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 statementsare 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.
This is false because the confidence interval does not estimate the sample proportion – it is an estimate of the entire population.
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.
This is true. The confidence interval is meant to estimate a population proportion. With a 3% margin of error around the sample proportion of 46%, we get an interval of 43—49%.
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%.
Yes, this is what the confidence interval signifies.
The margin of error at a 90% confidence level would be higher than 3%.
Yes. At a lower confidence error, the margin of error increases.
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.
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.
95% Confidence Interval:
\(0.48 \pm 1.96 \times 0.014 \approx (-0.45, 0.51)\)
sqrt((0.48*(1-0.48))/1259)## [1] 0.01408022A 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 sample size is larger than 30 samples.
We can assume that the observations are independent.
There are at least 10 successes and failures.
Number of successes = 604.32
Number of failures = 654.68
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?
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?
\(ME = z^* \times \sqrt{ \frac{p(1-p)}{n} }\)
\(0.02 = 1.96 \times \sqrt{ \frac{0.48(1-0.48)}{n} }\)
\(n = 2397.1584\)
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.
# Samples
n.CA <- 11545
n.OR <- 4961
# Proportions
p.CA <- 0.08
p.OR <- 0.088
# Point Estimate
pe.28 <- p.OR-p.CA
# Standard Error
SE.28 <- sqrt(((p.CA * (1-p.CA))/n.CA) + ((p.OR * (1-p.OR))/n.OR))
# 95% Confidence Interval
CI.lower28 <- pe.28 - (1.96 * SE.28)
CI.upper28 <- pe.28 + (1.96 * SE.28)
CI.lower28## [1] -0.001307924CI.upper28## [1] 0.01730792Microhabitat 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 grass plot, and 61 as deciduous forests. The table below summarizes these data.
d.total <- 426
d1 <- 4/d.total
d2 <- 16/d.total
d3 <- 61/d.total
d4 <- 345/d.total
d.x <- (1/4)*d.total
chi.44 <- ((d1-0.048)^2/0.048) + ((d2-0.147)^2/0.147) + ((d3-0.396)^2/0.396) + ((d4-0.409)^2/0.40)
chi.44## [1] 0.6756495Researchers 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.
\(H_0:\) The coffee consumption of women with clinical depression is not different from the coffee consumption of women without clinical depression.
\(H_A:\) The coffee consumption of women with clinical depression is different from the coffee consumption of women without clinical depression.
Suffer from depression \(\approx 0.051\)
Do not suffer from depression \(\approx 0.949\)
Expected count = \((1/5) \times 2607 = 521.4\)
Contribution = \(\frac{(373-521.4)^2}{521.4} \approx 41.24\)