Data 606 Lab 5
S. Tinapunan
March 25, 2018
setwd("C:/Users/stina/Documents/CUNY SPS Data Science/Spring 2018 Classes/DATA 606 - Probability and Statistics/Lab5")
load("./Lab5/more/nc.RData")library(dplyr)## Warning: package 'dplyr' was built under R version 3.4.3##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionExercise 1:
What are the cases in this data set? How many cases are there in our sample?
The observations in this dataset are a subset from birth records released by North Carolina in 2004. This dataset has 1000 observations and 13 variables.
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
##
##
##
## Exercise 2:
Make a side-by-side boxplot of habit and weight.
What does the plot highlight about the relationship between these two variables?
It appears that the median birth weight of newborns of mothers who smoke is lower, and the maximum weight is also lower.
The mean birth weight of newborns to nonsmoker mothers: 7.144273 The mean birth weight of newborns to smoker mothers: 6.82873
boxplot(nc$weight ~ nc$habit)
by(nc$weight, nc$habit, mean)## nc$habit: nonsmoker
## [1] 7.144273
## --------------------------------------------------------
## nc$habit: smoker
## [1] 6.82873Exercise 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.
nc\(habit: nonsmoker ---> 873 nc\)habit: smoker —> 126
The sample size is 1000, which is less than 10% the size of the population. I am going to assume that the 1000 observations were randomly selected. It seems reasonable to assume that the observations are independent.
Looking at the histogram below, the distribution looks approximately normal. There is some skew. It looks moderate to me. The sample size of 873 and 126 are sufficient I think for this skew.
by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126smoker_weight <- dplyr::filter(nc,habit=="smoker") %>% select(weight)## Warning: package 'bindrcpp' was built under R version 3.4.3nonsmoker_weight <- dplyr::filter(nc,habit=="nonsmoker") %>% select(weight)
smoker_weight <- unlist(smoker_weight)
nonsmoker_weight <- unlist(nonsmoker_weight)
par(mfrow=c(1,2))
hist(smoker_weight)
hist(nonsmoker_weight)
par(mfrow=c(1,1))Exercise 4:
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
H_0: There is no difference in the average weights of babies born to smoking and non-smoking mothers.
H_A: There is a difference in the average weights of babies born to smoking and on-smoking mothers.
The mean weight of babies of smoking mothers: 6.82873 The mean weight of babies of nonsmoking mothers: 7.144273 smoker - nonsmoking: -0.3155425
Based on the hypothesis testing done by inference function, the p-value is 0.32 or there is a 32% chance that the difference in average weight observed in this particular case could be due to chance. So, we would fail to reject the null hypothesis.
mean(smoker_weight) - mean(nonsmoker_weight)## [1] -0.3155425inference(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
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.
Standard error = 0.1338 95 % Confidence interval = ( -0.5777 , -0.0534 )
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
(a) 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.
Single mean
Summary statistics: mean = 38.3347 ; sd = 2.9316 ; n = 998
Standard error = 0.0928
95 % Confidence interval = ( 38.1528 , 38.5165 )
This means that if we do 100 samples from the population, we expect that 95 out of the 100 samples will have a mean pregnancy length that falls within this range.
inference(y = nc$weeks, est = "mean", type = "ci", null = 0,
alternative = "twosided", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )(b) Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.
Single mean
Summary statistics: mean = 38.3347 ; sd = 2.9316 ; n = 998
Standard error = 0.0928
90 % Confidence interval = ( 38.182 , 38.4873 )
inference(y = nc$weeks, est = "mean", type = "ci", conflevel=.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 )(c) Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
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
The p-value is 0.1686. At the significance level of 0.05, we would fail to reject the null hypothesis. There is about a 17% chance that an observed difference that is found in this sample is due to chance.
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
(d) 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.
Younger moms: 13 to 34 Mature mom: 35 to 50
mature_mom_age <- nc$mage[nc$mature=="mature mom"]
young_mom_age <- nc$mage[nc$mature=="younger mom"]
c(min(young_mom_age), max(young_mom_age))## [1] 13 34c(min(mature_mom_age), max(mature_mom_age))## [1] 35 50(e) 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.
nc\(weight: numerical nc\)gender: categorical
mean weight of female newborns: 6.902883 mean weight of male newborns: 7.301509 difference of mean (male-female): 0.398626
Premature babies have lower birth weights
Ho: There is no difference in weight between female and male newborns. HA: There is a difference in weight between female and male newborns.
Result:
H0: mu_male - mu_female = 0 HA: mu_male - mu_female != 0 Standard error = 0.095 Test statistic: Z = 4.211 p-value = 0
The p-value of 0 is strong evidence against the null hypothesis. There is a difference in the mean weight of male and female newborns. I had no idea.
by(nc$weight, nc$gender, mean)## nc$gender: female
## [1] 6.902883
## --------------------------------------------------------
## nc$gender: male
## [1] 7.3015097.301509 - 6.902883## [1] 0.398626inference(y = nc$weight, x = nc$gender, est = "mean", type = "ht", null = 0,
alternative = "twosided", method = "theoretical", order=c("male", "female"))## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_male = 497, mean_male = 7.3015, sd_male = 1.5168
## n_female = 503, mean_female = 6.9029, sd_female = 1.4759## Observed difference between means (male-female) = 0.3986
##
## H0: mu_male - mu_female = 0
## HA: mu_male - mu_female != 0
## Standard error = 0.095
## Test statistic: Z = 4.211
## p-value = 0