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. CI are application to population parameter. The sample proportion are within the CI interval.
  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, there CI is applicable to the americans that support decision of the US on the 2010 healthcare law and it provides the interval of findng this population parameter.
  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 - Sample proportion will be within the confidence interval.
  4. The margin of error at a 90% confidence level would be higher than 3%. False - As the confidence level increases, the margin of error increases in value. So 90% confidence level will have margin of error lesser (z * SE)

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. Since this is a survey, this is a point estimate and not a population parameter.

  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% confident that the population parameter or proportion of us residents that think Marijuana should be legal is between 0.45 and 0.50

phat <- .48
n <- 1259
se <- sqrt((phat* (1-phat)/n))
me <- 1.96 * se
me
## [1] 0.02759723
clower <- .48 - me
cupper <- .48 + me

clower
## [1] 0.4524028
cupper
## [1] 0.5075972
1-.48
## [1] 0.52
  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.

For normality check we check 1) That the observations are independent, since this is a survey, if we assume random survey then we have independent observations. 2) we also need to check np and n (1-p) is atleast 10, We have 1259 * .48 > 10 and we have 1259 * 0.52 > 10 so both have satisfied so we can apply central limit theorem and normal distribution formula.

  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? Our confidence interval is between .45 and .51, whch indicates that we are more likely to find our population parameter 95% of the time within this interval. Majority would imply more than half of Americans think it should be legalized which doesn’t fit the confidence interval.

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?

We need to survey 2398 (rounded) Americans

phat <- .48
n <- 1259
se <- sqrt((phat* (1-phat)/n))
me <- 1.96 * se

# 0.02 = 1.96  * sqrt((.48 * (1-.48))/n)

x <- (.02/1.96)^2
n <- (.48 * (1-.48))/x

SE <- sqrt(((0.08*0.92)/11545)+((0.088*0.0912)/4691))

margin <- 1.96 * SE

ci1 <- 0.008 - margin
ci2 <- 0.008 + margin

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.

The confidence interval is lower than 0 and it is (-0.002426598,-0.0135734)

SEcomb <- sqrt(((0.08*0.92)/11545)+((0.088*0.0912)/4691))
SEcomb
## [1] 0.002843573
me <- 1.96 * SEcomb
pdiff <- .08 - .088

pdiff 
## [1] -0.008
ci1 <- pdiff - me
ci2 <- pdiff + me

ci1
## [1] -0.0135734
ci2
## [1] -0.002426598

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. h0 - barking deer does not prefer certain habitats h1 - barking deer prefer some habitats over others

  2. What type of test can we use to answer this research question? CHI-SQUARE TEST FOR ONE-WAY TABLE

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

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.

  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.

We reject the hypothesis because looking at the p-value is less than 0.05 we reject the null hypothesis and accept alternative that barking deer do prefer some more than others

n <- 426
probWoods <- 0.048 * 426
probgrassplot <- 0.147 * 426 
probDeciduous <- 0.396 * 426
probOther <-  (1 - (0.048+0.147+0.396))* 426

probWoods
## [1] 20.448
probgrassplot
## [1] 62.622
probDeciduous 
## [1] 168.696
probOther
## [1] 174.234
chisq.test(x = c(4, 16, 67, 345), p = c(0.048, 0.147, 0.396, 0.409))
## 
##  Chi-squared test for given probabilities
## 
## data:  c(4, 16, 67, 345)
## X-squared = 272.69, df = 3, p-value < 2.2e-16

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-SQUARE TEST FOR ONE-WAY TABLE - we can use since we need to determine association Suppose we are to evaluate whether there is convincing evidence that a set of observed counts O1, O2, …, Ok in k categories are unusually different from what might be expected under a null hypothesis.

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

H0 : There is no association between coffee intake and depression H1 : There is association between coffee intake and depression

  1. Calculate the overall proportion of women who do and do not suffer from depression.
2607 / 50739 # women who suffer depression
## [1] 0.05138059
48132 / 50739
## [1] 0.9486194
  1. Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
expected_ct <- (2607/50739) * 6617
expected_ct
## [1] 339.9854
(373 - expected_ct)^ 2 / expected_ct
## [1] 3.205914
#COMPUTING EXPECTED COUNTS IN A TWO-WAY TABLE
#To identify the expected count for the i
#th row and j
#th column, compute
#Expected Countrow i, col j =
#(row i total) × (column j total)
#table total
  1. The test statistic is \(\chi^2=20.93\). What is the p-value? For Chi test we need right tail probability
chi <- 20.93
#df = (number of rows minus 1) × (number of columns minus 1)
df <- (2 - 1) * (5 - 1)
pchisq(chi, df,lower.tail=FALSE)
## [1] 0.0003269507
  1. What is the conclusion of the hypothesis test?

Since p value is lower than 0.05 we reject the null hypothesis and take up the alternate is that there is association between coffee intake and depression in women.

  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.

The chi square test on two way table establishes if there is dependence or association so there is an association here and the author making the statement based on this study is not unreasonable, however, radomized experiments will need to be created to prove if there is depression as a result of coffee intake.