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

Answers:
a. False, confidence level is calculated for the population, not for the sample.
b. True, as it’s given that 46% of the Americans sampled support the decision, and there is a 3% margin of error.
c. False, we cannot make claims about other random samples based on our confidence interval.
d. False, margin of error is a lower percentage as confidence level goes down (with the same standard error).

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.
  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.
  3. 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.
  4. 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?

Answers:
a. 48% is a sample statistic because it is a proportion of the sample.
b.

std_err <- sqrt((0.48*(1-0.48))/1259)
t = qt(0.975,1258)
margin_of_err <- std_err*t
0.48 + margin_of_err # upper tail
## [1] 0.5076233
0.48 - margin_of_err # lower tail
## [1] 0.4523767
  1. These data have a sufficient sample size and the observations are independent so the statistic most likely does follow the normal distribution closely enough.
  2. False, we cannot conclusively say the majority agree, even if the confidence interval’s uppertail goes above 50%. The hypothesis that the statistic is below 50% cannot be rejected.

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:

margin_of_err <- 0.02
z <- 1.96 # Z-score for 95% confidence
std_err <- margin_of_err/z
n <- (.48*(1-.48))/std_err^2
n
## [1] 2397.158

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_ore <- 0.088
n_ore <- 4691
p_cal <- 0.08
n_cal <- 11545
std_err <- sqrt((p_cal*(1-p_cal)/n_cal)+(p_ore*(1-p_ore)/n_ore))
z <- 1.96 # Z-score for 95% confidence
margin_of_err = z*std_err
-0.008 + margin_of_err
## [1] 0.001498128
-0.008 - margin_of_err
## [1] -0.01749813

0 is within the confidence interval so we fail to reject the null hypothesis – the difference in sleep deprivation is not significant.

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.
  2. What type of test can we use to answer this research question?
  3. Check if the assumptions and conditions required for this test are satisfied.
  4. 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.

Answers:
a.
H0: Deer sites are proportionately distributed over the habitats.
p1: 0.048
p2: 0.147
p3: 0.396
p4: 0.409

H1: Deer sites are not proportionately distributed over the habitats.
b. Chi-square goodness of fit test
c.
The data is obtained from a random sample.
The data is categorized.
The expected frequency of each category must be at least 5.
d.

woods <- 426*.048
grass <- 426*.147
forest <- 426*.396
other <- 426*.409
proportions <- c(woods,grass,forest,other)

habitat = c(4,16,67,345)
k = 4
dF = 3
chi = 0
for(x in 1:4){
  chi = chi + ((habitat[x]-proportions[x])^2/proportions[x])
}
p = pchisq(chi,dF,lower.tail = FALSE)
p
## [1] 1.144396e-59

We reject the null hypothesis in favor of the alternative hypothesis.

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?
  2. Write the hypotheses for the test you identified in part (a).
  3. Calculate the overall proportion of women who do and do not suffer from depression.
  4. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
  5. The test statistic is \(\chi^2=20.93\). What is the p-value?
  6. What is the conclusion of the hypothesis test?
  7. 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.

Answers:
a.
Chi-square test.
b.
H0: There is no association between caffeinated coffee consumption and depression.
H1: There is an association between caffeinated coffee consumption and depression.

  1. The overall proportion of women who do suffer from depression is 5.13%. The overall proportion of women who do not suffer from depression is 94.86%.
k <- 5
df <- k - 1
dep <- 2607/50739
highlighted_cell <- dep * 6617

expected <- highlighted_cell
contribution <- (373 - expected)^2 / expected
contribution
## [1] 3.205914

The highlighted cell contributes 3.2.

p_value <- pchisq(20.93, df=df,lower.tail=FALSE)
p_value
## [1] 0.0003269507
  1. Reject the null hypothesis due to the p-value below 0.05, conclude that there is an association between caffeinated coffee and depression.
  2. I agree with this statement because a statistical association does not necessary indicate a cure for depression – it better indicates that perhaps more studies on this topic could yield interesting results.