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.

46% of 1,012 agree margin of error: 3%

  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.

The 95% confidence interval: (0.43 0.49)

FALSE: We are 95% confident that the true proportion (not in the sample) of Americans who support the decision fall within this interval of 43 to 49 percent.

me <- 0.03
p <- .46
c(p - me, p + me)
## [1] 0.43 0.49
  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: We are 95% confident that the true proportion (not in the sample) of Americans who support the decision fall within this interval of 43 to 49 percent.

  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: So there is a 5% chance the sampled proportion will fall outside this range. In other words, if we take 100 different random samples of 1,012, we expect 95 of those samples to have a proportion that is within the range.

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

ME is z_critical * SE. Since 90% covers less of the distribution than 95%, then we expect the z_critical to be smaller. So the margin of error would be smaller.

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.

n = 1,259 p = .48 –> make legal

  1. Is 48% a sample statistic or a population parameter? Explain.

48% is a sample statistics and not a population parameter. This proportion was derived based on a sample of 1,259 US residents.

  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.

The 95% confidence interval is (0.45 0.51). We are 95% confident that true proportion of US residents who think marijuana should be made legal falls within 45% to 51%.

p <- 0.48
n <- 1259
se <- sqrt(p*(1-p)/n)
me <- 1.96 * se
c(p - me, p + me)
## [1] 0.4524028 0.5075972
  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.

Independence: We are going to assume that the 1,259 were randomly selected, and obviously the sample represents less than 10% of the US population.

Success-failure condition: pn and (1-p)n are both greater than 10. So the sample size is sufficient.

So it seems reasonable that these conditions are met for this data.

p <- 0.48
n <- 1259
p * n
## [1] 604.32
(1-p) * n
## [1] 654.68
  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?

The confidence interval at 95% is (0.45 0.51). So, 51% is included in this interval; however, the true proportion could also be a value that is less than 50%. For this news piece to be justified the starting range for the confidence interval should be 51 or higher. So, I think that this news piece is not justified.

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 ?

me <= .02 1.96 * se <= .02 1.96 * sqrt(p(1-p)/n) <= .02

Based on the calculations below (solving for n), n should at least 2,398 to have a margin of error that is no larger than 2% at the 95% confidence interval.

p <- .48
n <- 1259
# 1.96 * sqrt(p(1-p)/n) <= .02
#n should be equal or greater than: 
((.02/1.96)^2)^(-1)*(p*(1-p))
## [1] 2397.158

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.

Oregon:

California:

The 95% confidence interval for the difference between proportions (p1-p2) is (-0.0015, 0.0175).

n1 <- 4691
p1 <- .088
n2 <-  11545
p2 <-  .080

diff <- p1-p2
diff
## [1] 0.008
se_diff <- sqrt((p1*(1-p1)/n1) + (p2*(1-p2)/n2))
se_diff
## [1] 0.004845984
me <- 1.96 * se_diff
c(round(diff - me, 4), round(diff + me, 4))
## [1] -0.0015  0.0175