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.

Answer: False.95% confident range applies for the population, not for sample.

  1. 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.

Answer: True. 95% confident applies for the population.

  1. 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%.

Answer: False. Many random samples is still samples and the sample proportion might be very close to the range.

  1. The margin of error at a 90% confidence level would be higher than 3%.

Answer: False. The margin of error= Z * SD, and the smaller of the Z, the smaller of the margin of error. 90% confidence level has a smaller Z value, comparing 95% confidence level.

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.

  1. Is 48% a sample statistic or a population parameter? Explain.

Answer: The 48% is a sample statistic because it calculate by the 1259 US residents.

  1. 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.
z<-1.96
n<-1259
p<-0.48
sd = sqrt((p * (1 - p))/1259)

higer<-0.48+z*sd
lower<-0.48-z*sd
c(lower,higer)
## [1] 0.4524028 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.

Answer: The sampling distribution is normal when the sample’s observation are independent and meet the success-failure condition. This sample is a randomly sample. In addition, np=12590.48=604>10 and n (1-p) =1259(1-0.48) =655>10.The sample meets the success-failure test.

  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?

Answer: The 95% confident level is between 0.4524 and 0.5076. Although the higher bound exceed 0.5, we don’t know the exact weight. Based on the confidence interval, the news piece’s statement is not 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 ?

z<-1.96
p<-0.48
me<-0.02
n<-(p*(1-p))/((me/z)^2)
round(n,0)+1
## [1] 2398

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.

p1<-0.08
p2<-0.088
n1<-11545
n2<-4691
z<-1.96

#standard error of p1-p2
SE<-sqrt((p1*(1-p1)/n1)+(p2*(1-p2)/n2))


higher<-(p1-p2)+z*SE
lower<-(p1-p2)-z*SE
c(lower,higher)
## [1] -0.017498128  0.001498128

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.

Answer: \(H_0\)= Barking deer have no preference to forage in certain habitats over others. \(H_A\)= Barking deer prefer to forage in certain habitats over others

  1. What type of test can we use to answer this research question?

Answer: We can use chi-squre test to answer this research question.

  1. Check if the assumptions and conditions required for this test are satisfied.

Answer: First, Each case must be independent. Every observation is random and independent. Second, each particular scenario must have at least 5 expected cases. Every category is more than 5 cases, except woods.

  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.
n<-426
#Observation<-c(4,16,67,345)
#Expected<-c(n*0.048,n*0.147,n*0.396,n*0.409)
#Chisq<-sum((Observation-Expected)^2/Expected)
#df<-4-1
#1-pchisq(Chisq,df)

x<-matrix(c(4,n*0.048,16,n*0.147,67,n*0.396,345,n*0.409),nrow=2)
chisq.test(x)
## 
##  Pearson's Chi-squared test
## 
## data:  x
## X-squared = 138.72, df = 3, p-value < 2.2e-16

Answer: p value is so small and almost equal to 0. We will reject \(H_0\) and in favor of alternative hypothesis that barking deer prefer 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.

{

}

  1. What type of test is appropriate for evaluating if there is an association between coffee intake and depression?

Answer: We can use chi-square test.

  1. Write the hypotheses for the test you identified in part (a).

Answer: \(H_0\)= There is no association between coffee intake and depression. \(H_A\)= There is association between coffee intake and depression.

  1. Calculate the overall proportion of women who do and do not suffer from depression. Answer:women who do suffer from depression=2607/50739=0.05138059. And women who do not suffer from depression=48132/50739=0.9486194

  2. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).

#Identify the expected count for the highlighted cell
depression<-2607/50739
expected<-6617*depression
expected
## [1] 339.9854
#calculate the contribution of this cell to the test statistic
contribution<-(373-expected)^2/expected
contribution
## [1] 3.205914
  1. The test statistic is \(\chi^2=20.93\). What is the p-value?
pvalue <- pchisq(20.93, 4,lower.tail=FALSE)
pvalue
## [1] 0.0003269507
  1. What is the conclusion of the hypothesis test?

Answer: P value is lower than 0.05, so we can reject the null hypnosis and in favor of the alternative hypnosis. We can conclude that there is association between coffee intake and depression.

  1. 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.

Answer: I agree. Although from the chi-square test we can conclude an association between coffee intake and depression. We can rule out other variables which the study not provide.