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.
    Answer: False. A confidence interval is constructed to estimate the population proportion, not the sample proportion
  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.
    Answer: True. This statement is true based on the definition of confidence interval 46-3, 46+3 (43, 49).
  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%.
    Answer: False. A 95% confidence level indicates that 95% of the population will be in the range of the confidence interval
  4. The margin of error at a 90% confidence level would be higher than 3%. Answer: False. At a 90% confidence level, the margin of error will be less than 3% and confidence interval will be more narrow since we need to be less confident that the population probability is within the interval.

Legalization of marijuana, Part I. (6.10, p. 216) The 2010 General Social Survey asked 1,259 US res- idents: “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.
    Answer: 48% is a sample statistic that estimates the population parameter since it was derived from the sample data.
  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.
    Answer:
n = 1259
P = 0.48
SE = sqrt((P*(1-P))/n)
t = qt(0.975,n-1)
moe = t*SE
upperTail =P + moe
lowerTail = P - moe
c(lowerTail,upperTail)
## [1] 0.4523767 0.5076233

At 95% confidence level, the confidence interval is (0.4524028,0.5075972). We are 95% confident that the proportion of US residents who think marijuana should be made legal is between 45.24% and 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.
    Answer: We know from text book that the sampling distribution is nearly normal when the observations are independent, which they are in this case. Another condition is that we expect to see at least 10 success and 10 failures. This is also true in this case and we check below.
ratio <- n*P
ratio
## [1] 604.32
ratio.1 <- n*(1-P)
ratio.1
## [1] 654.68

Both are greater than 10 so we know that the success-failure condition is met.

  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?
    Answer: Our confidence interval is (.45, .51). This includes values greater than .50 (50%), so we can say that the statement above is valid, although we would be more confident if our interval spanned values greater than .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 ?
Answer:

p <- 0.48
ME <- 0.02

# ME = 1.96 * SE
SE <- ME / 1.96

# SE = sqrt((p * (1-p)) / n) 
# SE^2 = (p * (1-p)) / n
( n <- (p * (1-p)) / (SE^2) )
## [1] 2397.158

We would be about 2398 American to participate in the survey to reduce margin of error to 2%

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.

Answer: Ho : No difference in sleep deprivation
HA : Difference in sleep deprivation

P_Ore = .088
P_Cali = .08
n_Cali = 11545
n_Ore  = 4691
SE = sqrt((P_Cali*(1-P_Cali)/n_Cali)+(P_Ore*(1-P_Ore)/n_Ore))
ME = qnorm(0.975)*SE
upperTail = -0.008 + ME
lowerTail = -0.008 - ME
c(lowerTail,upperTail)
## [1] -0.017497954  0.001497954

The confidence interval contains 0, we fail to reject Ho. (null hypothesis). Sleep deprivation is not significantly different.

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.
    Answer:
    Ho: Barking deer has no preference of certain habitats for foraging.
    HA: Barking deer prefers some habitats over others for foraging.

  2. What type of test can we use to answer this research question?
    Answer: A chi-square test may be uesd to answer this question.
  3. Check if the assumptions and conditions required for this test are satisfied.
    Answer: There are two conditions that must be checked before performing a chi-square test: Independence: Each case that contributes a count to the table must be independent of all the other cases in the table. Sample size / distribution: Each particular scenario (i.e. cell count) must have at least 5 expected cases. Both conditions are satisfied here.

  4. 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.

observed <- c(4, 16, 61, 345, 426)
expected_prop <- c(0.048, 0.147, 0.396, 1-0.048-0.147-0.396, 1)
expected <- expected_prop * 426
deer <- rbind(observed, expected)
colnames(deer) <- c("woods", "grassplot", "forests", "other", "total")
deer
##           woods grassplot forests   other total
## observed  4.000    16.000  61.000 345.000   426
## expected 20.448    62.622 168.696 174.234   426
k <- 4
df <- k-1
chi2 <- sum(((deer[1,] - deer[2,])^2)/deer[2,])
( p_value <- 1 - pchisq(chi2, df) )
## [1] 0

our chi-square pvalue is 0 so we can conclude that the data does not support that barking deer forage in certain areas 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?
    Answer: We would use a chi-square test to evaluate if there is independence between depression and coffee intake.
  2. Write the hypotheses for the test you identified in part (a).
    Ho: There is no association between coffee intake and depression HA: There is an association between coffee intake and depression
  3. Calculate the overall proportion of women who do and do not suffer from depression.
    Answer:
depression <- 2607
no_depression <- 48132
total <- depression + no_depression
prop_depression <- depression/total
prop_nodepression <- no_depression/total
prop_depression
## [1] 0.05138059
prop_nodepression
## [1] 0.9486194

5.1% have depression and 94.86% do not have depression

  1. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
    Answer:
#using the proportions from above
expected <- prop_depression*6617
expected
## [1] 339.9854
observed <- 373
statistic <- ((observed-expected)^2)/expected
statistic
## [1] 3.205914

contribution of this cell is 3.21

  1. The test statistic is \(\chi^2=20.93\). What is the p-value?
    Answer:
pvalue <- pchisq(20.93, 4)
pvalue <- 1-pvalue
pvalue
## [1] 0.0003269507

our pvalue is .0003
(f) What is the conclusion of the hypothesis test?
Answer: Because our pvalue is less than .05, we can conclude that there is a association between caffinated coffee consumption and depression ie fail to reject the null hypothesis.

  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.64 Do you agree with this statement? Explain your reasoning.
    Answer: I agree with this statement. Although it looks like those who drank more coffee suffered less from depression, we were only looking at coffee. We don’t know what other factors could have lead to that finding.