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.
- Is 48% a sample statistic or a population parameter? Explain.
It’s a sample statistic. We can use inference to try to estimate population parameters.
- 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.
mj.phat <- .48 # point estimate
mj.n <- 1259
mj.z <- 1.96 # z-score based on 95% confidence interval
mj.se <- sqrt((mj.phat * (1 - mj.phat)) / mj.n) # meets conditions for nearly normal sampling distribution: 1) assume sample observations are independent; 2) with 1,259 observations and 48% successes, there are at least 10 successes and failures
mj.me <- mj.z * mj.se
mj.up <- mj.phat + mj.me # upper bound of confidence interval
mj.lo <- mj.phat - mj.me # lower bound of confidence interval
mj.ci <- c(mj.lo, mj.up)
round(mj.ci, 3)
## [1] 0.452 0.508
- 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.
In all likelihood, no. The sample is large, less than 10% of the population, and random - so we can assume indepedent observations. The success-failure condition is also met.
- 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’s not justified. As the confidence interval extends from .452 to .508, there’s a chance that more that the true proportion exceeds 50%, but there’s also a chance that it is lower.
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?
mj.me2 <- .02
mj.n2 <- (mj.z / mj.me2)^2 * (mj.phat * (1 - mj.phat))
round(mj.n2)
## [1] 2397
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 insufficient 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.
CA.phat <- .08
OR.phat <- .088
diff.phat <- abs(CA.phat - OR.phat)
CA.n <- 11545
OR.n <- 4691
slp.se <- sqrt(((CA.phat * (1 - CA.phat) / CA.n) + ((OR.phat * (1 - OR.phat)) / OR.n)))
slp.z <- 1.96
slp.me <- slp.z * slp.se
slp.lo <- diff.phat - slp.me
slp.up <- diff.phat + slp.me
slp.ci <- c(slp.lo, slp.up)
round(slp.ci, 5)
## [1] -0.0015 0.0175
The 95% confidence interval for the difference between CA and OR populations overlaps 0, so we cannot conclude that the populations are different from another.
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.
Woods Cultivated grassplot Deciduous forests Other Total 4 16 61 345 426
- Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.
H0: barking deer don’t exhibit preferences for certain habitats over others for foraging.
HA: barking deer do exhibit those preferences.
- What type of test can we use to answer this research question?
We can use a chi-squared test.
- Check if the assumptions and conditions required for this test are satisfied.
We can safely assume that each case contributes a count to the table that is independent of other cases. However, each cell count should containt at least 5 expected cases to meet the sample size / distribution case, which is not meat for woods habitats.
- 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.
deer.obs <- c(4, 16, 61, 345) # Observed distribution of foraging habitats
deer.obs.total <- 426
deer.exp <- c(.048, .147, .396, (1 - .048 - .147 - .396)) * deer.obs.total # Expected distribution of foraging habitats based on land distribution
deer.k <- length(deer.obs) # Number of habitat groups
deer.df <- deer.k - 1 # Degrees of freedom for chi-square test
# Loop to arrive at a test statistic.
deer.chi <- 0
for (i in 1:deer.k) {
deer.chi <- deer.chi + ((deer.obs[i] - deer.exp[i])^2 / deer.exp[i])
}
# Calculcate the p-value using pchisq function and computed test statistics.
deer.pval <- pchisq(deer.chi, df = deer.df, lower.tail = F)
deer.pval
## [1] 2.799724e-61
As the p-value is very small, we reject the null hypothesis anbd conclude that barking deer do prefer to forage in certain habitats over others.
6.48 Coffee and Depression. 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.
Caffeinated coffee consumption ???1 2-6 1 2-3 $ 4 cup/week cups/week cup/day cups/day cups/day Total Clinical Yes 670 373 905 564 95 2,607 depression No 11,545 6,244 16,329 11,726 2,288 48,132 Total 12,215 6,617 17,234 12,290 2,383 50,739
- What type of test is appropriate for evaluating if there is an association between coffee intake and depression?
** A two-table chi-squared test would be appropriate because … ANSWER**
- Write the hypotheses for the test you identified in part (a).
H0: The proportion of women who are depressed does not vary based on coffee consumption.
HA: The proportion of women who are depressed varies based on coffee consumption.
- Calculate the overall proportion of women who do and do not suffer from depression.
ydprs <- c(670, 373, 905, 564, 95)
ydprs.total <- sum(ydprs)
ndprs <- c(11545, 6244, 16329, 11726, 2288)
ndprs.total <- sum(ndprs)
cfe <- ydprs + ndprs
cfe.total <- ydprs.total + ndprs.total
ydprs.prop <- ydprs.total / cfe.total
ndprs.prop <- ndprs.total / cfe.total
paste0(round((100 * ydprs.prop), 1), "% of women suffer from depression")
## [1] "5.1% of women suffer from depression"
paste0(round((100 * ndprs.prop), 1), "% of women do not suffer from depression")
## [1] "94.9% of women do not suffer from depression"
- 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.
ydprs.2_6cupwk.exp <- cfe[2] * ydprs.prop
paste0("The expected count is ", round(ydprs.2_6cupwk.exp, 2))
## [1] "The expected count is 339.99"
ydprs.2_6cupwk.contrib <- (ydprs[2] - ydprs.2_6cupwk.exp)^2 / ydprs.2_6cupwk.exp
paste0("The contribution is to the test statistic is ", round(ydprs.2_6cupwk.contrib, 2))
## [1] "The contribution is to the test statistic is 3.21"
- The test statistic is 2 = 20.93. What is the p-value?
cfe.chi <- 20.93
cfe.k <- length(cfe)
cfe.df <- cfe.k - 1
cfe.pval <- pchisq(cfe.chi, df = cfe.df, lower.tail = F)
cfe.pval
## [1] 0.0003269507
- What is the conclusion of the hypothesis test?
Given the p-value of .0003 is less than .05, we reject the null hypothesis and conclude there is a relationship between caffeinated coffee consumption and depression.
- 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.
Yes, I agree - statistical significance is not the same as clinical statistical significance. While this study identifies a correlatioin, it does not imply causation.