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.
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. |
What are the cases in this data set? How many cases are there in our sample?
Ans:
There are 1000 cases.
## [1] 1000-
As 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.
Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?
Ans:
Based on the two boxplots, it appears that smokers tend to give birth to lower birth weight babies.
However, as there are more dots below the 1st quartile on the non-smokers side, we need to test their inference to determine if the relationship is significant.
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] 6.82873
There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test.
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.
Ans:
The samples are large (>30) enough and is certainly less than 10% of the population, therefore we assume the observations are chosen randomly and independent of each other.
## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126-
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
Ans:
H0: Weights of babies born to smoking and non-smoking mothers are the same.
H1: 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.
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.
Ans:
The 95% confidence interval for the difference between the weights of babies born to smoking and non-smoking mothers is (0.0534, 0.5777).
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 )
-
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.
Ans:
The 95% confidence interval is (38.1528, 38.5165).
inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )-
Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.
Ans:
The 90% confidence interval is (38.182, 38.4873).
inference(y = nc$weeks, est = "mean", type = "ci", conflevel = 0.90, null = 0,
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 )-
Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
Ans:
H0: the average weight gained by younger mothers and that of mature mothers are the same.
H1: the average weight gained by younger mothers is different than the average weight gained by mature mothers.
By the results, the difference on the weight gained between younger mom and mature mom is 1.77 pounds.
As the p-value = 0.1686 and is larger than 0.05, we do not have sufficient evidence to reject the null hypothesis. Therefore, we conclude that the weight gained between younger and mature mom is very similar.
inference(y = nc$gained, x = nc$mature, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical", order = c("younger mom", "mature mom"))## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_younger mom = 844, mean_younger mom = 30.5604, sd_younger mom = 14.3469
## n_mature mom = 129, mean_mature mom = 28.7907, sd_mature mom = 13.4824
## Observed difference between means (younger mom-mature mom) = 1.7697
##
## H0: mu_younger mom - mu_mature mom = 0
## HA: mu_younger mom - mu_mature mom != 0
## Standard error = 1.286
## Test statistic: Z = 1.376
## p-value = 0.1686
-
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.
Ans:
The age interval for younger mom is (13, 34). The age interval for mature mom is (35,50).
Therefore, the age cutoff for younger and mature mothers is 35. When the mother is 35, she is considered to be a mature mom.
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13.00 21.00 25.00 25.44 30.00 34.00
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 35.00 35.00 37.00 37.18 38.00 50.00-
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 inference function, report the statistical results, and also provide an explanation in plain language.
Ans:
Numerical variable: visits - number of hospital visits during pregnancy.
Categorical variable: lowbirthweight - whether baby was classified as low birthweight (low) or not (not low)
Research question: Is the number of hospital visits related to the baby born with low birth weight or not?
Assumptions:
The number of visits by mothers is independent of each other.
The sample size 1000 is large enough to assume normal distribution.
Hypothesis:
H0: There is no relationship between the number of hospital visits during pregnancy and whether the baby is borned with low birth weight or not. i.e. mean(number of visits) = mean(lowbirthweight)
H1: Number of hospital visits (low) is related to the baby classified as low birthweight. i.e. mean(number of visits) <> mean(lowbirthweight)
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 0.0 8.0 10.0 10.8 14.0 30.0 3
# summary of number of `visits` in not low birth weight
summary(nc$visits[nc$lowbirthweight=="not low"])## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 0.00 10.00 12.00 12.27 15.00 30.00 6
## low not low
## 111 889
# summary of weight considered low: lower than or equal to 5.5 pounds
summary(nc$weight[nc$lowbirthweight=="low"])## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.000 3.095 4.560 4.035 5.160 5.500
# summary of weight considered not low: higher than 5.5 pounds
summary(nc$weight[nc$lowbirthweight=="not low"])## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 5.560 6.750 7.440 7.484 8.130 11.750
# inference between `visits` and `lowbirthweight` in hypothesis type
inference(y = nc$visits, x = nc$lowbirthweight, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_low = 108, mean_low = 10.7963, sd_low = 4.8506
## n_not low = 883, mean_not low = 12.265, sd_not low = 3.8036
## Observed difference between means (low-not low) = -1.4687
##
## H0: mu_low - mu_not low = 0
## HA: mu_low - mu_not low != 0
## Standard error = 0.484
## Test statistic: Z = -3.035
## p-value = 0.0024
# inference between `visits` and `lowbirthweight` in 95% ci
inference(y = nc$visits, x = nc$lowbirthweight, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_low = 108, mean_low = 10.7963, sd_low = 4.8506
## n_not low = 883, mean_not low = 12.265, sd_not low = 3.8036
## Observed difference between means (low-not low) = -1.4687
##
## Standard error = 0.484
## 95 % Confidence interval = ( -2.4173 , -0.5201 )
By the statistics above, the mean visits by mothers of low birth weight is 10.8 visits, while the other of not low birth weight is 12.27 visits. Their 95% confidence interval is (-2.4173, -0.5201). The p-value of the two variables is 0.0024, which is considered very small (<0.05).
Therefore, we reject the null hypothesis. Hence, the number of hospital visits is related to babies’ low birth weight. In conclusion, the more the number of hospital visits, it tends to have the high birth weight (>5.5 pounds).
# inference between `weeks` and `lowbirthweight` in hypothesis type
inference(y = nc$weeks, x = nc$lowbirthweight, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_low = 110, mean_low = 33.4273, sd_low = 4.6991
## n_not low = 888, mean_not low = 38.9426, sd_not low = 1.8947
## Observed difference between means (low-not low) = -5.5153
##
## H0: mu_low - mu_not low = 0
## HA: mu_low - mu_not low != 0
## Standard error = 0.453
## Test statistic: Z = -12.188
## p-value = 0
Besides that, by looking at the third graph, it is very obvious that the lower the length of pregnancy in weeks tends to have lower birth weight. The p-value = 0 proves the relationship between the two variables. By looking at all three graphs, although we do not have significant statistical evidences, the smaller the number of hospital visits, may tend to have lower length of pregnancy, which may lead to lower birth weight.