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 set.
Exploratory analysis
Load the nc data set into our workspace.
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. |
- What are the cases in this data set? How many cases are there in our sample? JR Answer: There are 1000 cases of births in NC
## fage mage mature weeks
## Min. :14.00 Min. :13 mature mom :133 Min. :20.00
## 1st Qu.:25.00 1st Qu.:22 younger mom:867 1st Qu.:37.00
## Median :30.00 Median :27 Median :39.00
## 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
## premie visits marital gained
## full term:846 Min. : 0.0 married :386 Min. : 0.00
## premie :152 1st Qu.:10.0 not married:613 1st Qu.:20.00
## NA's : 2 Median :12.0 NA's : 1 Median :30.00
## Mean :12.1 Mean :30.33
## 3rd Qu.:15.0 3rd Qu.:38.00
## Max. :30.0 Max. :85.00
## NA's :9 NA's :27
## weight lowbirthweight gender habit
## Min. : 1.000 low :111 female:503 nonsmoker:873
## 1st Qu.: 6.380 not low:889 male :497 smoker :126
## Median : 7.310 NA's : 1
## Mean : 7.101
## 3rd Qu.: 8.060
## Max. :11.750
##
## whitemom
## not white:284
## white :714
## NA's : 2
##
##
##
## ## fage mage mature weeks premie visits marital gained weight
## 995 NA 41 mature mom 33 premie 13 not married 0 5.69
## 996 47 42 mature mom 40 full term 10 not married 26 8.44
## 997 34 42 mature mom 38 full term 18 not married 20 6.19
## 998 39 45 mature mom 40 full term 15 not married 32 6.94
## 999 55 46 mature mom 31 premie 8 not married 25 4.56
## 1000 45 50 mature mom 39 full term 14 not married 23 7.13
## lowbirthweight gender habit whitemom
## 995 not low female nonsmoker not white
## 996 not low male nonsmoker not white
## 997 not low female nonsmoker white
## 998 not low female nonsmoker white
## 999 low female nonsmoker not white
## 1000 not low female nonsmoker whiteAs a first step in the analysis, we should consider summaries of the data. This can be done using the summary command:
## fage mage mature weeks
## Min. :14.00 Min. :13 mature mom :133 Min. :20.00
## 1st Qu.:25.00 1st Qu.:22 younger mom:867 1st Qu.:37.00
## Median :30.00 Median :27 Median :39.00
## 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
## premie visits marital gained
## full term:846 Min. : 0.0 married :386 Min. : 0.00
## premie :152 1st Qu.:10.0 not married:613 1st Qu.:20.00
## NA's : 2 Median :12.0 NA's : 1 Median :30.00
## Mean :12.1 Mean :30.33
## 3rd Qu.:15.0 3rd Qu.:38.00
## Max. :30.0 Max. :85.00
## NA's :9 NA's :27
## weight lowbirthweight gender habit
## Min. : 1.000 low :111 female:503 nonsmoker:873
## 1st Qu.: 6.380 not low:889 male :497 smoker :126
## Median : 7.310 NA's : 1
## Mean : 7.101
## 3rd Qu.: 8.060
## Max. :11.750
##
## whitemom
## not white:284
## white :714
## NA's : 2
##
##
##
## As you review the variable summaries, consider which variables are categorical and which are numerical. For numerical variables, are there outliers? If you aren’t sure or want to take a closer look at the data, make a graph.
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.
## 'data.frame': 1000 obs. of 13 variables:
## $ fage : int NA NA 19 21 NA NA 18 17 NA 20 ...
## $ mage : int 13 14 15 15 15 15 15 15 16 16 ...
## $ mature : Factor w/ 2 levels "mature mom","younger mom": 2 2 2 2 2 2 2 2 2 2 ...
## $ weeks : int 39 42 37 41 39 38 37 35 38 37 ...
## $ premie : Factor w/ 2 levels "full term","premie": 1 1 1 1 1 1 1 2 1 1 ...
## $ visits : int 10 15 11 6 9 19 12 5 9 13 ...
## $ marital : Factor w/ 2 levels "married","not married": 1 1 1 1 1 1 1 1 1 1 ...
## $ gained : int 38 20 38 34 27 22 76 15 NA 52 ...
## $ weight : num 7.63 7.88 6.63 8 6.38 5.38 8.44 4.69 8.81 6.94 ...
## $ lowbirthweight: Factor w/ 2 levels "low","not low": 2 2 2 2 2 1 2 1 2 2 ...
## $ gender : Factor w/ 2 levels "female","male": 2 2 1 2 1 2 2 2 2 1 ...
## $ habit : Factor w/ 2 levels "nonsmoker","smoker": 1 1 1 1 1 1 1 1 1 1 ...
## $ whitemom : Factor w/ 2 levels "not white","white": 1 1 2 2 1 1 1 1 2 2 ...JR Answer: In the weight histogram - outliers are 1 lbs and 11.85 lbs int and num are numberical: fage, mage, weeks,visits, gained, weight factors are categorical hist($fage)
- Make a side-by-side boxplot of
habitandweight. What does the plot highlight about the relationship between these two variables?
# Boxplot of habit and weight
boxplot(weight~habit,data=nc, main="Mother's Habit vs Baby's Weight",
ylab="Baby Weight", xlab="Mother Smoker/Non-Smoker")
The box plots show how the medians of the two distributions compare, but we can also compare the means of the distributions using the following function to split the weight variable into the habit groups, then take the mean of each using the mean function.
## nc$habit: nonsmoker
## [1] 7.144273
## --------------------------------------------------------
## nc$habit: smoker
## [1] 6.82873There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test .
Inference
- Check if the conditions necessary for inference are satisfied. Note that you will need to obtain sample sizes to check the conditions. You can compute the group size using the same
bycommand above but replacingmeanwithlength.
## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126JR Answer: Sample observations are independent as are the sameple groups. The sample sizes are less then 10% of the population size. Sample size is sufficient as to not worry about skew.
- Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different. HJR Answer: H0 : Average weight of baby from non-smoking = average weight of baby from smoking. Ha : Average weight of baby from non-smoking not = average weight of baby from smoking.
Next, we introduce a new function, inference, that we will use for conducting hypothesis tests and constructing confidence intervals.
inference(y = nc$weight, x = nc$habit, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862## Observed difference between means (nonsmoker-smoker) = 0.3155
##
## H0: mu_nonsmoker - mu_smoker = 0
## HA: mu_nonsmoker - mu_smoker != 0
## Standard error = 0.134
## Test statistic: Z = 2.359
## p-value = 0.0184
Let’s pause for a moment to go through the arguments of this custom function. The first argument is y, which is the response variable that we are interested in: nc$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$habit. The third argument, est, is the parameter we’re interested in: "mean" (other options are "median", or "proportion".) Next we decide on the type of inference we want: a hypothesis test ("ht") or a confidence interval ("ci"). When performing a hypothesis test, we also need to supply the null value, which in this case is 0, since the null hypothesis sets the two population means equal to each other. The alternative hypothesis can be "less", "greater", or "twosided". Lastly, the method of inference can be "theoretical" or "simulation" based.
- Change the
typeargument to"ci"to construct and record a confidence interval for the difference between the weights of babies born to smoking and non-smoking mothers.
inference(y = nc$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 )On your own
- Calculate a 95% confidence interval for the average length of pregnancies (
weeks) and interpret it in context. Note that since you’re doing inference on a single population parameter, there is no explanatory variable, so you can omit thexvariable from the function. JR Answer:
95 % Confidence interval = ( 38.1528 , 38.5165 )
## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )- Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function:
conflevel = 0.90.
JR Answer: 90 % Confidence interval = ( 38.182 , 38.4873 )
## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 90 % Confidence interval = ( 38.182 , 38.4873 )- Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
JR Answer: H0: mu_mature mom - mu_younger mom = 0 HA: mu_mature mom - mu_younger mom != 0
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_mature mom = 133, mean_mature mom = 7.1256, sd_mature mom = 1.6591
## n_younger mom = 867, mean_younger mom = 7.0972, sd_younger mom = 1.4855## Observed difference between means (mature mom-younger mom) = 0.0283
##
## H0: mu_mature mom - mu_younger mom = 0
## HA: mu_mature mom - mu_younger mom != 0
## Standard error = 0.152
## Test statistic: Z = 0.186
## p-value = 0.8526
p-value is 0.8526. So, .8526 > .05 so we fail to reject the null hypothesis which was that there was no difference in the weights
- Now, a non-inference task: Determine the age cutoff for younger and mature mothers. Use a method of your choice, and explain how your method works. JR Answer: The cutoff is 34 fro younger mother.
older <- subset(nc, mature == "mature mom")
younger <- subset(nc, mature == "younger mom")
summary(older$mage)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 35.00 35.00 37.00 37.18 38.00 50.00## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13.00 21.00 25.00 25.44 30.00 34.00- Pick a pair of numerical and categorical variables and come up with a research question evaluating the relationship between these variables. Formulate the question in a way that it can be answered using a hypothesis test and/or a confidence interval. Answer your question using the
inferencefunction, report the statistical results, and also provide an explanation in plain language.
## fage mage mature weeks
## Min. :14.00 Min. :13 mature mom :133 Min. :20.00
## 1st Qu.:25.00 1st Qu.:22 younger mom:867 1st Qu.:37.00
## Median :30.00 Median :27 Median :39.00
## 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
## premie visits marital gained
## full term:846 Min. : 0.0 married :386 Min. : 0.00
## premie :152 1st Qu.:10.0 not married:613 1st Qu.:20.00
## NA's : 2 Median :12.0 NA's : 1 Median :30.00
## Mean :12.1 Mean :30.33
## 3rd Qu.:15.0 3rd Qu.:38.00
## Max. :30.0 Max. :85.00
## NA's :9 NA's :27
## weight lowbirthweight gender habit
## Min. : 1.000 low :111 female:503 nonsmoker:873
## 1st Qu.: 6.380 not low:889 male :497 smoker :126
## Median : 7.310 NA's : 1
## Mean : 7.101
## 3rd Qu.: 8.060
## Max. :11.750
##
## whitemom
## not white:284
## white :714
## NA's : 2
##
##
##
## JR Answer: H0: mu_fage from a premie - mu_fage from a full term = 0 HA: mu_fage from a premie - mu_fage from a full term != 0
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_full term = 714, mean_full term = 30.2423, sd_full term = 6.6329
## n_premie = 114, mean_premie = 30.3158, sd_premie = 7.5859## Observed difference between means (full term-premie) = -0.0735
##
## H0: mu_full term - mu_premie = 0
## HA: mu_full term - mu_premie != 0
## Standard error = 0.753
## Test statistic: Z = -0.098
## p-value = 0.9222