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; we are 100% sure that 46% of americans IN THE SAMPLE support the law. We know how to count.

  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; this pertains to the population, and our CI and ME are applicable

  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; this is the sampling distribution.

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

False; it would be lower, as the ME is proportional to the Z Score (which is lower at 90% than 95%).

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.

It is a sample statistic, as it corresponds to a sample of US residents (us residents being the population).

  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.
p = 0.48
n = 1259
ME = 1.96 * sqrt(p*(1-p)/n)
CI = c(-ME,ME)+p
CI
## [1] 0.4524028 0.5075972

There is a 95% chance that proportion of US residents who think marijuana should be made legal is between 45% and 51%.

  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.

Yes; although this is categorical data, the range of possible proportions should follow a normal distribution centered around the mean population proportion.

  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?

No, it is not justified. For it to be justified (with 95% confidence), the CI would have to exclude any values below 50% for it to be considered a ‘majority’. It’s possible they were reffering to a ‘plurality’, but that would depend on how many people had no opinion and didn’t think it should be legalized.

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?

ME = 0.02
z = 1.96
p = 0.48

n = (p*(1-p)*z^2)/(ME^2)
ceiling(n)
## [1] 2398

You would need to survey at least 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.

pcal = 0.080
pore = 0.088
ncal = 11545
nore = 4691
z = 1.96

SE = sqrt((pore*(1-pore)/nore)+(pcal*(1-pcal)/ncal))
ME = z * SE
CI = pore-pcal+c(-ME,ME)
CI
## [1] -0.001498128  0.017498128

There is a 95% probability that the difference between the proportions of Californians and Oregonians who are sleep deprived is between -0.15% and 1.75%.

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.

Null Hypothesis: The proportion of forage sites to the proportion of types of land is equivalant; the deer do not perfer foraging on any specific type of land. Alternate Hyptohesis: The proportion of forage sites is different from the proportion of the types of land; the deer do have a preference for some types of land over others.

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

This would be a good place to use a Chi^2 test.

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

The first condition -each case is independent- is met. Can’t be two types of land at the same time.
The second condition -that each case needs an EXPECTED value of at least 5- is met. We expect at least 20 of the woods regions to be foraged, even though the actual is only 4.

  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.
#first need to calculate the number of expected foraging spots, and make another vector of observed foraging spots
expected <- 426 * c(0.048, 0.147, 0.396, 1-(0.048+0.147+0.396))
observed <- c(4, 16, 61, 345)

#Make a table of the two

foraging <- as.table(setNames(observed,expected))
foraging
##  20.448  62.622 168.696 174.234 
##       4      16      61     345
#Using stat's chisq.test function to test hypothesis
library(stats)
chisq.test(foraging)
## 
##  Chi-squared test for given probabilities
## 
## data:  foraging
## X-squared = 729.1, df = 3, p-value < 2.2e-16

The p value here is very small, so we can reject the null hypothesis and instead say that deer do perfer to forage in certain habitats (in this case, ‘other’ sites not mentione) 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?

A Chi^2 test would be appropriate here.

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

Null Hypothesis: Those with clinical depression drink the same amount of coffee as those without; there isnt a relationship. Alternate Hypothesis: Those with clinical depression drink a different amount of coffee from those those without; there is a relationship.

  1. Calculate the overall proportion of women who do and do not suffer from depression.
nTotal <- 50739
nDepressed <- 2607
nNotDepressed <- 48132 
pDepressed <- nDepressed/nTotal
pNotDepressed <- nNotDepressed/nTotal

pNotDepressed
## [1] 0.9486194
pDepressed
## [1] 0.05138059

94.9% are not depressed, and 5.1% are depressed.

  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 count
6617 * pDepressed
## [1] 339.9854

We a count of about 340 for that cell.

#Contribution to the test statistic
(373-340)^2/340
## [1] 3.202941

This cell contributes 3.2 to the test statistic.

  1. The test statistic is \(\chi^2=20.93\). What is the p-value?

With 4 degrees of freedom (5 - 1 bins) the p-value is 0.000327 .

  1. What is the conclusion of the hypothesis test?

As the p-value is small, the observed values are bery likely to be different than the expected ones. Thus, we can reject the null hypothesis and say that coffee consumption does have a relationship with 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. Do you agree with this statement? Explain your reasoning.

I do agree with this statement. While the test does suggest there is a relationship, it says nothing about whether or not changing one’s coffee consumption will keep one from becoming depressed. It could well be that depression excludes coffee consumption, or for any number of other reasons.