Confidence Interval for One Population Proportion

According to the Centers for Disease Control and Prevention, the single best way to protect against the flu is to be vaccinated. Traditionally, flu vaccines are manufactured using viruses grown in chicken eggs. A new method allows growing viruses in cell cultures instead. To test the effectiveness of such vaccines, in a recent clinical trial n = 3900 randomly selected healthy subjects aged 18 – 49 received a cell-derived influenza vaccine by injection. During a follow-up period of approximately 28 weeks, each subject was assessed whether he or she developed influenza, which happened for 26 of the 3900 subjects.

x <- 26 # had influenza
n <- 3900 # total number of subjects
  1. What is the population of interest? What is the parameter of interest?

The population of interest is healthy subjects aged 18 to 49. The parameter of interest is the proportion of healthy subjects aged 18 to 49 who develop the flue even though they have the vaccine.

  1. Find the point estimate of the population proportion that were vaccinated with the vaccine but still developed the flu.
PE <- x/n
PE
## [1] 0.006666667

P hat (Point Estimate) of the proportion of people with the vaccine but still got the flu is 0.0067.

  1. Find the standard error of this estimate.
SE <- sqrt(PE * (1 - PE) / n)
SE
## [1] 0.001303075

We are okay if our point estimate is off by 0.0013, as that is our standard error.

  1. Find the margin of error for a 99% confidence interval.
z <- qnorm(0.995)
ME <- z * SE
ME
## [1] 0.0033565

The margin of error for our confidence interval is 0.0034

  1. Construct the 99% confidence interval for the population proportion. Interpret the interval.
lower_bound <- round(PE - ME, 5)
upper_bound <- round(PE + ME, 5)
CI <- paste("(", lower_bound, ",", upper_bound, ")", sep = "")
print(CI)
## [1] "(0.00331,0.01002)"

We are 99 percent confident that the true proportion of healthy subjects from ages 18 to 49 who get the flu after being vaccinated is between 0.00331 and 0.01002.

  1. Can you conclude that fewer than 1% of all people vaccinated with the vaccine will develop the flu? Explain by using results from part (e).

We cannot conclude that fewer than 1 percent of all people vaccinated with the vaccine will develop the flu due to the fact that 0.01002 contains 1 percent (0.01).

Hypothesis Testing for Two Population Means

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.

library(tidyverse)
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download.file("http://www.openintro.org/stat/data/nc.RData", destfile = "nc.RData")
load("nc.RData")
nc <- na.omit(nc)

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.

  1. 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?
library(ggplot2) 
ggplot(data = nc, aes(y = weight, fill = habit)) +
  geom_boxplot()

Visually, these box plots show that the median weight of the children born to each type of parent does not appear to be significantly different. The median weight of the non smoking baby is slightly larger, but not worth noting. The non smoking mothers had a large amount of outliers in their babies, which shows a large skew left. This may possibly be explained by a larger sample size.

Numerically how do the means compare to each other?

by(nc$weight, nc$habit, mean)
## nc$habit: nonsmoker
## [1] 7.24676
## ------------------------------------------------------------ 
## nc$habit: smoker
## [1] 6.886429

Numerically, the non smoker babies are larger by weight than the babies of smoking mothers.

  1. Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different.
  1. 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$habit: nonsmoker
## [1] 716
## ------------------------------------------------------------ 
## nc$habit: smoker
## [1] 84

One of the conditions to check if inference is satisfied is if the sample is random and is less than 10 percent of the entire population. As seen above, 800 newborns were in this same. Because the data covers the entire state of North Carolina, we can assume that 800 is less than 10 percent of all newborns born in the state in 2004. The second test for inference is to make sure that the two samples are independent of each other. Because a mother is either a smoker or not, the two samples are independent of each other. Because both tests are passed, the data set is eligible to use.

  1. Use t.test to conduct a two sample test for the hypothesis in part 2 at a significance level \(\alpha = 0.1\).
smoker <- subset(nc, nc$habit == 'smoker')

nonsmoker <- subset(nc, nc$habit == 'nonsmoker')
t_test <- t.test(smoker$weight, nonsmoker$weight, alternative = "two.sided", conf.level = 0.9)

t_test
## 
##  Welch Two Sample t-test
## 
## data:  smoker$weight and nonsmoker$weight
## t = -2.3625, df = 108.54, p-value = 0.01994
## alternative hypothesis: true difference in means is not equal to 0
## 90 percent confidence interval:
##  -0.6133701 -0.1072923
## sample estimates:
## mean of x mean of y 
##  6.886429  7.246760
  1. What is the conclusion of the hypothesis test from part 4?

Based on the t-test, the found p-value of 0.01994 is less than the significance level of 0.1. Because of this, we reject the null hypothesis. There is evidence that the mean difference between newborns whose mothers smoke vs. those who do not smoke is different.

  1. If the conclusion from part 5 is an error what type of error would this be? What does this error mean in the context of the problem.

If there was an error in from part 5, it would be a type one error. In context of this problem, a type one error would be rejecting the null hypothesis when there is actually no significant difference in the average birth weight depending on the mothers’ smoking status.