2010 Healthcare Law. (6.48, p. 248) 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. 46% of 1,012 Americans represent the population and not sample.
True. The confidence interval is for 46% population with 3% margin error.
True. 95% of the random samples will be between 43% and 49%
False. In this case, the margin of error will be lower than 3% since the confidence level is smaller.
Legalization of marijuana, Part I. (6.10, p. 216) The 2010 General Social Survey asked 1,259 US res- idents: “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 and not population parameter since it was a survey with 1,259 US residents and its not viable to survey entire population.
std_err <- sqrt(0.48*(1-0.48)/1259)
std_err## [1] 0.01408022lower <- 0.48 - 1.96*std_err
upper <- 0.48 + 1.96*std_err
paste(lower, upper)## [1] "0.452402769762903 0.507597230237097"Assuming samples are random and 1259 is less than 10% of all US residents.
# n*p
1259*0.48## [1] 604.32# n*(1-p)
1259*(1-0.48)## [1] 654.68Both are greater than 10 so the normal model is a good approximation.
Since the confidence interval is between 45% and 51% which covers half of the population, it wouldnt be completely justified to say majority supports leagalization.
Legalize Marijuana, Part II. (6.16, p. 216) As discussed in Exercise above, 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 <- 0.02
# z * sqrt(p*(1-p)/n) = me
n <- 0.48*(1-0.48) / (0.02 / 1.96)^2
n## [1] 2397.158So in this case, we need to survey 2398 Americans.
Sleep deprivation, CA vs. OR, Part I. (6.22, p. 226) According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insuffient 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.
p_ca <- 0.08
p_or <- 0.088
n_ca <- 11545
n_or <- 4691
z <- 1.96
se <- sqrt((p_ca*(1-p_ca))/n_ca + (p_or*(1-p_or))/n_or)
se## [1] 0.004845984me <- z*se
me## [1] 0.009498128# 95% confidence interval for the difference between the proportions of Californians and Oregonians
(p_ca - p_or) - me## [1] -0.01749813(p_ca - p_or) + me## [1] 0.001498128So the 95% confidence interval is (-0.0175, 0.0015).
Barking deer. (6.34, p. 239) 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.
\(H_0\):barking deer does not have preference to forage in certain habitats over others \(H_A\):barking deer prefers to forage in certain habitats over others
We can use chi-square test.
obs_woods <- 4
obs_grass <- 16
obs_forest <- 61
obs_other <- 345
observed <- c(obs_woods, obs_grass, obs_forest, obs_other)
exp_woods <- 20.448
exp_grass <- 62.622
exp_forest <- 168.696
exp_other <- 426 - (exp_woods + exp_grass + exp_forest)
expected <- c(exp_woods, exp_grass, exp_forest, exp_other)
k <- 4
df <- k-1
chisq <- sum(((observed - expected)^2)/expected)
chisq## [1] 284.0609p_val <- 1 - pchisq(chisq, df)
p_val## [1] 0The test results shows evidence to reject null hypothesis \(H_0\):barking deer does not have preference to forage in certain habitats over others
Coffee and Depression. (6.50, p. 248) 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.
{}
chi-square test could be appropriate for evaluating if there is an association between coffee intake and depression.
\(H_0\): There is no association between coffee intake and depression. \(H_A\): There is a relation between coffee intake and depression.
#who do suffer
2607/50739## [1] 0.05138059#who donot suffer
48132/50739## [1] 0.9486194obs <- 373
exp <- (2607/50739)*6617
# contribution of cell
cont <- (obs-exp)^2/exp
cont## [1] 3.205914df <- (2-1)*(5-1)
p_val <- 1 - pchisq(20.93, df)
p_val## [1] 0.0003269507We reject the null hypothesis.
I agree. Even though we reject the null hypothesis, we cant say much about the relation between coffee consumption and depression. It might be strong or wea.