Lab Week 8
Mary Glantz
10/19/2021
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 set.
Exploratory analysis
Load the NC dataset.
download.file("http://www.openintro.org/stat/data/nc.RData", destfile = "nc.RData")
load("nc.RData")
# also load the tidyverse
library(tidyverse)## ── Attaching packages ─────────────────────────────────────────────────────── tidyverse 1.3.0 ──## ✓ ggplot2 3.3.2 ✓ purrr 0.3.4
## ✓ tibble 3.0.3 ✓ dplyr 1.0.0
## ✓ tidyr 1.1.0 ✓ stringr 1.4.0
## ✓ readr 1.3.1 ✓ forcats 0.5.0## ── Conflicts ────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()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?
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 variables, are there outliers? If you aren’t sure or want to take a closer look at the data, make a graph.
I’m going to guess the categorical variables are those without the 5-number summary. They have two to three responses that seem categorical. To test this, I will plot the premie category. (I should come up with a simple bar graph.)
plot(nc$premie)
Consider the possible relationship between a mother’s smoking habit and the weight of her baby. Plotting the data is a useful first step because it helps us quickly visualize trends, identify strong associations, and develop research questions.
# Create plot of smoker vs smoker and baby birthweight
ggplot(nc, mapping = aes(habit, weight))+
geom_boxplot()+
ggtitle("Birthweight by Mother's Smoking Habit")
### Exercise 1
Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?
See above. The median birthweight of infants born to smokers is slightly lower than that of infants born to nonsmokers. There is more spread among nonsmokers though.
Exercise 2
The box plots show how the medians of the two distributions compare, but we can also compare the means of the distributions using the following function to split the weight variable into the habit groups, then take the mean of each using the mean function.
by(nc$weight, nc$habit, mean)## nc$habit: nonsmoker
## [1] 7.144273
## ------------------------------------------------------------
## nc$habit: smoker
## [1] 6.82873The mean weight for nonsmokers is also higher than smokers.
There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test .
Inference
Exercise 3
Check if the conditions necessary for inference are satisfied. Note that you will need to obtain sample sizes to check the conditions. You can compute the group size using the same by command above but replacing mean with length.
by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## ------------------------------------------------------------
## nc$habit: smoker
## [1] 126For inference, we need to make sure the samples are independent and roughly normally distributed (though I thought that with a sample this large, that’s not necessary).
Exercise 4
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
Null hypothesis: There is no difference between the average weights of babies born to smoking and non-smoking mothers. Alternative hypothesis: There is a difference between the average weights of babies born to smoking and non-smoking mothers. (Two-tailed hypothesis—we are not arguing that one weight is higher or lower only.)
Next, we introduce a new function, inference, that we will use for conducting hypothesis tests and constructing confidence intervals.
inference(y = nc$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
(ooo….that is very cool)
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.
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 (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$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical",
order = c("smoker","nonsmoker"))## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )The 95 % Confidence interval for mean length of pregnancy is ( 38.1528 , 38.5165 ) weeks.
- Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.
inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical",
order = c("smoker","nonsmoker"), conflevel = .90)## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 90 % Confidence interval = ( 38.182 , 38.4873 )The 90 % Confidence interval is ( 38.182 , 38.4873 ) weeks.
- 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$weight, x = nc$mature, est = "mean", type = "ht", null = 0,
alternative = "greater", method = "theoretical")## 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 = 0.4263
The observed difference between means (mature mom-younger mom) is 0.0283 pounds. The p-value is 0.4263, which means there is a 42.6% probability that a random sample of mature and younger moms would come up with the same or greater weight difference. In other words, we do not have enough evidence to reject the null hypothesis that there is no difference in the mean weight of children born to 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.
We know that there are 133 mature moms and 867 younger moms. Looking at the mage category, we can figure out what the top 133 ages are.
summary(nc$mage)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13 22 27 27 32 50The top 25% (3rd Qu), would be the top 250 people. So we know the mature category is older than 32.
#Create a vectore of momage
momage <- nc$mage
#Pull out the top 250 moms by age
maturemom <- momage[momage>32]
#summarize
summary(maturemom)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 33.0 34.0 35.0 35.7 37.0 50.0Now at least we know the top 125 moms are 35 years or over. I’m going to leave it here, that mature moms are between 34 and 35 years old. (If you can explain to me how to get this cutoff, I’d be grateful!)
- 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.
Is there a relationship between the gender and birth weight of babies?
# Create plot of sex and baby birthweight
ggplot(nc, mapping = aes(gender, weight))+
geom_boxplot()+
ggtitle("Birthweight by Gender")
It would appear that the median weight for male babies is higher than that for females. A hypothesis test would reveal if this is what’s the word, not significant, detectable?
inference(y = nc$weight, x = nc$gender, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_female = 503, mean_female = 6.9029, sd_female = 1.4759
## n_male = 497, mean_male = 7.3015, sd_male = 1.5168## Observed difference between means (female-male) = -0.3986
##
## H0: mu_female - mu_male = 0
## HA: mu_female - mu_male != 0
## Standard error = 0.095
## Test statistic: Z = -4.211
## p-value = 0
The observed difference between means (female-male) is -0.3986 pounds. The p-value is 0. This indicates that there is enough evidence to reject the null hypothesis that there is no difference in birth weight between male and female infants.