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. Solution: 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. Solution: TRUE. This statement exlains 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%. Solution: FALSE. The true proportion is unknown.
The margin of error at a 90% confidence level would be higher than 3%. Solution: 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 res- idents: “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. Solution: 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. Solution:
n <- 1259
p <- .48
z <- 1.96
SE <- sqrt((p*(1-p))/n) # Standard Error
Lower_CI <- p - (z * SE)
Upper_CI <- p + (z * SE)
CI <- c(Lower_CI, Upper_CI)
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. Solution: There are 2 conditions for the sampling distribution to be nearly normal.1. Observations are independentand 2. Success-failure condition. These conditions are both met in this case.
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? Solution: Yes, it is likely justified considering the upper limit of the CI is about 51% which is above 50%.
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 # Margin of Eror of 2%
Z = qnorm(0.975)
ZE = ME/Z
n = (P*(1-P))/ZE^2
round(n,0)+1## [1] 2398We should survey about 2,398 Americans.
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_Cal <- 11545
n_Ore <- 4691
mean_Cal <- .08
mean_Ore <- .088
meandiff <- mean_Cal - mean_Ore
SE <- sqrt((((mean_Cal)*(1-mean_Cal)/n_Cal)+mean_Ore)*(1-mean_Ore)/(n_Ore))z <- 1.96
Lower_CI <- meandiff-(z*SE)
Upper_CI <- meandiff+(z*SE)
CI <- c(Lower_CI, Upper_CI)
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.
Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others. Solution: H0: There’s no difference between each forage category. HA: There’s difference with at least one of the forage categories.
What type of test can we use to answer this research question? Solution: A Chi-Square test.
Check if the assumptions and conditions required for this test are satisfied. Solution: 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 but the sample distribution criteria is not met as the woods category can only occur 4 times.
Do these data provide convincing evidence that barking deer pre- fer to forage in certain habitats over others? Conduct an appro- priate hypothesis test to answer this research question. Solution:
N<-426 # Total from Table
Observation<-c(4,16,67,345)
Expected<-c(N*0.048,N*0.147,N*0.396,N*0.409)
Chisq<-sum((Observation-Expected)^2/Expected)
DF<-4-1
pchisq(Chisq,DF,lower.tail=FALSE)## [1] 1.144396e-59Coffee 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.
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What type of test is appropriate for evaluating if there is an association between coffee intake and depression? Solution: Chi Square test is appropriate.
Write the hypotheses for the test you identified in part (a). Solution: H0: There is no association between coffee consumption and depression. HA: There is an association between coffee consumption and depression.
Calculate the overall proportion of women who do and do not suffer from depression. Solution:
pdepressed <- (2607/50739)
pdepressed## [1] 0.05138059pnon_depressed <- (48132/50739)
pnon_depressed## [1] 0.9486194About 5% suffer from depression while about 95% do not suffer from depression.
Expected <- pdepressed*6617
Expected## [1] 339.9854Observed <- 373
Test_Stat <- ((Observed-Expected)^2)/Expected
Test_Stat## [1] 3.205914The contribution of the cell is 3.2
pvalue <- pchisq(20.93, 4)
pvalue <- 1-pvalue
pvalue## [1] 0.0003269507The p-value is 0.0003
What is the conclusion of the hypothesis test? Solution: Since the p-value of 0.0003 is less than 0.05, I will reject the Null Hypothesis and accept the alternative hypothesis that there is an association between coffee consumption and depression.
One of the authors of this study was quoted on the NY Times 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. Solution: I agree.There may be other conditions/variables which are not included in this study.