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.

  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, confidence interval shows us the proportion of the population is within that range based on the spot estimate and margin of error of the sample.

  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, confidence interval tells us about the population not just the sample.

  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, 95% of the sample’s confidence intervals would capture the true proportion of the population.

  4. The margin of error at a 90% confidence level would be higher than 3%. Answer: False, the critical value for a 90% confidence interval is smaller than that for a 95% confidence interval. 90% confidence interval would have less chance of capturing the true proportion.

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.

  1. Is 48% a sample statistic or a population parameter? Explain. Answer: It is a sample statistics. It is a survey conducted which is basically a small sample representing the big population.

  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:

#calculate the standard error first
SError = (0.48*(1-0.48)/1258)^0.5
SError
## [1] 0.01408581
# calculate margin of error
MError = 1.96*SError
MError
## [1] 0.0276082
CInterval = vector()
CInterval[1] = 0.48-MError
CInterval[2] = 0.48+MError
CInterval
## [1] 0.4523918 0.5076082

Here, sample is random, so there is change of successes and failures. In this case it meets the criteria for a confidence interval.:

  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: It doesn’t mention how the survey was conducted. It its randomnly selected and unbiased then yes it is a valid statement but it its not a randomnly selected survey then the statement is not valid.

  2. 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: According to the numbers, it is not a good idea to legalized the Marijuana practice.

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 ?

Answer:

MError2 = 0.48*0.52/(0.02/1.96)^2
round(MError2)
## [1] 2397

About 2398 people have to be surveyed.

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. Answer:

OR = 0.08
CA = 0.088
PE = CA - OR
PE
## [1] 0.008
#to calculate standard error
SError3 = ((CA*0.912/4691)+(OR*0.92/11545))^0.5
SError3
## [1] 0.004845984
#to calcualte margin of error
MError3 = 1.96*SError3
MError3
## [1] 0.009498128
#to calculate margin of error
ConfInterval = vector()
ConfInterval[1] = PE - MError3
ConfInterval[2] = PE + MError3
ConfInterval
## [1] -0.001498128  0.017498128

Here, the confidence interval is close to 0, there is no statistically significance difference in the proportion of sleep deprivation of the two populations at the 95% confidence level.

6.44 Barking deer. 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. Woods Cultivated grassplot Deciduous forests Other Total 4 16 67 345 426

  1. Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others. Answer: Ho is there is no difference in the proportion of deer that forage in different habitats.

HA is there is a difference in the proportion of deer that forage in different habitats.

  1. What type of test can we use to answer this research question? Answer: We can use chi-squared test x^2 for the research question

  2. Check if the assumptions and conditions required for this test are satisfied. Answer:

  1. Quantitative data - TRUE

  2. One or more categories - TRUE

  3. Independent observations - TRUE as long as 461 is less than 10% total population.

  4. Adequate sample size (at least 10) - TRUE

  5. Simple random sample - its not mentioned but we will assume it as a simple random sample

  6. Data in frequency form - TRUE

  7. All observations must be used - TRUE

  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:
#to check if the hypothesis is correct or not
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

According to the p-value we can say that there is a statistically significant difference in the proportion of where the deer forage.

6.48 Coffee and Depression. 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. Caffeinated coffee consumption <= ???1 2-6 1 2-3 >= 4 cup/week cups/week cup/day cups/day cups/day Total Clinical Yes 670 373 905 564 95 2607 depression No 11545 6244 16329 11726 2288 48132 Total 12215 6617 17234 12290 2383 50739

  1. What type of test is appropriate for evaluating if there is an association between coffee intake and depression? Answer: chi-squared test will be a good test for two tables.

  2. Write the hypotheses for the test you identified in part (a). Answer: Ho: There is no difference between the amount of coffee intake between the population of women with and without depression.

HA: There is a difference between the amount of coffee intake between the population of women with and without depression.

  1. Calculate the overall proportion of women who do and do not su???er from depression. Answer:
p_dep <-  2607/50739
p_dep
## [1] 0.05138059
p_hth <- 48132/50739
p_hth
## [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:
expected <- 6617*p_dep
expected 
## [1] 339.9854
ts <- (373-expected)^2/expected
ts
## [1] 3.205914

Test statistics is 3.20

  1. The test statistic is #2 = 20.93. What is the p-value? Answer:
df = (2-1)*(5-1)
df
## [1] 4
p = 1-pchisq(20.93,4)
p
## [1] 0.0003269507

Pvalue is 0.0003269

  1. What is the conclusion of the hypothesis test? Answer: Since, p < 0.05 we reject the null hypothesis and accept that there is a difference in the amount of coffee intake by women with depression compared to women without depression.

  2. 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: Yes I agree, the study only establishes statistical significance.It might not be sure that what the numbers are showing is correct in real life. There might be other factors also which triggers them to intake more coffee.