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.

  1. We are 95% confident that between 43% and 49% of Americans in this sample support the decision of the U.S. Supreme Court on the 2010 healthcare law. False The confidence interval estimates the population proportion not the sample.
  2. We are 95% confident that between 43% and 49% of Americans support the decision of the U.S. Supreme Court on the 2010 healthcare law. True The confidence interval estimate the population proportion.
  3. If we considered many random samples of 1,012 Americans, and we calculated the sample proportions of those who support the decision of the U.S. Supreme Court, 95% of those sample proportions will be between 43% and 49%. True by definition of the sampling model
  4. The margin of error at a 90% confidence level would be higher than 3%. False A 90% confidence interval has less proportion of population than a 95% confidence interval. then the margin error will also be less than 3%.

Legalization of marijuana, Part I. (6.10, p. 216) 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.

  1. Is 48% a sample statistic or a population parameter? Explain.
    • 48% is the sample statistic that estimates the population parameter.
  2. Construct a 95% confidence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of the data.
    • confidence interval = point estimate \(\pm\) 1.96 \(\times\) SE and point estimate is p,\(SE=\sqrt{\frac{p(1-p)}{n}}\). The confidence interval for the proportion of U.S. residents who think marijuana should be legalize is \((45%, 51%)\)
p <- .48
n <-  1259
SE <- sqrt(p*(1-p)/n)
cil <- .48 - 1.96*SE
cir <- .48 + 1.96*SE
cil
## [1] 0.4524028
cir
## [1] 0.5075972
  1. A critic points out that this 95% confidence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. Is this true for these data? Explain.
    • The sample has 1259 observations that represent less than 10% of the U.S. residents and observations are independent. It follows that \(np = 604>10\) and \(n(1-p)=654>10\). So the normal model is a good approximation.
n*p
## [1] 604.32
n*(1-p)
## [1] 654.68
  1. A news piece on this survey’s findings states, “Majority of Americans think marijuana should be legalized.” Based on your confidence interval, is this news piece’s statement justified?

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 ? - About 2398 U.S. residents would need to be survey if we wanted to limit the margin error at 2%.

ME <- 0.02
n <- (1.96**2)*p*(1-p)/ME**2
n
## [1] 2397.158

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.

pCA <- 0.080
pOR <- 0.088
nCA <- 11545
nOR <- 4691
p <- pCA - pOR
vaCA <- pCA*(1-pCA)/nCA
vaOR <- pOR*(1-pOR)/nOR
ME2 <- 1.96*sqrt(vaCA + vaOR)
CI2L <- p - ME2
CI2R <- p + ME2
CI2L*100
## [1] -1.749813
CI2R*100
## [1] 0.1498128

Because 0% is in the interval, we are 95% sure that the difference between the proportions of Californians and Oregonians who are sleep deprived is (-1.15%, 0.15%).

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.

  1. Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.
    • \(H_0\): Barking deer do not prefer to forage in certain habitats over others. \(H_A\):Barking deer prefer to forage on certain habitats over others.
  2. What type of test can we use to answer this research question? = We can use the chi-square test.
  3. Check if the assumptions and conditions required for this test are satisfied.
    • We can admit that all the 426 cases are independent
    • There were at least 5 categories count because the Others categories were reduced to one.
    • We create a test statistic for one way tables.
microhabitat <- c("Woods", "CuGP", "DeFo", "Others")
others <- 100 -(4.8 + 14.7 + 39.6)
observedData <- c(4,16,67,345)
expectedData <- c(4.8,14.7,39.6,others)
expectedData <- expectedData*426/100
BaDeer <- data.frame(microhabitat, observedData, expectedData)
BaDeer
  1. Do these data provide convincing evidence that barking deer pre- fer to forage in certain habitats over others? Conduct an appro- priate hypothesis test to answer this research question.
    • Chi-square test
df <- 3
Xsquare <- sum((BaDeer$observedData-BaDeer$expectedData)**2/BaDeer$expectedData)
pValue <- 1 - pchisq(Xsquare,df)
pValue
## [1] 0

The p value is pratically null. We reject the Null Hypothesis. There is convincing evidence that the Barking Deer prefer to forage in certain habitats over the 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.

{

}

  1. What type of test is appropriate for evaluating if there is an association between coffee intake and depression?
    • The chi-square test can be perform to evaluate if there is an association between coffee intake and depression.
  2. Write the hypotheses for the test you identified in part (a).
    • \(H_0\): There is no association between Women depression and the amount of coffee consumption. \(H_A\): The women depresion depends of the amount of coffee intaked.
  3. Calculate the overall proportion of women who do and do not suffer from depression.
    • 5.1% of women suffer from depression
    • 94.9% of women do not suffer from depression
sfd <- 2607/50739
nsfd <- 48132/50739
sfd*100
## [1] 5.138059
nsfd*100
## [1] 94.86194
-
  1. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
ec <- 2607*6617/50739
ec
## [1] 339.9854
csq <- (373 - ec)**2/ec
csq
## [1] 3.205914
- The expected count for the highlited cell is 339.9854
- The contribution of this cell to the test statistic is 3.2059
  1. The test statistic is \(\chi^2=20.93\). What is the p-value?
p_value <- 1 - pchisq(20.93, 4)
p_value
## [1] 0.0003269507
  1. What is the conclusion of the hypothesis test?
    • the p value is very low. We reject the null hypothesis. There is convincing evidence that the risk of depression in women depends of the amount of caffeine intake.
  2. One of the authors of this study was quoted on the NYTimes as saying it was “too early to recommend that women load up on extra coffee” based on just this study.64 Do you agree with this statement? Explain your reasoning.
    • This was an observational study; it does not establih causation. We need a traitment study for appropiate conclusion.