6.6 2010 Healthcare Law

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, 46% of the sample support the decision. Confidence interval applies to the population.

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. True

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%. True

4. The margin of error at a 90% confidence level would be higher than 3%. False, margin of error is calculated as SE x Z, as a result, with a smaller z-score, the margin of error will be smaller

6.12 Legalization of Marijuana, pt 1

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.
   This is a sample statistic as it is based on the responses of 1,259 residents

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.
   we are 95% certain that the true population proportion is between .466 and .493

p<- .48
se <- p*.52/sqrt(1259)
low <- p - se * 1.96
high <- p + se * 1.96
(conf_int <- c(low,high))

## [1] 0.4662124 0.4937876

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.
   Since the sample is less than 10% of the population, the independent observation criteria is met. As there are more than 10 observations for p and 1-p, we can also conclude that the sampling distribution may be well approximated by the normal model

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?
   Based on the confidence interval, this statement is wrong and is not justified

p<- .48
me <- .02
round(n <- 1.96^2 * p*(1-p)/me^2)

## [1] 2397

6.28

pooled <- (.08*11545 + .088*4691)/(11545 + 4691)
se <- sqrt(pooled/11545 + pooled/4691)
margin <- 1.96 * se
lower <- .008 - margin
upper <- .008 + margin
(ci<-c(lower,upper))

## [1] -0.001736344  0.017736344

6.44 Barking Deer

1. \(H_0\): The proportion of forage sites mirrors the distribution of the different habitats
   \(H_A\): The proportion of forage sites does not mirror the distribution of the different habitats

2. We can use a chi-square test

3. It is reasonable to assume the observations are independent and each observation has more than 5 cases

4. Since the p-value is near-zero (and below the significance level), we can reject the null hypothesis

obs <- c(4,16,67,345)
ev <- round(426*c(.048,.147,.396,.409),2)
chi <- sum((obs-ev)^2/ev)
(p<- 1 - pchisq(chi, 3))

## [1] 0

vals <- c(154, 132, 180, 126, 104, 131)
responses <- c("Support", "Oppose", "Do not know")
groups <- c("Grad", "nGrad")
groups_rsp <- list(responses, groups)
offshore_drilling <- matrix(vals, 3, 2, byrow = TRUE, dimnames = groups_rsp)

chisq.test(offshore_drilling)

## 
##  Pearson's Chi-squared test
## 
## data:  offshore_drilling
## X-squared = 11.461, df = 2, p-value = 0.003246