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, confidence interval (CI) is for the population as a whole and not 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.

TRUE, CI is for the entire 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%.

TRUE

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

FALSE, at 90% the margin of error would 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 a sample statistic as it describes the samples 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.

Answer

n <- 1259
p <- 0.48
SE <- sqrt((p * (1-p))/n)
ME <- 1.96 * SE
ME
## [1] 0.02759723
l <- p-ME
h <-p+ME
l
## [1] 0.4524028
h
## [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. We see both pxn and (1-p)xn are both higher than 10 so normal model is a good approximation.
p*n
## [1] 604.32
(1-p)*n
## [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?

THE CI is in between 45% ad 51%, which is 50% of the population. Hence this news piece 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?

Answer

p <- 0.48
ME <- 0.02
SE <- ME / 1.96
n <- (p * (1-p)) / (SE^2) 
paste("95% CI with 2% error margin requires ",n,"Americans in the survey.")
## [1] "95% CI with 2% error margin requires  2397.1584 Americans in the survey."

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.

Answer

CI is (-0.01749813, 0.001498128)

p_ca <- 0.08
p_or <- 0.088

p <- p_ca - p_or

n_ca <- 11545
n_or <- 4691

SE_ca <- (p_ca * (1-p_ca)) / n_ca
SE_or <- (p_or * (1-p_or)) / n_or

SEprop <- sqrt(SE_ca + SE_or)
ME <- 1.96 * SEprop

p-ME
## [1] -0.01749813
p+ME
## [1] 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.

(\(H0\)): Barking deer has no preference of habitats for foraging. (\(HA\)): Barking deer prefer foraging in a specific type of habitat.

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

We can use chi-square goodness of fit test to this hypothesis

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

The behavior of the barking deer are likely independent and each expected value is above 5.

observed <- c(4, 16, 61, 345, 426)
expected_prop <- c(0.048, 0.147, 0.396, 1-0.048-0.147-0.396, 1)
expected <- expected_prop * 426
expected
## [1]  20.448  62.622 168.696 174.234 426.000
  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.

The p value is 0 hence we reject the null hypothesis that deer have no preference.

k <- 4
df <- k-1
chisquaretest <- sum(((observed - expected)^2)/expected)
( p_value <- 1 - pchisq(chisquaretest, df) )
## [1] 0

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

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

(\(H0\)) : There is no difference in rates of depression in women based on caffeine consumption.

(\(H1\)) : There is difference in rates of depression in women based on caffeine consumption.

  1. Calculate the overall proportion of women who do and do not suffer from depression.
p_dpressed <- 2607/50739
p_normal <-48132/50739
paste("Proportion of women suffering from depression",round(p_dpressed*100,digits=2),"%")
## [1] "Proportion of women suffering from depression 5.14 %"
paste("Proportion of women not suffering from depression",round(p_normal*100,digits=2),"%")
## [1] "Proportion of women not suffering from depression 94.86 %"
  1. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
expected<-6617*.0514
observed <-373
(observed-expected)^2/expected
## [1] 3.179824
  1. The test statistic is \(\chi^2=20.93\). What is the p-value?
chisqr <-20.93
df<- (2-1)*(5-1)
1-pchisq(20.93,df)
## [1] 0.0003269507
  1. What is the conclusion of the hypothesis test?

Since the p values is very small 0.0003, we reject the null hypothesis

  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. Do you agree with this statement? Explain your reasoning.

This was an observational study and we cannot infer causation, I agree with the author.