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.conflevel = 0.90.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.
setwd("C:/Users/26291/Documents/Data606")
download.file("http://www.openintro.org/stat/data/nc.RData", destfile = "nc.RData")
load("nc.RData")
View(nc)
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.
habit and weight. What does the plot highlight about the relationship between these two variables?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] 6.82873
There is an observed difference, but is this difference statistically significant? In order to answer this question we will conduct a hypothesis test .
by command above but replacing mean with length.by(nc$weight, nc$habit, length)
## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126
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
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.
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.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 )
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$weeks, est="mean",type="ci",null=0,alternative="twosided",method="theoretical")
## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )
conflevel = 0.90.inference(y=nc$weeks, est="mean", type="ci", null=0, alternative="twosided", method="theoretical", 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 )
#Null Hypothesis: Average mother's weight gain after birth is independent of age
#Alternative Hypothesis: Average mother's weight gain after birth is depedent of age
age = ifelse(nc$mage<=mean(nc$mage),"Younger",'Older')
inference(y=nc$weeks, x=age, est="median", type="ht", null=0, alternative="twosided", method="simulation")
## Response variable: numerical, Explanatory variable: categorical
## Difference between two medians
## Summary statistics:
## n_Older = 455, median_Older = 39, n_Younger = 543, median_Younger = 39,
## Observed difference between medians (Older-Younger) = 0
##
## H0: median_Older - median_Younger = 0
## HA: median_Older - median_Younger != 0
## p-value = 1.9998
# With a P value of we can reject the null hypothesis and accept the alternative hypothesis that average mother's weight gain after birth is dependent of age.
#using a subset
#younger mother age range
c(min(subset(nc, mature == 'younger mom')$mage),max(subset(nc, mature == 'younger mom')$mage))
## [1] 13 34
#mature mother age range
c(min(subset(nc, mature != 'younger mom')$mage),max(subset(nc, mature != 'younger mom')$mage))
## [1] 35 50
research question evaluating the relationship between these variables. Formulate the question in a way that it can be answered using a hypothesis test and/or a confidence interval. Answer your question using the inference function, report the statistical results, and also provide an explanation in plain language.
Null Hypothesis: Average weight of a mother is independent of marital status Alternative Hypothesis: Average weight of a mother depends on marital status
inference(y=nc$weight,x=nc$marital,est='mean',type='ht',null = 0,alternative = "twosided",method = 'theoretical')
## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_married = 386, mean_married = 6.8007, sd_married = 1.6118
## n_not married = 613, mean_not married = 7.2958, sd_not married = 1.4027
## Observed difference between means (married-not married) = -0.4951
##
## H0: mu_married - mu_not married = 0
## HA: mu_married - mu_not married != 0
## Standard error = 0.1
## Test statistic: Z = -4.966
## p-value = 0
the chances of it being independent of marital status was 0. Therefore we will reject our null hypothesis and acceppt our alternativehypothesis that the average weight of a mother depends on her marital status.