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.

False: this sample has a 46% approval rate. We know that the US population approval rate with 95% confidence intervals is between 43% and 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: This aligns with definiton of confidence interval. The sample is less than 10% of the population.

  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%.

False: 95% will contain the population proportion.

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

False: The margin of error at a 90% confidence level, since we are lowering our confidence.

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.

48%, was derived from a sample of 1259 US residents, and NOT from the US population.

  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.
# 95% Confidence Interval -> alpha of 0.05, z = 1.96
z <- 1.96
p <- .48 #   proportion is 0.48
n <- 1259 
me <- z * sqrt(p*(1-p)/n)
ci.lower <- (p - me) * 100
ci.upper <- (p + me) * 100
ci.lower
## [1] 45.24028
ci.upper
## [1] 50.75972

The 95% confidence interval for the proportion of US residents who think marijuana should be made legal is from 45.240277 and 50.759723.

  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.

If observations are independent: < 10% of the population and sample size is sufficent this will hold true.

  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?

False. The 95% confidence interval falls between 45.24% to 50.76%. The chances are high that it will be < 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?

#margin of err =2%
mj.p <- .48
se <- .02/1.96
# Standard of Error = sqrt(p * (1-p) / n)
mj.n <- (mj.p * (1-mj.p))/(se^2)
mj.n
## [1] 2397.158

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.

ca.n <- 11545
ca.p <- 0.08
or.n <- 4691
or.p <- 0.088

z <- 1.96 # For 95% Confidence Interval
se.ca <- sqrt((ca.p)*(1-ca.p)/ca.n) 
me.ca <- z * se.ca # Margin of Error at 95% confidence interval
se.or <- sqrt((or.p)*(1-or.p)/or.n)
me.or <- z * se.or # Margin of Error at 95% confidence interval
round(me.ca * 100, 2)
## [1] 0.49
ca.lower <- ca.p - me.ca
ca.upper <- ca.p + me.ca
or.lower <- or.p - me.or
or.upper <- or.p + me.or

ca.lower
## [1] 0.07505122
ca.upper
## [1] 0.08494878
or.lower
## [1] 0.07989296
or.upper
## [1] 0.09610704
se <- sqrt((ca.p)*(1-ca.p)/ca.n + (or.p)*(1-or.p)/or.n) # Calculating a new SE for the differences
state.me <- z * se
state.me
## [1] 0.009498128
#95% CI
diff <- or.p - ca.p
diff.lower <- diff - state.me
diff.upper <- diff + state.me
diff.lower
## [1] -0.001498128
diff.upper
## [1] 0.01749813

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.

Hypothesis H0: the barking deer have no preference to certain habitats and that they have equal preference among them all. Ha: The barking deer have a preference to certain habitats

  1. What type of test can we use to answer this research question?

The Chi-square test can be used here.

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

Check conditions for inference

1.The deer are not fluencing each otehr and are independent of each other 2.Sample size and distribution: .048 * 426 = 20.448, which is greater or equal to 5

deer <- c(4, 16, 61, 345)
percnt <- c(0.048, 0.147, 0.396, 0.409)
chisq.test(x = deer, p = percnt)
## 
##  Chi-squared test for given probabilities
## 
## data:  deer
## X-squared = 284.06, df = 3, p-value < 2.2e-16

Since the p-value is less than 0.05, we reject the H0 and conclude that barking deer do perfer some habitats over others

  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.
library(visualize)
visualize.chisq(stat= 284.06, df = 3, section = "upper")

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?

Chi-squared test

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

H0: The coffee consumption and depression is not related. Ha: The coffee consumption and depression is related.

  1. Calculate the overall proportion of women who do and do not suffer from depression.
deprs <- 2607
not.deprs <- 48132
total <- deprs + not.deprs
deprs/total * 100
## [1] 5.138059
not.deprs/total * 100
## [1] 94.86194
  1. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
6617 * 2607 / 50739
## [1] 339.9854
(373 - 339.9854)^2 / 339.9854
## [1] 3.205914
  1. The test statistic is \(\chi^2=20.93\). What is the p-value?

p.value = 0.0003267

test.stat<- data.frame(Yes = c(670,373,905,564,95),
                    No =c(11545,6244,16329,11726,2288)
                    )
chisq.test(test.stat)
## 
##  Pearson's Chi-squared test
## 
## data:  test.stat
## X-squared = 20.932, df = 4, p-value = 0.0003267
library(visualize)
visualize.chisq(stat=20.932, df = 4, section = "upper")

  1. What is the conclusion of the hypothesis test?

We reject null Hypothesis(H0)

  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.

I Agree that it is too early to make this recommendation.The chisquare test only shows that there is a relationship in the study, not exactly what that relationship is. Correlation does not necessarily mean there’s causation.