Lab 5
Olga Shiligin
22/11/2018
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.
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?
dim(nc)## [1] 1000 13Answer: There are 13 variables and 1000 cases in the example.
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
## 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.
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)
Answer: Box plot shows that the average weights of babies born to smoking mothers are less than to non-smoking, but this difference is not large. Also baby’s weight spread is higher when mother is 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.
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 this question we will conduct a hypothesis test.
Excercise 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] 126hist(nc$weight[nc$habit == "smoker"], main = "Weight for Smoking Mothers", xlab = "Weight")
hist(nc$weight[nc$habit == "nonsmoker"], main = "Weight for Non-Smoking Mothers", xlab = "Weight")
- sample is random and I assume, sample values are independed of each other
- data slightly skewed, but size is large and skewness can be ignored
Conclusion: conditions necessary for inference are satisfied.
Excercise 4
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
H0 = the average weights of babies born to smoking and non-smoking mothers are the same
H1 = the average weights of babies born to smoking and non-smoking mothers are different
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.
Excercise 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 (μnonsmoker−μsmoker) . 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 the x variable from the function.
inference(y = nc$weeks, est = "mean", type = "ci",
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )We are 95% confident that the average length of pregnancy for the population is within 38.1528 and 38.5165 weeks.
- Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.
inference(y = nc$weeks, est = "mean", type = "ci", confleve = 0.90,
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 90 % Confidence interval = ( 38.182 , 38.4873 )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.
inference(y = nc$gained, x = nc$mature, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical",
order = c("mature mom", "younger mom"))## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_mature mom = 129, mean_mature mom = 28.7907, sd_mature mom = 13.4824
## n_younger mom = 844, mean_younger mom = 30.5604, sd_younger mom = 14.3469## Observed difference between means (mature mom-younger mom) = -1.7697
##
## H0: mu_mature mom - mu_younger mom = 0
## HA: mu_mature mom - mu_younger mom != 0
## Standard error = 1.286
## Test statistic: Z = -1.376
## p-value = 0.1686
As p-value > 0.05, we can not reject H0 and we can NOT conclude that a significant difference exists between the average weight gained by mature and younger mothers.
- 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.
young.mom <- subset(nc, nc$mature == "younger mom")
mature.mom <- subset(nc, nc$mature == "mature mom")
max(young.mom$mage)## [1] 34min(mature.mom$mage)## [1] 35Answer: younger mom <=34, mature mom >=35. Max age of younger mom and min age of mature mom can identify the cut off point between these two groups.
- 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 inference function, report the statistical results, and also provide an explanation in plain language.
Research Question: Do number of hospital visits of younger mom different from number of hospital visit of mature mom?
inference(y = nc$visits, x = nc$mature, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_mature mom = 131, mean_mature mom = 12.6107, sd_mature mom = 4.3793
## n_younger mom = 860, mean_younger mom = 12.0279, sd_younger mom = 3.8832## Observed difference between means (mature mom-younger mom) = 0.5828
##
## H0: mu_mature mom - mu_younger mom = 0
## HA: mu_mature mom - mu_younger mom != 0
## Standard error = 0.405
## Test statistic: Z = 1.439
## p-value = 0.15
As p-value > 0.05, we can not reject H0 and we can NOT conclude that a significant difference exists between number of hospital visits of younger and mature moms.