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.
(a) 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. We are certain that 46% of this sample supports the decision.
(b) 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. We make an infrence about the population based on this sample.
(c) 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.
(d) The margin of error at a 90% confidence level would be higher than 3%.
False. For 90% the critical value decreases.
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.44
(a) Is 48% a sample statistic or a population parameter? Explain. 48% is a sample statistic. The study looks at 1259 US residents out of the entire population.
(b) 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
ci <- 0.95
z <- 1.96
SE <- sqrt((p*(1-p)/n))
p + z * SE## [1] 0.5075972p - z * SE## [1] 0.4524028(c) 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.
Yes. The observations are independent, and the sample size is over 10 success and 10 failures.
(d) 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?
No. Our CI is between 45-50%. Barely 50% of americans think marijuana should be leaglized
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 ?
\(z* \sqrt{\frac{p(1-p)}{n}}\)
\(1.96^2 * {\frac{0.48(1-0.48)}{0.02^2}}\)
me <- 0.02
1.96^2 * p*(1-p)/me^2## [1] 2397.1582397 americans
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.
n1 <- 11545
n2 <- 4691
p1 <- 0.08
p2 <- 0.088
ci <- .95
z <- 1.96
SE <- sqrt(p1 * (1-p1)/n1 + p2 * (1-p2)/n2)
pe <- p2 - p1
round(pe + z * SE,4)## [1] 0.0175round(pe - z * SE,4)## [1] -0.00156.44 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
(a) Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.
\(H_0\): Barking deer don’t perfer different habitats over others \(H_a\): Barking deer perfer different habitats over others
(b) What type of test can we use to answer this research question?
Chi-sqare test
(c) Check if the assumptions and conditions required for this test are satisfied.
(d) 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.
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-16Since the p-value is less than 0.05, we reject the \(H_0\) and conclude that barking deer do perfer some habitats over others
library(visualize)
visualize.chisq(stat= 284.06, df = 3, section = "upper")
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.
(a) What type of test is appropriate for evaluating if there is an association between coffee intake and depression?
Chi-squared test - 2 way table.
(b) Write the hypotheses for the test you identified in part (a)
\(H_0\) There is no association between caffeine consuption and risk of depression in women \(H_a\) There is an association between caffeine consuption and risk of depression in women
(c) Calculate the overall proportion of women who do and do not suffer from depression.
depression <- 2607
no_depression <- 48132
total_no_dep <- no_depression + depression
depression/total_no_dep * 100## [1] 5.138059no_depression/total_no_dep * 100## [1] 94.86194(d) 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(e) The test statistic is ??2 = 20.93. What is the p-value?
(r-1) (c-1) = (2-1) (5-1)
df <- (2-1) * (5-1)
chisq <- 9.49less_1 <- c(670, 11545)
cups2.6 <- c(373, 6244)
cups1 <- c(905, 16329)
cups2.3 <- c(564, 11726)
cups4plus <- c(95, 2288)
table <- data.frame(less_1, cups2.6, cups1, cups2.3, cups4plus, row.names = c("Yes", "No"))
table## less_1 cups2.6 cups1 cups2.3 cups4plus
## Yes 670 373 905 564 95
## No 11545 6244 16329 11726 2288chisq.test(table)##
## Pearson's Chi-squared test
##
## data: table
## X-squared = 20.932, df = 4, p-value = 0.0003267The p-value is 0.0003267
Calculate the chisq critical value
qchisq(0.05, df=4, lower.tail = F)## [1] 9.487729library(visualize)
visualize.chisq(stat=20.932, df = 4, section = "upper")
(f) What is the conclusion of the hypothesis test?
Since p-value is low, null must go, meaning we reject the null hypothesis.
(g) 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 statment. The chisquare test only shows that there is a relationship in the study, not exactly what that relationship is. The study participants weren’t randomly assigned to treatments groups, and there might be other unknown factors not taken into account.