Graded: 6.6, 6.12, 6.20, 6.28, 6.44, 6.48

6.6 2010 Healthcare Law. 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.39

  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.

Ans: False. A ci is contructed to est. the population proportion, not the sample proportion.

  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.

margin of error for a 95% confidence interval: ME = 1.96 × SE = 1.96 × sqrt(p(1-p)/n)

Ans:True. The 95% ci is (0.43, 0.49).

  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%.

Ans: True. It’s the definition of ci.

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

Ans: True.Because of the decresing ME(=z*SE).

6.12 Legalization of marijuana, Part I. 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.44

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

Ans: a sample statistic, because 48% of respondents in 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.

Ans: The ci is between 0.4524 and 0.5076.

  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.

Ans: True.When the sample is unbias its result can be a good approximation to the population.

  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?

Ans:No, less than 50% the interval lies majority.

6.20 Legalize Marijuana, Part II. 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 ?

2% = 1.96 * sqrt(0.48(1-0.48)/n)

Ans: 2400.

6.28 Sleep deprivation, CA vs. OR, Part I. According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insucient 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 di???erence 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
SEsq.ca<-p.ca*(1-p.ca)/n.ca
SEsq.or<-p.or*(1-p.or)/n.or
SE<-sqrt(SEsq.ca+SEsq.or)
diff<-abs(p.ca-p.or)
c(diff-SE*1.96, diff+SE*1.96)
## [1] -0.001498128  0.017498128

6.44 Barking deer. 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.62 Woods Cultivated grassplot Deciduous forests Other Total 4 16 61 345 426

  1. Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.

Ans: H0: Each habitats and others are equally prefer for barking deer. Ha: Each habitats and others are Not equally prefer for barking deer.

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

Ans: chi-square goodness of fit test

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

Ans:Two sample sets are independent.Sample size is large enought to ignor the testing for the sknewness. Each habitats has like and unlike, and p and n are given in this question.

  1. Do these data provide convincing evidence that barking deer prefer to forage in certain habitats over others? Conduct an appropriate hypothesis test to answer this research question.

Ans: For chi-square equal to 284.0609 and df=3, p-value is lesser than 0.001. We reject H0.

#Observations Of the 426 sites where the deer forage
O<-as.numeric(c(4,16,61,345))

#Expected number Of the 426 sites where the deer forage
E<-as.numeric(c(20.448,62.622,168.696,174.234))
S<-c(0,0,0,0)
for (i in 1:4){
S[i]<-(O[i]-E[i])^2/E[i]
}
X<-S[1]+S[2]+S[3]+S[4]
X
## [1] 284.0609

6.48 Co???ee and Depression. Researchers conducted a study investigating the relationship between ca???einated co???ee 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 ca???einated co???ee 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 ca???einated co???ee consumption.

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

Ans: Chi-square test.

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

Ans: H0: No coffee effect on women between physician-diagnosed depression and antidepressants, Ha: there is differecen effect from coffee.

  1. Calculate the overall proportion of women who do and do not su???er from depression.

Ans: 5.138% for whom do and 94.8619% for who do not.

  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.
expect <-0.05138*6617
x=(373-expect)^2/expect
x
## [1] 3.206716
  1. The test statistic is chi-square= 20.93. What is the p-value?

Ans: less than 0.001.

  1. What is the conclusion of the hypothesis test?

Ans: Reject H0, there is significan difference of depression in women from coffee.

  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 co???ee” based on just this study.64 Do you agree with this statement? Explain your reasoning.

Ans: Yes, since the test result shows there is significan difference of depression in women by amount of coffee.