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.
Answer: False; Confidence interval applies to the population, not just to the 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.
Answer: True; 43% to 49% confidence interval range is for the population.
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: True.
The margin of error at a 90% confidence level would be higher than 3%.
Answer: Fasle; As the confidence level goes down, the margin of error would go down as well.
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.
Answer: 48% is a sample statistic; as only 48% of the 1,259 US residents were in survey.
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.
# Standard Error:
Stand.Error <- sqrt((0.48 * (1 - 0.48))/1259)
# Lower and Upper limit:
Low.Limit <- 0.48 - (qnorm(0.975)*Stand.Error)
Upper.Limit <- 0.48 + (qnorm(0.975)*Stand.Error)
paste('95% confidence interval for the given sample is',
round(Low.Limit*100, 2) , ' and ', round(Upper.Limit*100, 2) )## [1] "95% confidence interval for the given sample is 45.24 and 50.76"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: Yes, they follow a normal distribution as (1) the observation here are independent and (2) the Success-failure rate condition for this is greater then 10.
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: No, As the upper limit value is only 51% here, we cannot say “Majority” here. We are just above the half way mark.
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?
Answer: we can rewrite the formulae for ME to solve for sample n:
ME / z* = sqrt( p(p-1)/n )
(ME / z*)^2 = p(p-1)/n
(ME / z*)^2 / (p(p-1)) = 1/n
n = p(p-1) / (ME / z*)^2
paste('Number of people we would have to sample is', ( (0.48*(1 - 0.48)) / ((0.02/1.96)^2) ) )## [1] "Number of people we would have to sample is 2397.1584"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 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.
# Standard Error for California:
Stand.Error.California <- sqrt((0.08 * (1 - 0.08))/11545)
# Standard Error for Oregon:
Stand.Error.Oregon <- sqrt((0.088 * (1 - 0.088))/4691)
# Standard Error for the difference in proportion between Oregon and California:
Stand.Error.Diff <- sqrt( Stand.Error.California + Stand.Error.Oregon )
# Lower and Upper limit:
Low.Limit <- (0.088 - 0.08) - (qnorm(0.975)*Stand.Error.Diff)
Upper.Limit <- (0.088 - 0.08) + (qnorm(0.975)*Stand.Error.Diff)
paste('95% confidence interval for the given sample is',
round(Low.Limit*100, 2) , ' and ', round(Upper.Limit*100, 2) )## [1] "95% confidence interval for the given sample is -15.2 and 16.8"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.
Answer: The hypotheses for testing are:
H(null): The barking deer does not prefer a certain habitat to forage.
H(Alternate): The barking deer has certain habitats that it prefer to forage.
What type of test can we use to answer this research question?
Answer: A chi-square test for independence can be used.
Check if the assumptions and conditions required for this test are satisfied.
Answer: We must check for two things (1) the observation are independent; which seems true here and (2) the sample size or distribtution for all habitats have at least 5 expected cases; 4.8 seems close to that and therefore this condition can be treated as satisfied.
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.
Answer: The \(\chi^2\) value is so large the p-value is effectively 0. We can conclude there is convincing evidence the barking deer forage in certain habitats over others..
habitats.data <- c(4, 16, 67, 345)
expected.data <- c(20.45, 62.62, 168.70, 174.23)
k <- length(habitats.data)
df <- k - 1
chi.Sq <- 0
for(i in 1:length(habitats.data))
{
chi.Sq <- chi.Sq + ((habitats.data[i] - expected.data[i])^2 / expected.data[i])
}
p.Value <- pchisq(chi.Sq, df=df, lower.tail=FALSE)
paste('p-value is ',p.Value )## [1] "p-value is 1.13581517968704e-59"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?
Answer: A chi-square test for independence can be used.
Write the hypotheses for the test you identified in part (a).
Answer: The hypotheses for testing are:
H(null): There is no relationship between coffee consumption and clinical depression.
H(Alternate): There is a relationship between coffee consumption and clinical depression.
Calculate the overall proportion of women who do and do not suffer from depression.
paste('Percentage of Women who are depressed are ', round((2607/50739) * 100,2) )## [1] "Percentage of Women who are depressed are 5.14"paste('Percentage of Women who are NOT depressed are ', round((48132/50739) * 100,2) )## [1] "Percentage of Women who are NOT depressed are 94.86"Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
Expected.Cnt <- round(6617 * 0.0514,0)
Observed.Cnt <- 373
paste('Value for the cell 373 is ', round( ((Observed.Cnt - Expected.Cnt)^2/Expected.Cnt) ,2 ))## [1] "Value for the cell 373 is 3.2"The test statistic is \(\chi^2=20.93\). What is the p-value?
p.Value <- pchisq(20.93, df=(5-1), lower.tail=FALSE)
paste('p-value is ',p.Value )## [1] "p-value is 0.000326950725917055"What is the conclusion of the hypothesis test?
Answer: With p-value of ~ 0.0003 being less than 0.05; we can reject null hypothesis. We can say 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. Do you agree with this statement? Explain your reasoning.
Answer: Yes, agree with his statement. Above study states that there is a very weak relationship between coffee consumption and depression among women.