Justin Herman Inference for numerical data
Exploratory analysis
load("more/nc.RData")
library(psych)
library(tidyverse)- What are the cases in this data set? How many cases are there in our sample?
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.
- Make a side-by-side boxplot of
habitandweight. What does the plot highlight about the relationship between these two variables?
boxplot(nc$fage ~ nc$habit)
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.
psych::describeBy(nc$weight, nc$habit,mat = TRUE)## item group1 vars n mean sd median trimmed mad
## X11 1 nonsmoker 1 873 7.144273 1.518681 7.31 7.277310 1.215732
## X12 2 smoker 1 126 6.828730 1.386180 7.06 6.939314 1.200906
## min max range skew kurtosis se
## X11 1.00 11.75 10.75 -1.1827980 2.879335 0.05139955
## X12 1.69 9.19 7.50 -0.9898247 1.666930 0.12349071by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126There 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.
psych::describeBy(nc$weight, nc$habit,mat = TRUE)## item group1 vars n mean sd median trimmed mad
## X11 1 nonsmoker 1 873 7.144273 1.518681 7.31 7.277310 1.215732
## X12 2 smoker 1 126 6.828730 1.386180 7.06 6.939314 1.200906
## min max range skew kurtosis se
## X11 1.00 11.75 10.75 -1.1827980 2.879335 0.05139955
## X12 1.69 9.19 7.50 -0.9898247 1.666930 0.12349071by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126- the sample size is over 30
- the individual samples are likely independent of each other
- Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
H~O: There is no difference between the average weights of babies born to smoking and non-smoking mothers are different\[U_{diff}=0\]H~A: There is a difference between the average weights of babies born to smoking and non-smoking mothers are different\[U_{diff} \neq0\]
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")## Warning: package 'BHH2' was built under R version 3.4.4## 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
- 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.
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.
inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )Answer: 95 % Confidence interval = 38.1528 - 38.5165 weeks
- Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function:
conflevel = 0.90.
inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical",conflevel = 0.90)## 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 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$gained, x = nc$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 = 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
#order = c("smoker","nonsmoker"))Answer our p value is .17 and we therefore fail to reject the null hypothesis
- 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.
summary(nc$mature)## mature mom younger mom
## 133 867moms <- nc %>%
group_by(mature) %>%
summarize(min_age = (min(mage)), max_age=(max(mage)))
moms## # A tibble: 2 x 3
## mature min_age max_age
## <fctr> <dbl> <dbl>
## 1 mature mom 35.0 50.0
## 2 younger mom 13.0 34.0- seems like the cutoff for out data was in between 34 and 35
- my method just goes through the data, groups by mature and young moms and then finds min and max values for each. Being that they dont overlap, it answers our question for us
- 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.
H~O: There is no difference between the mothers age of babies born to white versus nonwhite mothers \[U_{diff}=0\] H~A: There is a difference between the mothers age of babies born to white versus nonwhite mothers \[U_{diff} \neq0\]
inference(y = nc$mage, x = nc$whitemom, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical") ## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_not white = 284, mean_not white = 25.331, sd_not white = 6.435
## n_white = 714, mean_white = 27.6499, sd_white = 5.9898## Observed difference between means (not white-white) = -2.3189
##
## H0: mu_not white - mu_white = 0
## HA: mu_not white - mu_white != 0
## Standard error = 0.443
## Test statistic: Z = -5.237
## p-value = 0
inference(y = nc$mage, x = nc$whitemom, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical",
order = c("white","not white"))## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_white = 714, mean_white = 27.6499, sd_white = 5.9898
## n_not white = 284, mean_not white = 25.331, sd_not white = 6.435
## Observed difference between means (white-not white) = 2.3189
##
## Standard error = 0.4428
## 95 % Confidence interval = ( 1.451 , 3.1867 )Although I hate the wording of my hypothesis test, It would appear white mothers tend to have chidren at an older age than non-white mothers Our P value is near 0 and the 95% confidence interval is anywhere from 1.451 - 3.1867 years older
This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was adapted for OpenIntro by Mine Çetinkaya-Rundel from a lab written by the faculty and TAs of UCLA Statistics.