.080363e-65 ageg65-69 0.2502514 0.01884833 13.27711 3.142725e-40 ageg70-74 0.6572621 0.01862924 35.28122 1.139967e-272 ageg75-79 1.0948175 0.01787660 61.24306 0.000000e+00 ageg80-84 1.7198508 0.01754596 98.01978 0.000000e+00 ageg85+ 2.4513869 0.01705754 143.71277 0.000000e+00 kz 0.4794215 0.00627376 76.41694 0.000000e+00 > exp(coef(fit)["kz"]) kz 1.61514

Exponentiating the coefficient of Kazakhstan we see that on average age-specific mortality is 61.5% higher than in Sweden, a result very similar to the 61.2% that we obtained using direct standardization.

An advantage of the regression approach is that one can easily control for multiple confounders. Also, we can test for the presence of an interaction, which would question the validity of the additive model underlying direct standardization. In this case, adding the interaction would make the model saturated for this data, so we just test goodness of fit.

. estat gof Deviance goodness-of-fit = 2218.456 Prob > chi2(18) = 0.0000 Pearson goodness-of-fit = 2026.899 Prob > chi2(18) = 0.0000

> data.frame(deviance(fit), df=fit$df.residual) deviance.fit. df 1 2218.549 18

The deviance of 2218 on 18 d.f. is highly significant, so we conclude that the rates are not proportional. With data based on national counts, however, even small differences can be significant. This interaction, while significant, represents less than one percent of the combined effect of age and country

. di r(chi2_d)/e(chi2) .00702117

> deviance(fit)/(fit$null.deviance - deviance(fit)) [1] 0.007021467

If the discussion in the last section doesn't make much sense to you don't worry. Just make a note to come back and re-read this part after you take WWS509/POP507, the generalized linear models course.

Updated 22-Jan-2016