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.
False. We are 95% confident that between 43% and 49% of Americans (meaning population not the sample) support the decision of the U.S. Supreme Court on the 2010 healthcare law. While we are 100% confident that 46% of Americans in this sample support the decision of the U.S. Supreme Court on the 2010 healthcare law.
True.
False. 95% confidence interval provides the range of values for the true population proportions. The true population proportions will fall between 43% and 49% 95% of the time.
False. It would be lower because we are descreasing our confidence percentage.
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.
The 48% is a sample statistic. It was calculated based on a 1,259 sample of the total US population.
n <- 1259
p <- 0.48
#degree of freedom
df <- n-1
#confidence interval percentage
CI <- 0.95
#calculating standsard error
SE <- ((p * (1 - p)) / n) ^ 0.5
#calculating t-value
T <- qt(CI + (1 - CI)/2, df)
#margin of error
ME <- T * SE
#95% confidance interval range of values
CI_range <- c(p - ME, p + ME)
CI_range## [1] 0.4523767 0.5076233The proportion of US residents who think marijuana should be made legal will fall into interval from 45.24% to 50.76% 95% of the time.
True. p is normally distributed when sample observations are independent and the sample size is large enough such that n*p ≥ 10 and n⋅(1−p) ≥ 10.
False. We already found out that the proportion of US residents who think marijuana should be made legal will fall into interval from 45.24% to 50.76% 95% of the time. The upper bound of the interval is a little bit greater than 50%, while the lower bound is less that 50%. So that, we can’t reject the hypotheses that Majority of Americans think marijuana should be legalized, however, we can’t less 50% of Americans think marijuana should be legalized.
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?
ME = Z*SE
SE = sqrt(p(1−p)/n)
ME = Z*sqrt(p(1−p)/n)
sqr(ME/Z) = p(1−p)/n
n = p(1−p)/sqr(ME/Z)
p <- 0.48
ME <- 0.02
CI <- 0.95
Z <- qnorm(CI + (1 - CI) / 2)
n <- (p*(1-p))/(ME/Z)^2
round(n,0)## [1] 2397We should survey 2397 Americans.
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.
p_CA <- 0.08
n_CA <- 11545
p_OR <- 0.088
n_OR <- 4691
#confidence interval
CI <- 0.95
p_diff <- p_OR - p_CA
#standard error for the proportion difference
SE <- ( ((p_CA * (1 - p_CA)) / n_CA) + ((p_OR * (1 - p_OR)) / n_OR)) ^ 0.5
Z <- qnorm(CI + (1 - CI) / 2)
#margin of error for the proportion difference.
ME <- Z * SE
# 95% confidence interval.
CI_values <- c(p_diff - ME, p_diff + ME)
CI_values## [1] -0.001497954 0.017497954The 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived is from -0.0015 to 0.0175.
Since the interval contains 0 we can state with a 95% confidence that the proportions of Californians and Oregonians are not statistically different.
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.
Null Hypothesis: Barking deer prefers to forage in certain habitats over others.
Alternative Hypothesis: Barking deer doesn’t prefer to forage in certain habitats over others.
We can use a Chi-square test for one-way table.
First, we have to assume that the sampling method is simple random sampling.
Second, as independence is not provided in the conditions we assume that variables are independent.
Third, the expected value of the number of sample observations in each level of the variable is at least 5. We assume that woods habitat has only almost 5 (it has 4.8)
Null Hypothesis: Barking deer prefers to forage in certain habitats over others.
Alternative Hypothesis: Barking deer doesn’t prefer to forage in certain habitats over others.
n <- 4 # there are 4 habitats
#degree of freedom
df <- n - 1
#calculate the chi2 test statistic for woods
chi_woods <- (0.9 - 4.8) ^ 2 / 4.8
#calculate the chi2 test statistic for cultivated grassplot
chi_cul_grass <- (3.8 - 14.7) ^ 2 / 14.7
#calculate the chi2 test statistic for deciduous forests
chi_forests <- (14.3 - 39.6) ^ 2 / 39.6
#calculate the chi2 test statistic for other
chi_other <- (81 - 40.9) ^ 2 / 40.9
chi_sum <- chi_woods + chi_cul_grass + chi_forests + chi_other
#calculate p-val
p_val <- 1 - pchisq(chi_sum, df = df)
p_val## [1] 2.14273e-14Since p-value is less that 5% we have to reject null hypotheses. So, there is no convincing evidence that barking deer prefer to forage in certain habitats over others.
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.
The Chi-squared test for two-way tables is appropriate for evaluating if there is an association between coffee intake and depression.
Null Hypothesis: There is NO association between caffeinated coffee consumption and depression.
Alternative Hypothesis: There is an association between caffeinated coffee consumption and depression.
dep<- 2607
no_dep <- 48132
dep_prop <- dep/(dep+no_dep)
round(dep_prop*100,2)## [1] 5.14no_dep_prop <- no_dep/(dep+no_dep)
round(no_dep_prop*100,2)## [1] 94.86The overall proportion of women who do suffer from depression is 5.14%. The overall proportion of women who do not suffer from depression is 94.86%
exp_count = 6617 * 0.0514
round(exp_count,0)## [1] 340obs_count <- 373
(obs_count - exp_count)^2/exp_count## [1] 3.179824The expected count for the highlighted cell is 3.179824
n <- 5 #numeber of variables
k <- 2 #types of cases - yes/no
#degree of freedom
df <- (n-1)*(k-1)
chi <- 20.93
p_value <- 1 - pchisq(chi, df)
p_value## [1] 0.0003269507The p-value is 0.0003269507
As the p-value is below 0.05, we reject the null hypothesis which states that there is no relationship between depression and coffee consumption among women. We can conclude that there is a relationship between depression and coffee consumption among women.
I agree with this statements since the chi-square test showed that there is a relationship between depression and coffee consumption among women.