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"lm" is for "linear model" - facet_grid(~sex) + # adding this function will generate a separate graph for each category of sex + facet_wrap(~sex) + # adding this function will generate a separate graph for each category of sex scale_x_continuous(name = "Age" , limits = c(0,100)) + # Format the x-axis to a range between 0 and 100 @@ -162,7 +162,7 @@ tbe %>% geom_histogram() + - facet_grid(~hyper) + # add facet_grid() to get a graph for each hyper status + facet_wrap(~hyper) + # add facet_wrap() to get a graph for each hyper status labs( x = "Length of hospitalization in days", @@ -177,7 +177,7 @@ tbe %>% geom_histogram(aes(y = ..density..)) + #here we are telling ggplot2 to display the density and not the freq count - facet_grid(~hyper) + # add facet_grid() to get a graph for each hyper status + facet_wrap(~hyper) + # add facet_wrap() to get a graph for each hyper status #the function below will add the normal curve. stat_function(fun = dnorm, #The fun = argument we are specifying that 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Characteristic

0, N = 188

+1

1, N = 103

+1
sex
+
+
    086 (46%)53 (51%)
    1102 (54%)50 (49%)
age_group
+
+
    0-1987 (48%)50 (50%)
    20+96 (52%)50 (50%)
    Unknown53
1 +

n (%)

+ + + + +


+ +
+QUESTION: What percentage of ill cases was female? +
+ +
+
+ + +
+

Step 5: Develop and test hypothesis

+

In the previous step we have described the number of ill attendees by date of onset, and we have described how sex and age were distributed according to illness status. The next step in an outbreak investigation would be to develop and test hypothesis about the possible source of the outbreak.

+
+

Step 5.1: Compute food-specific attack rates and % of cases exposed

+

Task: Create a list with exposed and unexposed cases and calculate the attack rates.

+
+ +

Click to read a hint

+ +


+

You could use the same approach we used before with tbl_summary() from {gtsummary}, but to calculate attack rates, you’ll need to change from column to row percentages.

+
+
+ +

Click to see a solution (try it yourself first!)

+ +


+
+
tira_clean %>% 
+  
+  select(tira, wmousse, dmousse, wmousse, beer, 
+         redjelly, fruitsalad, tomato, mince, salmon,
+         horseradish, chickenwin, roastbeef, pork, ill) %>% # we select the columns we're interested in
+  
+  tbl_summary(by = ill, percent = "row") %>% #we assign the grouping column and define where we want the percentages by row
+  
+  add_overall()    # we add the overall counts
+
+
+ + + ++++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

Characteristic

Overall, N = 291

+1

0, N = 188

+1

1, N = 103

+1
tira
+
+
+
    0165 (100%)158 (96%)7 (4.2%)
    1121 (100%)27 (22%)94 (78%)
    Unknown532
wmousse
+
+
+
    0205 (100%)156 (76%)49 (24%)
    172 (100%)23 (32%)49 (68%)
    Unknown1495
dmousse
+
+
+
    0174 (100%)148 (85%)26 (15%)
    1113 (100%)37 (33%)76 (67%)
    Unknown431
beer
+
+
+
    0165 (100%)96 (58%)69 (42%)
    1106 (100%)76 (72%)30 (28%)
    Unknown20164
redjelly
+
+
+
    0212 (100%)154 (73%)58 (27%)
    179 (100%)34 (43%)45 (57%)
fruitsalad
+
+
+
    0220 (100%)163 (74%)57 (26%)
    171 (100%)25 (35%)46 (65%)
tomato
+
+
+
    0208 (100%)140 (67%)68 (33%)
    183 (100%)48 (58%)35 (42%)
mince
+
+
+
    0204 (100%)133 (65%)71 (35%)
    187 (100%)55 (63%)32 (37%)
salmon
+
+
+
    0183 (100%)120 (66%)63 (34%)
    1104 (100%)67 (64%)37 (36%)
    Unknown413
horseradish
+
+
+
    0217 (100%)145 (67%)72 (33%)
    172 (100%)42 (58%)30 (42%)
    Unknown211
chickenwin
+
+
+
    0207 (100%)137 (66%)70 (34%)
    184 (100%)51 (61%)33 (39%)
roastbeef
+
+
+
    0262 (100%)167 (64%)95 (36%)
    129 (100%)21 (72%)8 (28%)
pork
+
+
+
    0169 (100%)115 (68%)54 (32%)
    1120 (100%)72 (60%)48 (40%)
    Unknown211
1 +

n (%)

+ +
+
+
+


+
+
+QUESTION: What is the attack rate in those who ate withe mousse? +
+ +
+
+
+
+

Step 5.2: Estimate the relative risk of the different exposures

+

Task: First, estimate the relative risk for the exposure of having eaten white mousse at the dinner

+
+ +

Click to read a hint

+ +


+

One way yo estimate the relative risk in these situations is to use the function riskratio() from the package {epitools}. You need to first give the exposure and then the predictor.

+
+
+ +

Click to see a solution (try it yourself first!)

+ +


+
+
# Relative risk wmousse
+riskratio(tira_clean$wmousse, tira_clean$ill)
+
+
$data
+         Outcome
+Predictor   0  1 Total
+    0     156 49   205
+    1      23 49    72
+    Total 179 98   277
+
+$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0 1.000000       NA       NA
+        1 2.847222 2.128267 3.809049
+
+$p.value
+         two-sided
+Predictor   midp.exact fisher.exact   chi.square
+        0           NA           NA           NA
+        1 4.219824e-11 5.825494e-11 1.576275e-11
+
+$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+
+


+
+
+QUESTION: What is the risk of illness among those who ate white mousse compared to those who did not have this food ? +
+ +
+
+

Task: Calculate now the risk ratio for all the other exposures to develop hypothesis of the possible source

+
+ +

Click to read a hint

+ +


+

You could write the riskratio() function changing the exposure variable one by one. However, to save time, avoid making errors and being more efficient, you could try to use approaches that allow you to apply the same function to many different objects (e.g., multiple columns) simultaneously. There are different options for this such as loops, lapply or purrr. Here we give the solution with purrr, so if you want to explore further purrr have a look at the dedicated section in the EpiRhandbook.

+
+
+ +

Click to see a solution (try it yourself first!)

+ +


+
+
tira_clean %>% 
+  select(-uniquekey, -age, -age_group, -sex, -dateonset, -ill) %>% # we first remove the columns we're not interested + the outcome column
+  map(.f = ~riskratio(.x, tira_clean$ill))     # here we call the function for each column. ".x" represents all the columns in the database that have not been removed previously
+
+
$tira
+$tira$data
+         Outcome
+Predictor   0   1 Total
+    0     158   7   165
+    1      27  94   121
+    Total 185 101   286
+
+$tira$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0  1.00000       NA       NA
+        1 18.31169 8.814202 38.04291
+
+$tira$p.value
+         two-sided
+Predictor midp.exact fisher.exact   chi.square
+        0         NA           NA           NA
+        1          0 1.794084e-41 9.939537e-38
+
+$tira$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$tportion
+$tportion$data
+         Outcome
+Predictor   0   1 Total
+    0     158   7   165
+    1      21  44    65
+    2       4  38    42
+    3       2  12    14
+    Total 185 101   286
+
+$tportion$measure
+         risk ratio with 95% C.I.
+Predictor estimate     lower    upper
+        0  1.00000        NA       NA
+        1 15.95604  7.581600 33.58069
+        2 21.32653 10.261806 44.32172
+        3 20.20408  9.488562 43.02073
+
+$tportion$p.value
+         two-sided
+Predictor   midp.exact fisher.exact   chi.square
+        0           NA           NA           NA
+        1 0.000000e+00 8.032814e-24 1.810033e-25
+        2 0.000000e+00 8.285272e-30 1.099996e-33
+        3 2.752243e-12 2.733856e-12 2.064029e-21
+
+$tportion$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$wmousse
+$wmousse$data
+         Outcome
+Predictor   0  1 Total
+    0     156 49   205
+    1      23 49    72
+    Total 179 98   277
+
+$wmousse$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0 1.000000       NA       NA
+        1 2.847222 2.128267 3.809049
+
+$wmousse$p.value
+         two-sided
+Predictor   midp.exact fisher.exact   chi.square
+        0           NA           NA           NA
+        1 4.219824e-11 5.825494e-11 1.576275e-11
+
+$wmousse$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$dmousse
+$dmousse$data
+         Outcome
+Predictor   0   1 Total
+    0     148  26   174
+    1      37  76   113
+    Total 185 102   287
+
+$dmousse$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0 1.000000       NA       NA
+        1 4.501021 3.086945 6.562862
+
+$dmousse$p.value
+         two-sided
+Predictor midp.exact fisher.exact   chi.square
+        0         NA           NA           NA
+        1          0 1.167009e-19 1.474249e-19
+
+$dmousse$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$mousse
+$mousse$data
+         Outcome
+Predictor   0   1 Total
+    0     144  22   166
+    1      42  81   123
+    Total 186 103   289
+
+$mousse$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0 1.000000       NA       NA
+        1 4.968958 3.299403 7.483336
+
+$mousse$p.value
+         two-sided
+Predictor midp.exact fisher.exact   chi.square
+        0         NA           NA           NA
+        1          0 1.257341e-20 2.669081e-20
+
+$mousse$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$mportion
+$mportion$data
+         Outcome
+Predictor   0  1 Total
+    0     144 22   166
+    1      17 38    55
+    2      17 30    47
+    3       4  7    11
+    Total 182 97   279
+
+$mportion$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0 1.000000       NA       NA
+        1 5.213223 3.399895 7.993687
+        2 4.816248 3.087198 7.513688
+        3 4.801653 2.655107 8.683592
+
+$mportion$p.value
+         two-sided
+Predictor   midp.exact fisher.exact   chi.square
+        0           NA           NA           NA
+        1 1.421085e-14 1.319062e-14 7.031442e-16
+        2 2.832690e-11 2.617491e-11 1.035313e-12
+        3 3.562368e-04 3.317748e-04 1.230337e-05
+
+$mportion$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$beer
+$beer$data
+         Outcome
+Predictor   0  1 Total
+    0      96 69   165
+    1      76 30   106
+    Total 172 99   271
+
+$beer$measure
+         risk ratio with 95% C.I.
+Predictor  estimate     lower   upper
+        0 1.0000000        NA      NA
+        1 0.6767842 0.4757688 0.96273
+
+$beer$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1 0.02439079   0.02806394 0.02413045
+
+$beer$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$redjelly
+$redjelly$data
+         Outcome
+Predictor   0   1 Total
+    0     154  58   212
+    1      34  45    79
+    Total 188 103   291
+
+$redjelly$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0  1.00000       NA       NA
+        1  2.08206 1.555917 2.786123
+
+$redjelly$p.value
+         two-sided
+Predictor   midp.exact fisher.exact   chi.square
+        0           NA           NA           NA
+        1 4.364656e-06 4.415074e-06 2.646618e-06
+
+$redjelly$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$fruitsalad
+$fruitsalad$data
+         Outcome
+Predictor   0   1 Total
+    0     163  57   220
+    1      25  46    71
+    Total 188 103   291
+
+$fruitsalad$measure
+         risk ratio with 95% C.I.
+Predictor estimate    lower    upper
+        0 1.000000       NA       NA
+        1 2.500618 1.886773 3.314171
+
+$fruitsalad$p.value
+         two-sided
+Predictor   midp.exact fisher.exact   chi.square
+        0           NA           NA           NA
+        1 6.152039e-09 9.998203e-09 2.572527e-09
+
+$fruitsalad$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$tomato
+$tomato$data
+         Outcome
+Predictor   0   1 Total
+    0     140  68   208
+    1      48  35    83
+    Total 188 103   291
+
+$tomato$measure
+         risk ratio with 95% C.I.
+Predictor estimate     lower    upper
+        0 1.000000        NA       NA
+        1 1.289865 0.9379601 1.773799
+
+$tomato$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1   0.131876    0.1368934  0.1269162
+
+$tomato$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$mince
+$mince$data
+         Outcome
+Predictor   0   1 Total
+    0     133  71   204
+    1      55  32    87
+    Total 188 103   291
+
+$mince$measure
+         risk ratio with 95% C.I.
+Predictor estimate     lower    upper
+        0 1.000000        NA       NA
+        1 1.056824 0.7571858 1.475036
+
+$mince$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1  0.7455228    0.7893882  0.7467072
+
+$mince$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$salmon
+$salmon$data
+         Outcome
+Predictor   0   1 Total
+    0     120  63   183
+    1      67  37   104
+    Total 187 100   287
+
+$salmon$measure
+         risk ratio with 95% C.I.
+Predictor estimate     lower    upper
+        0 1.000000        NA       NA
+        1 1.033425 0.7452531 1.433026
+
+$salmon$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1  0.8427855    0.8976422  0.8440915
+
+$salmon$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$horseradish
+$horseradish$data
+         Outcome
+Predictor   0   1 Total
+    0     145  72   217
+    1      42  30    72
+    Total 187 102   289
+
+$horseradish$measure
+         risk ratio with 95% C.I.
+Predictor estimate     lower    upper
+        0 1.000000        NA       NA
+        1 1.255787 0.9008457 1.750578
+
+$horseradish$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1  0.1973981    0.2026013  0.1916218
+
+$horseradish$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$chickenwin
+$chickenwin$data
+         Outcome
+Predictor   0   1 Total
+    0     137  70   207
+    1      51  33    84
+    Total 188 103   291
+
+$chickenwin$measure
+         risk ratio with 95% C.I.
+Predictor estimate     lower    upper
+        0 1.000000        NA       NA
+        1 1.161735 0.8376217 1.611261
+
+$chickenwin$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1  0.3803364    0.4176598  0.3766386
+
+$chickenwin$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$roastbeef
+$roastbeef$data
+         Outcome
+Predictor   0   1 Total
+    0     167  95   262
+    1      21   8    29
+    Total 188 103   291
+
+$roastbeef$measure
+         risk ratio with 95% C.I.
+Predictor  estimate     lower    upper
+        0 1.0000000        NA       NA
+        1 0.7607985 0.4129094 1.401795
+
+$roastbeef$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1   0.365577    0.4172932  0.3540318
+
+$roastbeef$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+$pork
+$pork$data
+         Outcome
+Predictor   0   1 Total
+    0     115  54   169
+    1      72  48   120
+    Total 187 102   289
+
+$pork$measure
+         risk ratio with 95% C.I.
+Predictor estimate     lower    upper
+        0 1.000000        NA       NA
+        1 1.251852 0.9176849 1.707703
+
+$pork$p.value
+         two-sided
+Predictor midp.exact fisher.exact chi.square
+        0         NA           NA         NA
+        1  0.1620037    0.1708777  0.1583527
+
+$pork$correction
+[1] FALSE
+
+attr(,"method")
+[1] "Unconditional MLE & normal approximation (Wald) CI"
+
+
+


+
+
+QUESTION: Which exposure has the highest risk ratio? +
+ +
+
+
+QUESTION: Which of the following seems like a protective factor? +
+ +
+
+
+QUESTION: Which of the following has a non-significant association? +
+ +
+
+
+QUESTION: Is there a dose-response in the association between illness and tiramisu? +
+ +
+
+

Several food items seemed to be associated with the occurrence of illness; tiramisu, dark and white chocolate mousse, fruit salad, and red jelly. They can potentially explain up to 94, 76, 49, 46, and 45 of the 103 cases, respectively. Investigators decided to identify their respective roles in the occurrence of illness. There also seems to be a dose-response in the amount of tiramisu eaten, with those eating 2 or more portions having a higher risk than those eating only 1 (who had a higher risk than those who did not eat tiramisu)

+

From the crude analysis, epidemiologists noticed that the occurrence of gastroenteritis was lower among those attendants who had drunk beer. They also decided to assess if beer had a protective effect on the occurrence of gastroenteritis.

+
+
+

Step 5.3: Confounding and effect modification

+

The finding that beer had a protective effect on GI illness is surprising. One of your colleagues suggested that tiramisu might be confounding this association. Given that tiramisu has the highest risk ratio, it might be that the dessert is associated with the outcome (causes it) and to the exposure (those drinking beer might have preferred not to ruin its sour flavour with the sweetness of tiramisu). Another colleague disagrees and thinks that, actually, tiramisu might be an effect modifier in the relation between beer and illness.

+

QUESTION: How would you check if tiramisu is a confounder or an effect modifier in the association between beer and illness?

+
+ +

Click to see a solution (try it yourself first!)

+ +


+

To check both things: confounding and effect modification, we should stratify our data by tiramisu consumption. Once stratified, we can calculate risk ratios in both strata and an adjusted risk ratio: the adjusted Mantel–Haenszel risk ratio. If this risk ratio is very different from the crude risk ratio (the overall risk ratio between beer and illness), then we can say that tiramisu is confounding the association between beer and illness. We normally define as “very different” when: \((RRcrude/RRadjusted - 1)*100\) is larger than 10-20% difference (no strict threshold, your interpretation and expertise counts!).

+

If there is confounding but no effect modification, the adjusted M-H risk ratio should be different from the crude rr, but the risk ratios in each strata -tiramisu yes and tiramisu no- should not be significantly different. If they are significantly different then there is effect modification. How do we know if they are significantly different? We can run a test to tell us that called the Mantel–Haenszel test of homogeneity.

+
+

Task: Check if tiramisu is a confounder or an effect modifier in the association between beer and illness

+

Try doing this task with the function epi.2by2() from the {epiR} package. Read very carefully the documentation of the function. Pay special attention at how the data needs to be structured. Read the hint if you are struggling

+
+ +

Click to read a hint

+ +


+

Unfortunately epi.2by2() is a very “picky” function. Although you can pass data into epi.2by2() in different formats, the easiest way is to create a new data frame grouped by the three columns that we will use: tira, beer and ill.  You could pass this new dataframe into epi.2by2() and it may work, but be careful. epi.2by2() will consider as exposed and as ill the first levels of your factor data, as per the image below Now run this line of code riskratio(tira_clean$beer, tira_clean$ill) and look at the first part of the output. You should see the output of the image below

+

As you see, in the top left (96) we have unexposed who did not become ill, but in the previous image we saw that epi.2by2() will consider those in the top left as exposed and ill. How to overcome this? The trick is to change the levels of the factor variables. By default, they have “0” as first level and “1” as second level. So, whilst grouping the data we can add a line that changes the order of the levels (See the solution if you don’t know what its meant)

+


+
+
+ +

Click to see a solution (try it yourself first!)

+ +


+
+
## Group the dataframe
+group_data <- tira_clean %>% 
+  mutate(across(.cols = c(tira, beer, ill), .fns = ~factor(.x, levels = c(1,0)))) %>% # we change the order of the levels
+  group_by(tira, beer, ill) %>% # it's important to group them in this order: strata, exposure, outcome
+  summarise(n = n()) 
+
+## Run the function to obtain M-H estimates
+strat_analysis <-  
+  epi.2by2(dat = group_data, method = "cohort.count", digits = 2, 
+           conf.level = 0.95, units = 100, interpret = FALSE, outcome = "as.columns")
+#Print the results
+strat_analysis
+
+
             Outcome +    Outcome -      Total                 Inc risk *
+Exposed +           30           74        104     28.85 (20.38 to 38.55)
+Exposed -           67           95        162     41.36 (33.69 to 49.35)
+Total               97          169        266     36.47 (30.67 to 42.56)
+
+
+Point estimates and 95% CIs:
+-------------------------------------------------------------------
+Inc risk ratio (crude)                         0.70 (0.49, 0.99)
+Inc risk ratio (M-H)                           0.80 (0.62, 1.03)
+Inc risk ratio (crude:M-H)                     0.87
+Inc odds ratio (crude)                         0.57 (0.34, 0.97)
+Inc odds ratio (M-H)                           0.48 (0.22, 1.05)
+Inc odds ratio (crude:M-H)                     1.19
+Attrib risk in the exposed (crude) *           -12.51 (-24.06, -0.97)
+Attrib risk in the exposed (M-H) *             -7.53 (-15.59, 0.52)
+Attrib risk (crude:M-H)                        1.66
+-------------------------------------------------------------------
+ M-H test of homogeneity of IRRs: chi2(1) = 0.136 Pr>chi2 = 0.713
+ M-H test of homogeneity of ORs: chi2(1) = 1.512 Pr>chi2 = 0.219
+ Test that M-H adjusted OR = 1:  chi2(1) = 3.590 Pr>chi2 = 0.029
+ Wald confidence limits
+ M-H: Mantel-Haenszel; CI: confidence interval
+ * Outcomes per 100 population units
+
+
+


+
+
+QUESTION: Look at the results, what’s the magnitude of the difference between the adjusted MH risk ratio and the crude one? +
+ +
+
+
+QUESTION: Look at the results, is tiramisu an effect modifier in the association between beer and illness? +
+ +
+
+

You got several results. Let’s now review them to make sense of them:

+

The first part is a table in which you get the overall table for the association between beer and illness. The second part, titled Point estimates and 95% CIs: has different estimates depending on whether you’re interested in risk ratios, odds ratios or attributable risks. As we are carrying out a cohort study, we’re interested in the ones for risk ratios.

+

Inc risk ratio (crude): The crude incidence risk ratio, comparing the risk of the outcome between exposed and unexposed groups, with 95% confidence intervals. Inc risk ratio (M-H): The Mantel-Haenszel adjusted incidence risk ratio, accounting for stratification by tiramisu, with 95% confidence intervals. Inc risk ratio (crude:M-H): The ratio of the crude incidence risk ratio to the Mantel-Haenszel adjusted incidence risk ratio.

+

This is all we need to evaluate confounding. As we said before, we need to evaluate if the crude and the adjusted risk ratios are very different. We said that we can do that with the formula \((RRcrude/RRadjusted - 1)*100\). Here, the Inc risk ratio (crude:M-H): corresponds to the first part: \(RRcrude/RRadjusted\). So we need to substract 1 and multiple by 100: (0.87-1)*100. If you do that you’ll get 13%, which is borderline, but we can say that tiramisu is not massively confounding the association between beer and illness.

+

To know whether there is effect modification, you can check the M-H test of homogeneity of IRRs in the lower part of the results. As it is not significant (p = 0.713), we can conclude there is no effect modification of tiramisu on the association between beer and illness.

+
diff --git a/pages/stengen-en.epiet.qmd b/pages/stengen-en.epiet.qmd index 506b96d..31598c0 100644 --- a/pages/stengen-en.epiet.qmd +++ b/pages/stengen-en.epiet.qmd @@ -231,8 +231,9 @@ Participants are expected to be familiar with data management and basic analysis 3. Inside the folder "stengen_en": Subfolder "data" contains a raw data file named *tira.csv*. This is the only data file you will use in - this case study. - + this case study. In the same folder you can find the **data dictionary** with + a description of the dataframe variables. + 4. Subfolder scripts should be used to save any scripts related to the analysis. Inside "backup" you will find a solution script with the code of the case study named *stengen_mva_backup.R*. @@ -331,10 +332,10 @@ Using the updated list of attendants, a retrospective cohort study including all - Choose relevant characteristics to describe the population **Identify the outbreak vehicle if any** -- Chose the appropriate measure of association -- Chose the appropriate statistical tests and significance level - Calculate food-specific attack rates - Look at the proportions of cases exposed +- Chose the appropriate measure of association +- Chose the appropriate statistical tests and significance level - Calculate the percentages of cases exposed to each exposure - Search for any dose-response relationship if appropriate - Interpret the results @@ -411,7 +412,7 @@ Sys.setlocale("LC_ALL", "English") #### Step 1.3: Install/load packages Install and load the following packages: rio, skimr, janitor, gtsummary, -broom, rstatix, ggfortify and tidyverse. +rstatix, epitools, epiR, epikit and tidyverse. You can find more about installing/loading packages in the [Packages](https://epirhandbook.com/en/new_pages/basics.html#packages) @@ -467,7 +468,9 @@ pacman::p_load( janitor, # a package for data cleaning gtsummary, # to create frequency tables rstatix, # to generate summary statistics - EpiStats, # for univariable and stratified analysis + epitools, # for univariable analysis + epikit, # for creating age categories + epiR, # for the adjusted risk ratios across strata tidyverse # data management and visualization ) @@ -480,7 +483,7 @@ pacman::p_load( ### **Step 2: Import and explore data** ```{r, include=FALSE} -tira <- import("../cs/ENG/stengen_mva/data/tira.csv") +tira_raw <- import("../cs/ENG/stengen_mva/data/tira.csv") ``` #### Step 2.1: Import the data @@ -524,7 +527,7 @@ it yourself first!)
-```{r , echo=TRUE, results = 'hide', eval=FALSE} +```{r , echo=TRUE, eval=FALSE} # Import the dataset 'tira.csv' tira_raw <- import("data/tira.csv") ``` @@ -535,6 +538,7 @@ tira_raw <- import("data/tira.csv") + #### Step 2.2: Explore the data **Task**: Explore the data trying to answer the following questions: @@ -595,6 +599,23 @@ cat("QUESTION: What is the class of the column 'salmon'?", longmcq(opts)) ``` ::: +::: {.webex-check} + +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "83%", + "70%", + answer = "29%", + "6%" +) + + +cat("QUESTION: What percentage of attendees ate tomato?", longmcq(opts)) + +``` +::: ::: {.webex-check} @@ -620,14 +641,14 @@ cat("QUESTION: What is the class of the column 'dateonset'?", longmcq(opts)) pacman::p_load(webexercises) opts <- c( - answer = "106", - "20", - "474", - "30" + "As 'NA'", + "As empy spaces", + "With an 'Unknown' cateogry", + answer = "With the number 9" ) -cat("QUESTION: How many cases drank beer at the party?", longmcq(opts)) +cat("QUESTION: How are missing categorised in the variables 'salmon', 'horseradish' and 'pork'?", longmcq(opts)) ``` ::: @@ -644,6 +665,7 @@ cat("QUESTION: How many cases drank beer at the party?", longmcq(opts))
An efficient way to explore data is to use the function `skim()` from the {skimr} package, as it gives you all the information needed with only one command. Of course, there are several alternatives. You can also have a look at `glimpse()` from {dplyr}, for example. +An easy way to explore single variables that have categories is to use `tabyl()` from {janitor}
@@ -660,11 +682,14 @@ it yourself first!)
-```{r , echo=TRUE, results = 'hide', eval=FALSE} +```{r , echo=TRUE} # Inspect your dataset and generate basic statistics using the skim function skim(tira_raw) - +# Glimpse is an alternative to skim which gives more accurate information about the nature of the variables glimpse(tira_raw) +# To see the different categories of the variables +tabyl(tira_raw, tomato) + ``` @@ -673,11 +698,77 @@ glimpse(tira_raw) -### **Step 3 Clean the data** +#### Step 2.3: Explore age in more detail + +**Task** Explore the age of attendees by: + +- Getting summary statistics + +- Generating a boxplot of age by illness status. + +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+ +```{r , echo=TRUE} +## Generate summary statistics +summary(tira_raw$age) + +boxplot_age <- tira_raw %>% + + ggplot(mapping = aes(x = ill, + y = age, + fill = ill)) + + + geom_boxplot() + + + labs( + title = "Boxplot: Age distribution per disease status", + x = "Disease status", + y = "Age", + fill = "Illness (0=No;1=Yes)" + ) + + + theme_bw() + +boxplot_age +``` + +
+ +
+ +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "No ill", + "Ill", + answer = "Impossible to know from this boxplot" +) + + +cat("QUESTION: Which group (ill or not ill) has more variation around age's median?", longmcq(opts)) + +``` +::: + +We were not able to create the boxplot that we aimed for. In fact, you should have seen a warning message in your console saying, at the end, something like: "Did you forget to specify a `group` aesthetic or to convert a numerical variable into a factor?" + +Looks like all columns that consist of 1 (present) and 0 (absent) values are of class 'integer', including "ill". However, we want them to be categorial, otherwise we won't be able to carry out many of the analysis we want (including this boxplot). So now we'll sort this out alongside other cleaning functions. -You might notice that most of the columns are of class 'integer'. However, most variables consist out of 1 (present) and 0 (absent) values. For some of the following analyses, we will need R to treat these variables as factors (i.e., categorical variables). +### **Step 3: Clean the data** -**Task**: Change all present (1) / absent (0) variables from integer to factors. +#### Step 3.1: Transform categorical variables +**Task**: Change all present (1) / absent (0) variables from integer to factors. Do the same for the variables that record the number of portions eaten.
@@ -705,14 +796,14 @@ it yourself first!)
-```{r , echo=TRUE, results = 'hide', eval=FALSE} +```{r , echo=TRUE} tira_clean <- tira_raw %>% # tira_clean will be out clean dataframe after we have performed cleaning tasks on tira_raw # this would be changing one-by-one the columns into factors mutate(ill = as.factor(ill)) %>% # this would change all columns we are interested into 'factors' in only one command - mutate(across(.cols = !c(uniquekey, dateonset, age, mportion, tportion), # with the exclamation mark before c() we are telling R that we want all columns, except those in the c() list, to be transformed into factors + mutate(across(.cols = !c(uniquekey, dateonset, age), # with the exclamation mark before c() we are telling R that we want all columns, except those in the c() list, to be transformed into factors .fns = ~ as.factor(.x))) @@ -723,4 +814,685 @@ glimpse(tira_clean)
-
\ No newline at end of file + + +#### Step 3.2: Categorise "missing" as NA +We saw with the `skim()` overview -and in the data dictionary- that the variables *salmon*, *pork* and *horseradish* have 'missing' codified with number '9'. In most analysis, as for visualisations, it is very important that R considers missing data truly as "missing". To do that, missing data needs to be coded as NA (**without quotes**). Other codifications of missing data, such as empty spaces or the number 9 (as in this case) will mean that R will treat this data as a category or a number. For R, 9 is not different from number 1. + +**Task** Recode the columns salmon, pork and horseradish so that values of 9 are coded as NA. + +
+ + + +`r fontawesome::fa("lightbulb", fill = "gold")` Click to read a hint + + + +
+ +For this task you can use the `mutate()` function in combination with `recode()` or `if_else()`. The former is used specifically for recoding, whereas the latter is used to create/modify columns based on logical conditions. In any case, functions like `if_else()` and `recode()` may require you to specify the type of NA (e.g., NA_character_, NA_real_, NA_integer_) to ensure that the resulting vector maintains a consistent data type. This is important because R is strongly typed, meaning that each vector must contain elements of the same type. +If you want to know more about recoding, read the EpiRhandbook section on [Recoding](https://epirhandbook.com/new_pages/cleaning.html#re-code-values). Regardless of the approach that you choose, add it to the cleaning pipe command you started before, so that your code is tidy. +
+ +
+ + +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+ +```{r , echo=TRUE} +tira_clean <- tira_raw %>% # tira_clean will be out clean dataframe after we have performed cleaning tasks on tira_raw + + # this would be changing one-by-one the columns into factors + mutate(ill = as.factor(ill)) %>% + + # this would change all columns we are interested into 'factors' in only one command + mutate(across(.cols = !c(uniquekey, dateonset, age), # with the exclamation mark before c() we are telling R that we want all columns, except those in the c() list, to be transformed into factors + .fns = ~ as.factor(.x))) %>% + + # we can do it one by one. Here we use recode for this example + mutate(salmon = recode(salmon, + "9" = NA_character_)) %>% + + # we can do all columns at the same time with across(). Now we use the if_else approach + mutate(across(.cols = c(salmon, horseradish, pork), .fns = ~ if_else(.x == "9", NA_character_, .x))) + + +# Check that the changes have been succesful +tabyl(tira_clean, pork) +``` + + +
+ +
+ +#### Step 3.3: Create age groups + + +**Task**: Create a new variable with two categories (those aged < age's median and those >=age's median). **Add the code to create this new variable to the cleaning command you created above to ensure your code is tidy. Do not write multiple cleaning commands** + + +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+ +```{r , echo=TRUE} +tira_clean <- tira_raw %>% # tira_clean will be out clean dataframe after we have performed cleaning tasks on tira_raw + + # this would be changing one-by-one the columns into factors + mutate(ill = as.factor(ill)) %>% + + # this would change all columns we are interested into 'factors' in only one command + mutate(across(.cols = !c(uniquekey, dateonset, age), # with the exclamation mark before c() we are telling R that we want all columns, except those in the c() list, to be transformed into factors + .fns = ~ as.factor(.x))) %>% + + # we can do it one by one. Here we use recode for this example + mutate(salmon = recode(salmon, + "9" = NA_character_)) %>% + + # we can do all columns at the same time with across(). Now we use the if_else approach + mutate(across(.cols = c(salmon, horseradish, pork), .fns = ~ if_else(.x == "9", NA_character_, .x))) %>% + + # create age categories + mutate(age_group = age_categories(age, # the column to use to create the cateogories + lower = 0, # the lower value + upper = 20, # the upper value (we want a group of 20+) + by = 20)) # as we only want two groups, this is also 20 + + + +# Check that the changes have been succesful +tabyl(tira_clean, age_group) +``` + + +
+ +
+ + + + + +#### Step 3.4: Re-create the boxplot +**Task**: Create the boxplot we tried in **Step 2.3** using the clean version of the data. Save the boxplot in the subfolder "outputs". + +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+```{r, echo=TRUE} +## Boxplot of age by illness status +boxplot_age <- tira_clean %>% + + ggplot(mapping = aes(x = ill, + y = age, + fill = ill)) + + + geom_boxplot() + + + labs( + title = "Boxplot: Age distribution per disease status", + x = "Disease status", + y = "Age", + fill = "Illness (0=No;1=Yes)" + ) + + + theme_bw() + +boxplot_age +``` + +```{r, echo=TRUE, eval=FALSE} + +### Save the boxplot +#### A static way of saving it +ggsave(filename = "outputs/boxplot_age.png", plot = boxplot_age) +#### A dynamic way of doing it so that it contains the date of the analysis +ggsave(filename = str_glue("outputs/boxplot_age_", #First we write the beginning of the file name until the date + str_replace_all(Sys.Date(), "-", ""), #Sys.Date() gives us the date of analysis. Str_replace_all() removes all hyphens + ".png"), # Finally we add the ending which is the file type + plot = boxplot_age) +``` +
+ +
+ +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "No ill", + answer = "Ill", + "Impossible to know from this boxplot" +) + + +cat("QUESTION: Which group (ill or not ill) has more variation around age's median?", longmcq(opts)) + +``` +::: + +### **Step 4: Descriptive analysis** + + +#### Step 4.2: Analysis by time + +**Task**: Generate an epidemic curve with the dates of onset. Display in the x-axis each day of the study period. + +
+ + + +`r fontawesome::fa("lightbulb", fill = "gold")` Click to read a hint + + + +
+ +You should be able, by now, to create the histogram required for this epicurve. For the x-axis breaks, remember there are two distinct elements in a `ggplot()`. Inside `geom_histogram()` we can specify the `binwidth =` of the bins to the unit we want (e.g, 1 for daily, 7 for weekly etc.). This will change how the epicurve is displayed. +The second element is the scale. Rememember that scales change the way the axis or the colours are displayed. In this case, as we want to modify the X-axis which is a date, we should use the function `scale_x_date()` and specify inside that we want `date_breaks =` by "day". +If you are struggling, have a look at the [Epicurves](https://epirhandbook.com/new_pages/epicurves.html#epicurves-with-ggplot2) chapter of the EpiRhandbook, I'm sure it will solve all your doubts! +
+ +
+ +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+```{r, echo=TRUE} +tira_clean %>% + + ggplot(mapping = aes(x = dateonset)) + + + geom_histogram(binwidth = 1, # we assign the width of the bins to one day + color = "darkgreen", # color of lines around bars + fill = "lightgreen" ) + # color of fill within bars + + scale_x_date(date_breaks = "day") + # we want each day to be shown in the x-axis + + labs( + title = "Epicurve of cases by date of onset", + y = "Number of cases", + x = "" # we leave x-axis title blank because we already say in title it is date of onset + ) + + + theme_bw() + # predefined theme + + theme(axis.text.x = element_text(angle = 90)) # we change the angle of the days in the x-axis so it is readble. Important this happens at the end + + +``` + +
+ +
+ + +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "48h", + "A week later", + answer = "24h", + "The day of the party" +) + + +cat("QUESTION: How long after the party was peak day in cases?", longmcq(opts)) + +``` +::: + +#### Step 4.3: Analysis by person + +**Task**: Generate 2-by-2 tables with sex and age group by illness status + +
+ + + +`r fontawesome::fa("lightbulb", fill = "gold")` Click to read a hint + + + +
+ +There are several approaches to generate descriptive tables and cross-tabulations. You could use `tabyl()` from {janitor}, which is a very fast way; or you could use `group_by()` in combination with `summarise()`, from {dplyr} which is the most flexible. In this case, however, we show you the solution with a very efficient approach: `tbl_summary()` from {gtsummary} +
+ +
+ +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+```{r, echo=TRUE} +tira_clean %>% + + select(sex, age_group, ill) %>% # we select the columns we're interested in + + tbl_summary(by = ill, percent = "column") # we assign the grouping column and define where we want the percentages by column +``` + +
+ +
+ +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + answer = "51%", + "49%", + "45%", + "80%" +) + + +cat("QUESTION: What percentage of ill cases was female?", longmcq(opts)) + +``` +::: + +### **Step 5: Develop and test hypothesis** + +In the previous step we have described the number of ill attendees by date of onset, and we have described how sex and age were distributed according to illness status. The next step in an outbreak investigation would be to develop and test hypothesis about the possible source of the outbreak. + +#### Step 5.1: Compute food-specific attack rates and % of cases exposed + +**Task**: Create a list with exposed and unexposed cases and calculate the attack rates. + +
+ + + +`r fontawesome::fa("lightbulb", fill = "gold")` Click to read a hint + + + +
+ +You could use the same approach we used before with `tbl_summary()` from {gtsummary}, but to calculate attack rates, you'll need to change from column to row percentages. +
+ +
+ +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+```{r, echo=TRUE} +tira_clean %>% + + select(tira, wmousse, dmousse, wmousse, beer, + redjelly, fruitsalad, tomato, mince, salmon, + horseradish, chickenwin, roastbeef, pork, ill) %>% # we select the columns we're interested in + + tbl_summary(by = ill, percent = "row") %>% #we assign the grouping column and define where we want the percentages by row + + add_overall() # we add the overall counts +``` + +
+ +
+ +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "39%", + "25%", + "78%", + answer = "68%" +) + + +cat("QUESTION: What is the attack rate in those who ate withe mousse?", longmcq(opts)) + +``` +::: + + +#### Step 5.2: Estimate the relative risk of the different exposures + +**Task**: First, estimate the relative risk for the exposure of having eaten white mousse at the dinner + +
+ + + +`r fontawesome::fa("lightbulb", fill = "gold")` Click to read a hint + + + +
+ +One way yo estimate the relative risk in these situations is to use the function `riskratio()` from the package {epitools}. You need to first give the exposure and then the predictor. +
+ +
+ +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+```{r, echo=TRUE} +# Relative risk wmousse +riskratio(tira_clean$wmousse, tira_clean$ill) +``` + +
+ +
+ +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "2.1", + "179", + answer = "2.8", + "1.0" + ) + + +cat("QUESTION: What is the risk of illness among those who ate white mousse compared to those who did not have this food ?", longmcq(opts)) + +``` +::: + +**Task**: Calculate now the risk ratio for all the other exposures to develop hypothesis of the possible source + +
+ + + +`r fontawesome::fa("lightbulb", fill = "gold")` Click to read a hint + + + +
+ +You could write the `riskratio()` function changing the exposure variable one by one. However, to save time, avoid making errors and being more efficient, you could try to use approaches that allow you to apply the same function to many different objects (e.g., multiple columns) simultaneously. There are different options for this such as *loops*, *lapply* or *purrr*. Here we give the solution with *purrr*, so if you want to explore further purrr have a look at the dedicated [section](https://epirhandbook.com/new_pages/iteration.html#iter_purrr) in the EpiRhandbook. +
+ +
+ + +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+```{r, echo=TRUE} +tira_clean %>% + select(-uniquekey, -age, -age_group, -sex, -dateonset, -ill) %>% # we first remove the columns we're not interested + the outcome column + map(.f = ~riskratio(.x, tira_clean$ill)) # here we call the function for each column. ".x" represents all the columns in the database that have not been removed previously +``` + +
+ +
+ +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "beer", + answer = "tiramisu", + "pork", + "dark mousse" + ) + + +cat("QUESTION: Which exposure has the highest risk ratio?", longmcq(opts)) + +``` +::: + + +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + answer ="beer", + "tiramisu", + "pork", + "dark mousse" + ) + +cat("QUESTION: Which of the following seems like a protective factor?", longmcq(opts)) + +``` +::: + +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + "beer", + "tiramisu", + answer = "pork", + "dark mousse" + ) + +cat("QUESTION: Which of the following has a non-significant association?", longmcq(opts)) + +``` +::: + +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + answer = "Yes", + "No" + ) + +cat("QUESTION: Is there a dose-response in the association between illness and tiramisu?", longmcq(opts)) + +``` +::: + +Several food items seemed to be associated with the occurrence of illness; **tiramisu, dark and white chocolate mousse, fruit salad, and red jelly**. They can potentially explain up to 94, 76, 49, 46, and 45 of the 103 cases, respectively. Investigators decided to identify their respective roles in the occurrence of illness. There also seems to be a **dose-response** in the amount of tiramisu eaten, with those eating 2 or more portions having a higher risk than those eating only 1 (who had a higher risk than those who did not eat tiramisu) + +From the crude analysis, epidemiologists noticed that the occurrence of gastroenteritis was lower among those attendants who had drunk beer. They also decided to assess if beer had a protective effect on the occurrence of gastroenteritis. + +#### Step 5.3: Confounding and effect modification + +The finding that beer had a protective effect on GI illness is surprising. One of your colleagues suggested that tiramisu might be confounding this association. Given that tiramisu has the highest risk ratio, it might be that the dessert is associated with the outcome (causes it) and to the exposure (those drinking beer might have preferred not to ruin its sour flavour with the sweetness of tiramisu). Another colleague disagrees and thinks that, actually, tiramisu might be an effect modifier in the relation between beer and illness. + +QUESTION: How would you check if tiramisu is a confounder or an effect modifier in the association between beer and illness? + + +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+ +To check both things: confounding and effect modification, we should stratify our data by tiramisu consumption. Once stratified, we can calculate risk ratios in both strata and an adjusted risk ratio: **the adjusted Mantel–Haenszel risk ratio**. If this risk ratio is very different from the crude risk ratio (the overall risk ratio between beer and illness), then we can say that tiramisu is confounding the association between beer and illness. We normally define as "very different" when: $(RRcrude/RRadjusted - 1)*100$ is larger than 10-20% difference (no strict threshold, your interpretation and expertise counts!). + +If there is confounding but no effect modification, the adjusted M-H risk ratio should be different from the crude rr, but **the risk ratios in each strata -tiramisu yes and tiramisu no- should not be significantly different**. If they are significantly different then there is **effect modification**. How do we know if they are significantly different? We can run a test to tell us that called **the Mantel–Haenszel test of homogeneity**. +
+ +
+ + + + +**Task**: Check if tiramisu is a confounder or an effect modifier in the association between beer and illness + +Try doing this task with the function `epi.2by2()` from the {epiR} package. **Read very carefully the documentation of the function**. Pay special attention at how the data needs to be structured. Read the hint if you are struggling + +
+ + + +`r fontawesome::fa("lightbulb", fill = "gold")` Click to read a hint + + + +
+ +Unfortunately `epi.2by2()` is a very "picky" function. Although you can pass data into `epi.2by2()` in different formats, the easiest way is to create a new data frame grouped by the three columns that we will use: tira, beer and ill. +You could pass this new dataframe into `epi.2by2()` and it may work, but be careful. `epi.2by2()` will consider as exposed and as ill the first levels of your factor data, as per the image below +![](../images/stengen/epi.2by2_data.png) +Now run this line of code `riskratio(tira_clean$beer, tira_clean$ill)` and look at the first part of the output. You should see the output of the image below +![](../images/stengen/beer_data_before.png) + +As you see, in the top left (96) we have unexposed who did not become ill, but in the previous image we saw that `epi.2by2()` will consider those in the top left as exposed and ill. How to overcome this? The trick is to change the levels of the factor variables. By default, they have "0" as first level and "1" as second level. So, whilst grouping the data we can add a line that changes the order of the levels (See the solution if you don't know what its meant) + + +
+ +
+ + + +
+ + + +`r fontawesome::fa("check", fill = "red")`Click to see a solution (try +it yourself first!) + + + +
+```{r, echo=TRUE} +## Group the dataframe +group_data <- tira_clean %>% + mutate(across(.cols = c(tira, beer, ill), .fns = ~factor(.x, levels = c(1,0)))) %>% # we change the order of the levels + group_by(tira, beer, ill) %>% # it's important to group them in this order: strata, exposure, outcome + summarise(n = n()) + +## Run the function to obtain M-H estimates +strat_analysis <- + epi.2by2(dat = group_data, method = "cohort.count", digits = 2, + conf.level = 0.95, units = 100, interpret = FALSE, outcome = "as.columns") +#Print the results +strat_analysis +``` + +
+ +
+ + +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + answer = "13%", + "87%", + "30%", + "20%" + ) + +cat("QUESTION: Look at the results, what's the magnitude of the difference between the adjusted MH risk ratio and the crude one?", longmcq(opts)) + +``` +::: + +::: webex-check +```{r, results="asis", echo=FALSE} +pacman::p_load(webexercises) + +opts <- c( + answer = "No", + "Yes" + ) + +cat("QUESTION: Look at the results, is tiramisu an effect modifier in the association between beer and illness?", longmcq(opts)) + +``` +::: + +You got several results. Let's now review them to make sense of them: + +The first part is a table in which you get the overall table for the association between beer and illness. +The second part, titled *Point estimates and 95% CIs:* has different estimates depending on whether you're interested in risk ratios, odds ratios or attributable risks. As we are carrying out a cohort study, we're interested in the ones for **risk ratios**. + +**Inc risk ratio (crude)**: The crude incidence risk ratio, comparing the risk of the outcome between exposed and unexposed groups, with 95% confidence intervals. +**Inc risk ratio (M-H):** The Mantel-Haenszel adjusted incidence risk ratio, accounting for stratification by tiramisu, with 95% confidence intervals. +**Inc risk ratio (crude:M-H):** The ratio of the crude incidence risk ratio to the Mantel-Haenszel adjusted incidence risk ratio. + + +This is all we need to evaluate confounding. As we said before, we need to evaluate if the crude and the adjusted risk ratios are very different. We said that we can do that with the formula $(RRcrude/RRadjusted - 1)*100$. Here, the **Inc risk ratio (crude:M-H):** corresponds to the first part: $RRcrude/RRadjusted$. So we need to substract 1 and multiple by 100: (0.87-1)*100. If you do that you'll get 13%, which is borderline, but we can say that tiramisu is not massively confounding the association between beer and illness. + +To know whether there is effect modification, you can check the **M-H test of homogeneity of IRRs** in the lower part of the results. As it is not significant (p = 0.713), we can conclude there is no effect modification of tiramisu on the association between beer and illness. + + + diff --git a/pages/stengen-en.epiet_files/figure-html/unnamed-chunk-13-1.png b/pages/stengen-en.epiet_files/figure-html/unnamed-chunk-13-1.png new file mode 100644 index 0000000000000000000000000000000000000000..0089679ba413e8c984acc547a1e1fe66b8037185 GIT binary patch literal 18535 zcmeHv30zaxwl@Y86q%f&B2Y&v2vL!FNa}-XOM-o4*79?s8w&ffd1v-VnR z|JPc3pBtY!*{iMCxI#feLG9>~gU1yVmLe1s6n|Aw1g}(;Y#mfkSaRyKVb&J0dCOaRdn?FV$bJdnKko$aT;!d=_AV;&E&|U0*jUyS9fyv~MaR{lb9>{! z?$F=?UdWBB>y4`e608&wz-N0aBzS{oc^?Ty2}R(+PGExvkc)QAjYH?w_2%{hMcxWU z-d05k;Ne}AAg@&f{@6uq@BoScpf1j_E)HFni>|8!TL5(cz)rAYCwQ}qz$1YTR+ZHS ze*jpW4FJ8l=-#@x-n!i0Ua&eaI5-GC1MCHC2C$0&uwHL2&|3%ef;VN|gZ+Vx005{I z=$r=mx;)^>nIHuPooLzLl6cOTmx97a3P%rqaxyG^xc$P_bWTRbSkjrF*YDlEPu(G) z`!}cW_kN#J|9XS<>RQc)knYz>s~_TbD|tayeYEsvk<|Zq!->|y$@X@@WkiqX2Jzf)9Y%%nyh 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