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 (a): False, the confidence interval and hypothesis test are used to derive conclusions about the larger population. Here, we already know with certainty that 46% of the sample (1,012) Americans agree with the decision of the Supreme court.

  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.

##Answer (b): True, as stated in part (a) of this question, the confidence interval applies to the larger population in question.

  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%.

##Answer (c): False, the statement will apply to the population proportion and not to the sample proportions.

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

##Answer (d): MOE is CI/2 so if the confidence interval is narrowed, MOE would also decrease.

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.

##Answer (a): 48% is a sample statistic. The General Social Survey was administered to 1,259 US residents (the sample), hence the response from 48% of this sample is the sample statistic.

  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.

##Answer (b): This is a two-tailed test (Should marijuana be made legal or not?). We look up the z-score for 0.025 in the normal probability table which is 1.96. We use the z-score to calculate the MOE and also the confidence interval which is (0.4524, 0.5076). The interpretation would be that we are 95% sure that the true proportion of US residents who favor marijuana legalization is between 45.24% and 50.76%.

MOE = 1.96 * sqrt(0.48 * 0.52/1259); MOE
## [1] 0.02759723
LBCI = 0.48 - MOE; LBCI
## [1] 0.4524028
UBCI = 0.48 + MOE; UBCI
## [1] 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.

##Answer (c): Two conditions must be satisfied for us to be able to draw an inference, the observations should be drawn from a random sample and should be independent. The second condition is that we expect the number of successes and failures to be equal to or greater than 10. n * p = 1,259 * 0.48 = 604.32 and n * (1-p) = 654.68. Hence, both successes and failures are greater than 10, so the normal model is a good approximation.

  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 (d): False. The null hypothesis here is that there is no difference in the proportion of Americans who want marijuana legalized and those who don’t (Ho: Americans legalize - Americans don’t legalize = 0.5). Since, the CI calculated in the prior part contains the population proportion of 50%, we do not reject the null hypothesis. i.e. the proportion of Americans who favor legalization is equal to the number of Americans who don’t favor legalization= 50%. 50% is not a majority, hence the news piece’s statement is not justified.

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: Reaaranging the MOE formula and using a z-score value of 1.96 for a confidence level of 0.95, we calculate the sample size as ~2398.

n <- ((1.96 ^ 2) * (0.48 * 0.52) / (0.02^2)); 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.

##Answer: The 95% confidence interval is (-0.0175, 0.0015) which implies that the true proportion of California residents who are sleep deprived ranges between 1.75% lower to 0.15% higher compared to their Oregon counterparts.

PE <- 0.08 - 0.088; PE
## [1] -0.008
MOE <- 1.96 * (sqrt(((0.08 * 0.92) / 11545) + ((0.088 * 0.912) / 4691))); MOE
## [1] 0.009498128
LBCI <- PE - MOE; LBCI
## [1] -0.01749813
UBCI <- PE + MOE; UBCI
## [1] 0.001498128

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 (a): The null hypothesis is that barking deer don’t have a preference for any particular habotat. The alternative hypothesis is that barking deer prefer one or some habitats over other(s).

##In mathematical terms: Ho: p1 = p2 = p3 = p4, where p1 = 0.048 (Woods), p2 = 0.147 (Cultivated grassplot), p3 = 0.396 (Deciduous forests), p4 = 0.409 (Other, calculated as 1 - sum of other probabilities).

##Similarly, the alternative hypothesis is: Ha: Atleast one of p1, p2, p3, p4 differs from the others.

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

##Answer (b): Chi-square goodness of fit would be most appropriate.

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

##Answer (c): There are two conditions for the chi-square test - 1). Eah case that contributes a count to the table must be independent of all the other cases in the table, which is satisfied here assuming plots of land are independent and do not overlap. The second condition is that each particular scenario (i.e. cell count) must have at least 5 expected cases. This is satisfied as well.

  1. Do these data provide convincing evidence that barking deer prefer to forage in certain habitats over others? Conduct an appropriate hypothesis test to answer this research question.

##Answer (d): We need to calculate the Chi-square test statistic which is the sum of chi-square subtotals. First, we need to calculate expected values and the degrees of freedom which is k - 1 = 3. Looking up in the chi-square probability table, for 3 degrees of freedom, we see that the p-value < 0.001. Since the p-value is less than 0.05, we reject the null hypothesis which means these data provide convincing evidence that barking deer prefer to forage in certain habitats over others.

Ewoods <- 426 * 0.048; Ewoods
## [1] 20.448
ECGP <- 426 * 0.147; ECGP
## [1] 62.622
EDF <- 426 * 0.396; EDF
## [1] 168.696
EOther <- 426 * 0.409; EOther
## [1] 174.234
ChisquareW <-(((4 - 20.448) ^ 2) / 20.448);ChisquareW
## [1] 13.23047
ChisquareECGP <- (((16 - 62.622) ^ 2) / 62.622);ChisquareECGP
## [1] 34.71002
ChisquareEDF <- (((61 - 168.696) ^ 2) / 168.696);ChisquareEDF
## [1] 68.75343
ChisquareEOther <- (((345 - 174.234) ^ 2) / 174.234);ChisquareEOther
## [1] 167.367
Chisquare <- ChisquareW + ChisquareECGP + ChisquareEDF + ChisquareEOther; Chisquare   
## [1] 284.0609

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 (a): This is a two-way table that describes counts for combinations of outcomes for two variables. Here, the two variables are caffeinated coffee consumption and clinical depression. We use a chi-square test for independence to determine association between coffee intake and depression.

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

##Answer (b): The null (or boring) hypothesis is that there is no association between the two variables - coffee intake and depression. The alternative (interesting) hypothesis is that there is an association between the two variables.

##Ho: Variables are independent ##Ha: Variables are dependent

  1. Calculate the overall proportion of women who do and do not suffer from depression.

##Answer (c): The proportion of women who suffer from depression is 5.138% and the proportion of women who do not suffer from depression is 94.862%.

Depyes <- 2607 / 50739; Depyes
## [1] 0.05138059
Depno <- 48132 / 50739; Depno
## [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. (\(Observed - Expected)^2 / Expected\)).

##Answer (d): The expected count can be calculated as (Row 1 Total / Table Total) * (Column 1 Total). Contribution is 3.205914.

EV <- (2607 / 50739) * 6617; EV
## [1] 339.9854
Chisquare373 <- ((373 - 339.9854) ^ 2) / 339.9854; Chisquare373
## [1] 3.205914
  1. The test statistic is \(\chi^2=20.93\). What is the p-value?

##Answer (e): First we need to calculate the degrees of freedom which in this case is (rows - 1) * (columns - 1) = 4. Looking up the p-value in the Chi-square probability table, we find it to be < 0.001.

  1. What is the conclusion of the hypothesis test?

##Answer (f): Since the p-value calculated is < 0.05, we reject the null hypothesis that there is no association between coffee intake and depression.

  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.

##Answer (g): This is an observational study and an experimental study would need to be conducted before making such an inference (to load up on extra cups of coffee) as perhaps there is a confounding variable impacting the association.