inf_for_numerical_data.utf8.md

title: ‘Inference for numerical data’

author: ‘sufian’

output:

html_document:

css: ./lab.css

highlight: pygments

theme: cerulean

pdf_document: default

Rpub link:

http://rpubs.com/ssufian/543957

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",mode="wb")
load("nc.RData")

#load("more/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?

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.

ans:

Making box plots for the numerical data sets:

Female and male ages do not exhibit outliers
Weeks, visits, gained and weights all do have outliers

library(ggplot2)

ggplot(nc, aes(y=nc$fage, x=1))  +geom_boxplot()

## Warning: Removed 171 rows containing non-finite values (stat_boxplot).

ggplot(nc, aes(y=nc$mage , x=1))  +geom_boxplot()

ggplot(nc, aes(y=nc$weeks  , x=1))  +geom_boxplot()

## Warning: Removed 2 rows containing non-finite values (stat_boxplot).

ggplot(nc, aes(y=nc$visits , x=1))  +geom_boxplot()

## Warning: Removed 9 rows containing non-finite values (stat_boxplot).

ggplot(nc, aes(y=nc$gained , x=1))  +geom_boxplot()

## Warning: Removed 27 rows containing non-finite values (stat_boxplot).

ggplot(nc, aes(y=nc$weight , x=1))  +geom_boxplot()

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?

qplot(nc$habit, nc$weight, geom = "boxplot", na.rm = TRUE)

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 .

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 by command above but replacing mean with length.

#computing samples sizes of groups
by(nc$weight, nc$habit, length)

## nc$habit: nonsmoker
## [1] 873
## -------------------------------------------------------- 
## nc$habit: smoker
## [1] 126

nc_smoking_mothers <- nc[nc$habit == "nonsmoker",]# smoking mothers and their babies' weight
hist(nc_smoking_mothers$weight)

nc_non_smoking_mothers <- nc[nc$habit == "smoker",] #Non smoking mother and their babies' weight
hist(nc_non_smoking_mothers$weight)

ans:

The 2 conditions are satisfied:

Both the data sets - smomking and non-smking mothers are individually independent
The distributions are left skewed for both but the counts are large enough to offset the skewness,

and its quite normal without extreme outliers

Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.

Ho: The difference between the average weight of babies born to smoking and no-smoking mothers is Zero

Ha: The difference between the average weight of babies born to smoking and no-smoking mothers is Not

  Zero

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.

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 the x variable from the function.

ans:

95 % confidence level is: ( 38.1528 , 38.5165 ) That means that 95% of the women will have the average

length falling in this range.

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.

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 )

Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.

ans:

Ho: Avg weight gained by younger mothers and mature mothers is the same

Ha: Avg weight gained by younger mothers and mature mothers is Not the same

Now check for the 2 conditions for inference testing:

Independence and randomness

In this case, we assume its indpedent and random

Normality

See histograms below, the mature vs. younger moms have quite normal distribution and since their samples

are greater than 30, it can withstand the skewness

#sample n of younger and mature mothers
by(nc$gained,nc$mature, length)

## nc$mature: mature mom
## [1] 133
## -------------------------------------------------------- 
## nc$mature: younger mom
## [1] 867

#historgram to check for normality
nc_mature_mothers <- nc[nc$mature == "mature mom",] #histogram of mature mom and its weight gained
hist(nc_mature_mothers$gained)

nc_young_mothers <- nc[nc$mature == "younger mom",] #histogram of mature mom and its weight gained
hist(nc_young_mothers$gained)

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

ans:

Since the p value is > 0.05, Therefore, I fail to reject the null hypothesis that younger mothers and

mature mothers have the same mean weight gain.

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.

library(dplyr)

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

younger.mom.min.age <- nc$mage[nc$mature == "younger mom"] %>% min()
younger.mom.max.age <- nc$mage[nc$mature == "younger mom"] %>% max()
sprintf("Younger moms min age is: %s", younger.mom.min.age)

## [1] "Younger moms min age is: 13"

sprintf("Younger moms max age is: %s", younger.mom.max.age)

## [1] "Younger moms max age is: 34"

mature.mom.min.age <- nc$mage[nc$mature == "mature mom"] %>% min()
mature.mom.max.age <- nc$mage[nc$mature == "mature mom"] %>% max()
sprintf("Mature moms min age is: %s", mature.mom.min.age)

## [1] "Mature moms min age is: 35"

sprintf("Mature moms max age is: %s", mature.mom.max.age)

## [1] "Mature moms max age is: 50"

ans:

Since young mother’s max age is 34 while mature mom’s min age is 35, a good cut off is between 34 or 35

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:

Research qeustion - Is there a relationship between gender of child vs. age of mothers?

Ho: Avg lowbirthweight by younger mothers and mature mothers is the same

Ha: Avg lowbirthweight by younger mothers and mature mothers is Not the same

#sample n of younger and mature mothers and lowbirthweights
by(nc$mage,nc$lowbirthweight,length)

## nc$lowbirthweight: low
## [1] 111
## -------------------------------------------------------- 
## nc$lowbirthweight: not low
## [1] 889

#historgram to check for normality
nc_age_mothers <- nc[nc$lowbirthweight == "low",] 
hist(nc_age_mothers$mage)

nc_age_mothers <- nc[nc$lowbirthweight== "not low",] 
hist(nc_age_mothers$mage)

Again, checking for the 2 conditions for inference testing:

Independence and randomness

In this case, we assume its indpedent and random

Normality

See histograms below, the mature vs. younger moms have quite normal distribution and since their samples

are greater than 30, it can withstand the skewness

inference(y = nc$mage, x = nc$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 = 111, mean_low = 26.964, sd_low = 6.7755
## n_not low = 889, mean_not low = 27.0045, sd_not low = 6.1439

## Observed difference between means (low-not low) = -0.0405
## 
## H0: mu_low - mu_not low = 0 
## HA: mu_low - mu_not low != 0 
## Standard error = 0.675 
## Test statistic: Z =  -0.06 
## p-value =  0.9522

ans:

Since p > 0.05, failed to reject Ho, therefore low birth weight is the same across mothers ages