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.

Solution

False

  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.

Solution

True

  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%.

Solution

False

  1. The margin of error at a 90% confidence level would be higher than 3%. Solution: FALSE. As the confidence level decreases, Margin of Error will increase.

Solution

False

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.

Solution

48% would represent the sample statistic.

  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.

Solution

n <- 1259

p <- 0.48

CIP <- 0.95 

SE <- ((p * (1 - p)) / n) ^ 0.5

t <- qt(CIP + (1 - CIP)/2, n - 1)


ME <- t * SE

CI <- c(p - ME, p + ME)

CI
## [1] 0.4523767 0.5076233

The 95% confidence interval for the proportion of US residents who think marijuana should be made legal is from 45.24% to 50.76%.

  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.

Solution

True, only if the sample obervations are independent. Futhermore, if the sample size is of a large significance whereas \(n \cdot p \ge 10\) & \(n \cdot (1 - p) \ge 10\), \(\hat{p}\) is normally distributed.

  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?

Solution

We cannot reject the hypothesis that the proportion of Americans who think marijuana should be legalized is above 50%, however we also cannot reject the hypothesis that the proportion is below 50%.

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 ?

Solution

ME <- 0.02
n <- (p * (1 - p ) * t ^ 2) / ME ^ 2
n
## [1] 2401.689

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.

Solution

pCA <- 0.08
nCA <- 11545
pOR <- 0.088
nOR <- 4691
Cip <- 0.95 

PDiff <- pOR - pCA



SE <- ( ((pCA * (1 - pCA)) / nCA) +  ((pOR * (1 - pOR)) / nOR)) ^ 0.5
z <- qnorm(Cip + (1 - Cip) / 2)

me <-z * SE


ci <- c(PDiff - me, PDiff + me )

ci
## [1] -0.001497954  0.017497954

With the 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived is from ~ -0.0015 to ~ 0.0175

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.

imgage <- "C:/Users/jpsim/Documents/Stat & Probability for Data/beer.png"
include_graphics(imgage)

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

H0: There’s no difference between each forage category. HA: There’s difference with at least one of the forage categories.

(b)What type of test can we use to answer this research question?

#####Solution

A Chi-Square test

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

Solution

The two conditions that must be checked before performing a chi-square test are independence and distribution.

  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.

Solution

habitats <- c(4, 16, 67, 345)
region <- c(20.45, 62.62, 168.70, 174.23)
k <- length(habitats)
df <- k - 1

chi <- (habitats - region ) ^ 2 / region
chi <- sum(chi)


p_Val <- 1 - pchisq(chi, df = df)

p_Val
## [1] 0

The chi-Square value is large enough that the p-value is 0. Hence, we conclude that there is convincing evidence the barking deer forage in certain habitats 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.

imgage <- "C:/Users/jpsim/Documents/Stat & Probability for Data/coffee.png"
include_graphics(imgage)

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

Solution

The Chi-squared test for two-way tables is appropriate for evaluating if there is an association between coffee intake and depression.

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

Solution

\(H_0:\) There is no association between caffeinated coffee consumption and depression.

\(H_A:\) There is an association between caffeinated coffee consumption and depression.

  1. Calculate the overall proportion of women who do and do not suffer from depression.

Solution

yes_dep <- 2607
no_dep <- 48132
total_dep <- yes_dep + no_dep
total_dep
## [1] 50739

Women who do suffer from depression is 5.14%, women who do not suffer from depression is 94.86%

  1. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic.

Solution

twocup <- 6617
ek <- (yes_dep * twocup) / total_dep
chipart <- ((373 - ek) ^ 2) / ek
  1. The test statistic is \(\chi ^ 2 = 20.93\). What is the p-value?

Solution

n <- 5
k <- 2

df <- (n-1)*(k-1)
chi2 <- 20.93

p_value <- 1 - pchisq(chi2, df)
p_value
## [1] 0.0003269507
  1. What is the conclusion of the hypothesis test?

Solution

Since the p-value is below 0.05, we cannot reject the null hypothesis that coffee doesn’t cause 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.

Solution:

According to the chi-square test, we found no link between coffee consumption and depression. Therefore, I would have to agree.