6.6, 6.12, 6.20, 6.28, 6.44, 6.48
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.
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 - Confidence intervals reflect population parameters, not sample parameters
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 - Definition of confidence interval
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 - Definition of confidence interval
The margin of error at a 90% confidence level would be higher than 3%.
False - At a lower confidence level margin of error will be lower than 3%
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.
Is 48% a sample statistic or a population parameter? Explain.
It’s a sample statistic because it’s based on the survey results
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.
n <- 1259
p <- 0.48
z <- 1.96
me <- z * sqrt(p * (1-p)/n)
p+me## [1] 0.5075972p-me## [1] 0.4524028#We are 95% confident that the porpotion of US residents who think marijuana should be made legal is between 0.4524028 and 0.5075972.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.
The sample size is 1259, which is less than 10% of the population and observations can be assumed to be independent. Since pn = 0.48 ∗ 1259 = 604.32 and (1−p)n = 0.52 ∗ 1259 = 654.68, which are both greater than 10, the normal model is a good approximation.
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?
Based on my confidence interval this statment is not justified. Although my confidence included value over 0.5, the majority of the confidence interval is less than 0.5.
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 ?
```r
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)
n
```
```
## [1] 2397.158
```
```r
#We would need to survey 2398 Americans in order to limit the margin of error of a 95% confidence interval 2%.
```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.
```r
p.ca <- 0.08
p.or <- 0.088
n.ca <- 11545
n.or <- 4691
p <- p.ca - p.or
se.ca <- (p.ca * (1-p.ca)) / n.ca
se.or <- (p.or * (1-p.or)) / n.or
se <- sqrt(se.ca + se.or)
me <- 1.96 * se
p-me
```
```
## [1] -0.01749813
```
```r
p+me
```
```
## [1] 0.001498128
```
```r
#We are 95% confident that the difference between the proportions of Californians and Oregonians who are sleep deprived is between -0.01749813 and 0.001498128. Since 0 is included in the interval, there is no difference between the proportions.
```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.
Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.
Null hypothesis: Barking deer do not prefer to forage in certain habitats over others
Alternative hypothesis: Barking deer do prefer to forage in certain habitats over others
What type of test can we use to answer this research question?
We can use a chi-square test
Check if the assumptions and conditions required for this test are satisfied.
Independence: We will have to assume it since is not given in the description.
Sample size/distribution: In our expected cases scenario, all habitats have at least 5 expected cases
Do these data provide convincing evidence that barking deer prefer to forage in certain habitats over others? Conduct an appro- priate hypothesis test to answer this research question.
habitats <- c(4, 16, 67, 345)
region <- c(20.45, 62.62, 168.70, 174.23)
k <- length(habitats)
df <- k - 1
# Compute the chi2 test statistic
chi <- (habitats - region ) ^ 2 / region
chi <- sum(chi)
# check the chi2 test statistic and find p-val
p <- 1 - pchisq(chi, df = df)
p## [1] 0#Since the p-value is less than 0.05, we reject the null hypothesis in favor of the alternative and conclude that barking deer do prefer to forage in certain habitats over others6.48 Coffee and Depression. Researchers conducted a study investigating the relationship between caffinated coffe 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 caffinated 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.
What type of test is appropriate for evaluating if there is an association between coffee intake and depression?
Chi-square test for the two-way table
Write the hypotheses for the test you identified in part (a).
Null hypothesis: There is no association between caffeinated coffee consumption and depression.
Alternative: There is an association between caffeinated coffee consumption and depression.
Calculate the overall proportion of women who do and do not suffer from depression.
#Proportion with depression
2607/50739## [1] 0.05138059#Proportion without depression
48132/50739## [1] 0.9486194Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (Observed - Expected)2/Expected.
n <- 6617
E <- 2607 * n / 50739
((373 - E)^2)/E## [1] 3.205914The test statistic is 2 = 20.93. What is the p-value?
n <- 5
k <- 2
df <- (n-1)*(k-1)
chi2 <- 20.93
p <- 1 - pchisq(chi2, df)
p## [1] 0.0003269507What is the conclusion of the hypothesis test?
We reject the null in favor of the alternative and conclude that there is an association between caffeinated coffee consumption and depression.
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.
I agree with this statement because the study was observational so it cannot be used to determine causal relationships.