X-Git-Url: https://code.communitydata.science/stats_class_2019.git/blobdiff_plain/dbde6a6880af749afd92848e06128fe161159d0b..07da118471a31405d2044b68b4eb30334dea9d8b:/problem_sets/week_03/ps3-worked_solution.html diff --git a/problem_sets/week_03/ps3-worked_solution.html b/problem_sets/week_03/ps3-worked_solution.html index 9927124..a909ded 100644 --- a/problem_sets/week_03/ps3-worked_solution.html +++ b/problem_sets/week_03/ps3-worked_solution.html @@ -464,7 +464,7 @@ w2.data <- log1p(w2.data) 
## [1] 9.643215 2.158358 1.396595 0.192623 1.752234 0.170634

Inspecting the first few values returned by head() gave you a clue. Rounded to six decimal places, the vectors match!

I can create a table comparing the sorted rounded values to check this.

-
table(sort(round(w2.data, 6)) == sort(round(w3.data$x, 6)))
+
table(round(w2.data,6) == round(w3.data$x,6))
## 
 ## TRUE 
 ##   95
@@ -546,7 +546,7 @@ head(w3.data) ## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's ## -4.42 3.19 7.81 9.96 14.61 33.14 5 
### Run this line again to assign the new dataframe to p
-p <- ggplot(w3.data, aes(x=x, y=y))
+p <- ggplot(data=w3.data, mapping=aes(x=x, y=y))
 
 p + geom_point(aes(color=j, size=l, shape=k))
## Warning: Using size for a discrete variable is not advised.
@@ -641,7 +641,7 @@ l68/length(d)
  1. Is random assignment to tents likely to ensure \(\leq1~arachnophobe\) per tent?
-

Random assignment and the independence assumption means that the answer to part c is the inverse of the outcome we’re looking to avoid: \(P(\gt1~arachnophobes) = 1-P(\leq1~arachnophobe)\). So, \(P(\gt1~arachnophobes) = 1-0.84 = 0.16 = 16\%\). Those are the probabilities, but the interpretation really depends on how confident the camp counselor feels about a \(16\%\) chance of having multiple arachnophobic campers in one of the tents.

+

Random assignment and the independence assumption means that the answer to part c is the complement of the outcome we’re looking to avoid: \(P(\gt1~arachnophobes) = 1-P(\leq1~arachnophobe)\). So, \(P(\gt1~arachnophobes) = 1-0.84 = 0.16 = 16\%\). Those are the probabilities, but the interpretation really depends on how confident the camp counselor feels about a \(16\%\) chance of having multiple arachnophobic campers in one of the tents.