On June 28, 2012 the U.S. Supreme Court upheld the much debated 2010 healthcare law, declaring it constitutional. A Gallup poll released the day after this decision indicates that 46% of 1,012 Americans agree with this decision. At a 95% confidence level, this sample has a 3% margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning.
False; the confidence interval doesn’t refer to the sample, but any random sample from the population.
True; this use of the confidence interval is appropriate with margin of error for random samples.
True; this outlines the concept of what a confidence interval is.
False; with a lower confidence level, the margin of error would decrease with the range of the confidence interval.
The 2010 General Social Survey asked 1,259 US residents: “Do you think the use of marijuana should be made legal, or not?” 48% of the respondents said it should be made legal.
48% is a sample statistic. If the statistic passes specific criteria, it may be applied to the population.
z <- 1.96
n <- 1259
p <- 0.48
SE <- ((p*(1 - p))/n)^0.5
low <- p - (z * SE)
high <- p + (z * SE)
cat("The confidence interval is (",low,",",high,")")## The confidence interval is ( 0.4524028 , 0.5075972 )Sampling distribution appears to be nearly normal and composed of independent observations. Proportions of “successes”" and “failures” are checked below.
#p & n from prior code chunk
r1 <- p*n
r2 <- (1-p)*n
cat("Since",r1,"&",r2,"are sufficiently large, the success-failure criterion is met.")## Since 604.32 & 654.68 are sufficiently large, the success-failure criterion is met.No, the statement is not entirely justified. Not all–not even most–randomly composed samples of the sample will feature a majority that approve of legalization.
As discussed in Exercise 6.12, the 2010 General Social Survey reported a sample where about 48% of US residents thought marijuana should be made legal. If we wanted to limit the margin of error of a 95% confidence interval to 2%, about how many Americans would we need to survey?
\[ ME = z \times SE\implies ME=z \times\sqrt{\bigg(\frac{p(1-p)}{n}\bigg)}\implies n=\frac{p\times(1-p)\times z^2}{ME^2}\]
#for 95% confidence interval, z-score is 1.96
z <- 1.96
#value of p does not change from before
p <- 0.48
#from above
ME<- 0.02
n <- (p*(1-p)*(z^2))/(ME^2)
cat("We would have to sample",ceiling(n),"people to limit the margin of error to 2%.")## We would have to sample 2398 people to limit the margin of error to 2%.According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.
z<-1.96
pCA<-.08
nCA<-11545
pOR<-.088
nOR<-4691
seCAsq<-(pCA*(1-pCA))/nCA
seORsq<-(pOR*(1-pOR))/nOR
#calculate the difference in standard error
se28<-sqrt(seCAsq + seORsq)
low <- pCA-pOR - (z * se28)
high <- pCA-pOR + (z * se28)
cat("The confidence interval is (",low,",",high,")")## The confidence interval is ( -0.01749813 , 0.001498128 )Microhabitat factors associated with forage and bed sites of barking deer in Hainan Island, China were examined from 2001 to 2002. In this region woods make up 4.8% of the land, cultivated grass plot makes up 14.7% and deciduous forests makes up 39.6%. Of the 426 sites where the deer forage, 4 were categorized as woods, 16 as cultivated grassplot, and 61 as deciduous forests. The table below summarizes these data.
\(H_0\): there is no significant difference; deer have no preference over others
\(H_A\): there is a significant difference; deer have a preference
Chi-square.
Independence is observed, as all habitats are independent. Sample sizes are marginally acceptable, with the lower limit of 5 challenged by inlcusion of 4 units of land classified as “woods”.
land<-c(4,16,61,345)
finpct<-1-.048-.147-.396
pct<-c(.048,.147,.396,finpct)
deer<-as.data.frame(cbind(land,pct))
deer$ratio<-deer$pct*426
deer$pct<-NULL
chisq.test(deer)##
## Pearson's Chi-squared test
##
## data: deer
## X-squared = 145.37, df = 3, p-value < 2.2e-16chi<-((deer$land-deer$ratio)^2)/deer$ratio
chisum<-sum(chi)
p_val <- 1 - pchisq(chisum,df=3)
cat("The secondary p-value agrees with the output above with p-value of",p_val)## The secondary p-value agrees with the output above with p-value of 0We reject the hull hypothesis and conclude that there is a significant difference in the foraging habitats, and the deer prefer some habitats over others.
Researchers conducted a study investigating the relationship between caffeinated coffee consumption and risk of depression in women. They collected data on 50,739 women free of depression symptoms at the start of the study in the year 1996, and these women were followed through 2006. The researchers used questionnaires to collect data on caffeinated coffee consumption, asked each individual about physician-diagnosed depression, and also asked about the use of antidepressants. The table below shows the distribution of incidences of depression by amount of caffeinated coffee consumption.
A chi-square test is appropriate.
\(H_0\): there is no association between coffee intake and depression
\(H_A\): there is an association between coffee intake and depression
fTOT <- 50739
fDEP <- 2607
fFINE <- 48132
fDEPpct<-fDEP/fTOT
fFINEpct<-fFINE/fTOT
cat("There are",fDEP,"(",round(fDEPpct*100,2),"%) depressed women and",fFINE,"(",round(fFINEpct*100,2),"%) women who are not depressed in the sample.")## There are 2607 ( 5.14 %) depressed women and 48132 ( 94.86 %) women who are not depressed in the sample.exp<-fDEPpct*6617
cat("The expected value for the highlighted cell is",exp,"\n")## The expected value for the highlighted cell is 339.9854obs<-373
st<-((obs-exp)^2)/exp
cat("The resulting contribution is",st,"as calculated above.")## The resulting contribution is 3.205914 as calculated above.p_val<-pchisq(20.93,df=4,lower.tail=FALSE)
cat("The p-value is",p_val)## The p-value is 0.0003269507We reject the null hypothesis and accept the alternative and conclude that there is an association between coffee intake and depression.
Yes, as an avid coffee drinker I recommend it to everyone, however, more data need to be collected and more trials conducted. This assumes that depression is equally likely among all women in the sample.