8_MarcelaRendon_Lab7
Marcela Rendon
10/24/2021
Inference for numerical data
North Carolina births
In 2004, the state of North Carolina released a large data set containing information on births recorded in this state. This data set is useful to researchers studying the relation between habits and practices of expectant mothers and the birth of their children. We will work with a random sample of observations from this data sets.
Exploratory analysis
Load the nc data set into our workspace.
download.file("http://www.openintro.org/stat/data/nc.RData", destfile = "nc.RData")
load("nc.RData")We have observations on 13 different variables, some categorical and some numerical. The meaning of each variable is as follows.
variable description fage father’s age in years. mage mother’s age in years. mature maturity status of mother. weeks length of pregnancy in weeks. premie whether the birth was classified as premature (premie) or full-term. visits number of hospital visits during pregnancy. marital whether mother is married or not married at birth. gained weight gained by mother during pregnancy in pounds. weight weight of the baby at birth in pounds. lowbirthweight whether baby was classified as low birthweight (low) or not (not low). gender gender of the baby, female or male. habit status of the mother as a nonsmoker or a smoker. whitemom whether mom is white or not white.
Exercise 1
What are the cases in this data set? how many cases are there in our sample?
The cases in this dataset are the number of newborns in North Carolina. There are 1000 cases in the samples.
As a first step in the analysis, we should consider summaries of the data. This can be done using the summary command:
summary(nc)## fage mage mature weeks premie
## Min. :14.00 Min. :13 mature mom :133 Min. :20.00 full term:846
## 1st Qu.:25.00 1st Qu.:22 younger mom:867 1st Qu.:37.00 premie :152
## Median :30.00 Median :27 Median :39.00 NA's : 2
## Mean :30.26 Mean :27 Mean :38.33
## 3rd Qu.:35.00 3rd Qu.:32 3rd Qu.:40.00
## Max. :55.00 Max. :50 Max. :45.00
## NA's :171 NA's :2
## visits marital gained weight
## Min. : 0.0 married :386 Min. : 0.00 Min. : 1.000
## 1st Qu.:10.0 not married:613 1st Qu.:20.00 1st Qu.: 6.380
## Median :12.0 NA's : 1 Median :30.00 Median : 7.310
## Mean :12.1 Mean :30.33 Mean : 7.101
## 3rd Qu.:15.0 3rd Qu.:38.00 3rd Qu.: 8.060
## Max. :30.0 Max. :85.00 Max. :11.750
## NA's :9 NA's :27
## lowbirthweight gender habit whitemom
## low :111 female:503 nonsmoker:873 not white:284
## not low:889 male :497 smoker :126 white :714
## NA's : 1 NA's : 2
##
##
##
## As you review the variable summaries, consider which variables are categorical and which are numerical. For numerical values, are there outliers, If you aren’t sure or want to take a closer look at the data, make a graph.
The following variables are numerical: fage, mage, weeks, visits, gained, and weight.
boxplot(nc$fage, nc$mage, nc$weeks, nc$visits, nc$gained, nc$weight,
main = "Numerical variables in nc",
xlab = "Numerical variables in nc",
ylab = "Number per variable",
names = c("fage", "mage", "weeks", "visits", "gained", "weight"))
It looks like all the numerical values have outliers, but the ones with the most outliers are weeks, visits, and gained.
Consider the possible relationship between a mother’s smoking habit and the weight of her baby. Plotting the data is a useful first step because it helps us quickly visualize trends, identify strong associations, and develop research questions.
Exercise 2
Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?
boxplot(nc$weight ~ nc$habit,
main = "Comparing Weight of Babies to Smoking Habits of Mothers",
xlab = "Mother Smoking Habits",
ylab = "Baby Weight")
The overall median and interquartile distributions of smoker and non-moker mothers are fairly similar, but the non-smoker group has far more outliers that the smoker group.
The boxplots show how the medians of the two distributions compare, but we can also compare the means of the distributions using the following function to split the weight variable into the habit groups, then take the mean of each using the mean function.
by(nc$weight, nc$habit, mean)## nc$habit: nonsmoker
## [1] 7.144273
## ------------------------------------------------------------
## nc$habit: smoker
## [1] 6.82873There is an observed difference, but is this difference statistically significant? In order to answer the question we will conduct a hypothesis test.
Inference
Exercise 3
Check if the conditions necessary for inference are satisfied. Note that you will need to obtain sample sizes to check the conditions. You can compute the group size using the same by command above but replacing mean with length.
by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## ------------------------------------------------------------
## nc$habit: smoker
## [1] 126Since both sample sizes are both over 30, which means that the conditions necessary for inference are satisfied.
Exercise 4
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
Null hypothesis: There is no difference in the average weight of babies born to mother who smoke or don’t smoke. Alternative hypothesis: There is a difference in the average weight of babies depending on whether mother smoked or did not smoke.
Next, we introduce a new function, inference, that we will use for conducting hypothesis tests and constructing confidence intervals.
inference(y = nc$weight, x = nc$habit, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862## Observed difference between means (nonsmoker-smoker) = 0.3155
##
## H0: mu_nonsmoker - mu_smoker = 0
## HA: mu_nonsmoker - mu_smoker != 0
## Standard error = 0.134
## Test statistic: Z = 2.359
## p-value = 0.0184
Let’s pause for a moment to go through the arguments of this custom function. The first argument is y, which is the response variable that we are in interested in: nc.weight. The second argument is the explanatory variable, x, which is the variable that splits the data into two groups, smokers and non-smokers: nc.habit. The third argument, est, is the parameter we’re interested in: “mean” (other options are “median”, or “proportion). Next we decide on the type of inference we want: a hypothesis test (”ht“) or a confidence interval (”ci“). When performing a hypothesis test, we also need to supply the null value, which is in this case 0, since the null hypothesis sets the two population means equal to each other. The alternative hypothesis can be”less“,”greater“, or”twosided“. Lastly, the method of inference can be”theoretical" or “simulation” based.
Exercise 5
Change the type argument to “ci” to construct and record a confidence interval for the difference between the weights of babies born to smoking and non-smoking mothers.
inference(y = nc$weight, x = nc$habit, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862
## Observed difference between means (nonsmoker-smoker) = 0.3155
##
## Standard error = 0.1338
## 95 % Confidence interval = ( 0.0534 , 0.5777 )By default, the function reports an interval for (\(\mu\)[nonsmoker] - \(\mu\)[smoker]). We can easily change this order by using the order argument:
inference(y = nc$weight, x = nc$habit, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical",
order = c("smoker", "nonsmoker"))## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187
## Observed difference between means (smoker-nonsmoker) = -0.3155
##
## Standard error = 0.1338
## 95 % Confidence interval = ( -0.5777 , -0.0534 )