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.
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. The ci is applicable to the population and not a sample
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. This is an explanation of the ci.
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. The true proportion is unknown.
The margin of error at a 90% confidence level would be higher than 3%. FALSE. As the confidence level decreases, margin of error will increase.
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.
Is 48% a sample statistic or a population parameter? Explain. 48% is a sample statistic
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.
r <- 1259
pro <- .48
z <- 1.96
se <- sqrt((pro*(1-pro))/r)
ci_l<- pro - (z * se)
ci_u <- pro + (z * se)
ci <- c(ci_l, ci_u)
ci## [1] 0.4524028 0.5075972A 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 condition for normal distribution were met, which is observations were independent and success-failure condition.
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? The upper limit of th ci 51%, so there is a posibility of majority if American thinking marijuana should be legalized.
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?
p<-0.48
me <- 0.02
Z<- qnorm(0.975)
ZE <- me/z
n<- (p * (1 - p)) / ZE^2
ceiling(n)## [1] 23982398 Americans will have to be surveyed for a 2% margin of error.
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.
n_ca <- 11545
n_or <- 4691
mean_ca <- .08
mean_or <- .088
mean_diff <- mean_ca - mean_or
se <- sqrt((((mean_ca)*(1-mean_ca)/n_ca)+mean_or)*(1-mean_or)/(n_or))
z <- 1.96
ci_l <- mean_diff-(z*se)
ci_u <- mean_diff+(z*se)
ci <- c(ci_l, ci_u)
ci## [1] -0.0161073298 0.0001073298There is no significant evidence that there is difference in the sleep deprivation between California and Oregon residents.
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.
H0: Barking deer has no preference of habitats for forage .
HA: Barking deer has preference of habitats for forage.
There are two conditions that must be checked before performing a chi-square test, Independence and Sample size / distribution.The cases seem to be independent of each other and the sample distribution criteria is met as it is expected to be above 5.
observations <- c(4, 16, 67, 345)
observations_ratio <- c(0.048*426, 0.147*426, 0.396*426, 0.409*426)
ch <- sum((observations- observations_ratio) ^ 2 / observations_ratio)
pv <- 1 - pchisq(ch, 3)
pv## [1] 0P-value is dropping to 0 means which the null hypothysis will be rejected that barking deer have no preference for habitats to forage.
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.
What type of test is appropriate for evaluating if there is an association between coffee intake and depression? A chi-square test.
Write the hypotheses for the test you identified in part (a). H0: There is no association between coffee intake and depression. HA: There is an association between coffee intake and depression.
Calculate the overall proportion of women who do and do not suffer from depression.
#The proportion of women NOT suffering from depression is about 95%:
p_w <- (48132/50739)
p_w## [1] 0.9486194# The proportion of women suffering from depression is about 5%:
p_w_d <- (2607/50739)
p_w_d## [1] 0.05138059Expected <- p_w_d*6617
Expected## [1] 339.9854Observed <- 373
test <- ((Observed-Expected)^2)/Expected
test## [1] 3.205914The contribution of the cell is 3.2
p<- pchisq(20.93, 4)
p <- 1-p
p## [1] 0.0003269507The p-value is 0.0003
What is the conclusion of the hypothesis test? The null hypothesis is rejected due to the p value less than 0.05 that there is no association between the coffee intake 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. Do you agree with this statement? Explain your reasoning. I agree, because there may be other conditions which are not aware of in this study.