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, the confidence interval tells you that the proportion of the population is within that range based on the spot estimate and margin of error of the sample.
True, as above the confidence interval tells you about the population not just the sample.
False, 95% of the sample’s confidence intervals would capture the true proportion of the population which the first sample claims is between 43% and 49%.
False, the critical value for a 90% confidence interval is smaller than that for a 95% confidence interval. The 90% confidence interval would have less chance of capturing the true proportion and would have to have a narrower range.
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.
It is a sample statistic. A population parameter would be calculated from a census.
We will have to assume the sample is random, but there are more than 10 successes and 10 failures. In this case it meets the criteria for a confidence interval.
SE = (0.48*(1-0.48)/1258)^0.5
SE## [1] 0.01408581ME = 1.96*SE
ME## [1] 0.0276082CI = vector()
CI[1] = 0.48-ME
CI[2] = 0.48+ME
CI## [1] 0.4523918 0.5076082Impossible to say because the exercise text mentions nothing about how the sample was selected. If it was not random or otherwise biased, then no it is not valid. If it is random an unbiased, then it is valid.
No, taking the upper boundary of a confidence interval as the true population proportion is bad practice.
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 ?
\[ ME = 1.96*\sqrt{\frac{p(1-p)}{n}} \\ n = \frac{p(1-p)}{(ME/1.96)^2} \]
ME2 = 0.48*0.52/(0.02/1.96)^2
ME2## [1] 2397.1582397 people rounded to the nearest person.
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.
\(PE = p_o-p_c =\)
PE = 0.088-0.08
PE## [1] 0.008\(SE = \sqrt{(\frac{p_o(1-p_o)}{n_o})+(\frac{p_c(1-p_c)}{n_c})} =\)
SE3 = ((0.088*0.912/4691)+(0.08*0.92/11545))^0.5
SE3## [1] 0.004845984\(ME = 1.96*SE =\)
ME3 = 1.96*SE3
ME3## [1] 0.009498128\(CI = PE \pm ME =\)
CI2 = vector()
CI[1] = PE- ME3
CI[2] = PE + ME3
CI## [1] -0.001498128 0.017498128Since the confidence interval spans 0, there is not a statistically significant difference in the proportion of sleep deprivation of the two populations at the 95% confidence level.
Micro-habitat 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.
| Woods | Cult Grass | Decid Forst | Other | Total |
|---|---|---|---|---|
| 4 | 16 | 67 | 345 | 426 |
\(H_o\) is there is no difference in the proportion of deer that forage in certain habitats.
\(H_A\) is there is a difference in the proportion of deer that forage in certain habitats.
A goodness of fit test using \(\chi^2\).
from: https://www.trentu.ca/academicskills/documents/chi-squareinstruction.pdf Chi-Square Test Requirements:
Quantitative data -true.
One or more categories -true.
Independent observations -true as long as 461 is less than 10% total population.
Adequate sample size (at least 10) -true.
Simple random sample - not stated, assumed to be true for the sake of this exercise.
Data in frequency form -true.
All observations must be used -true.
We can do this easily in R:
chisq.test(x=c(4,16,67,345),p=c(0.048,0.147,0.396,0.409))##
## Chi-squared test for given probabilities
##
## data: c(4, 16, 67, 345)
## X-squared = 272.69, df = 3, p-value < 2.2e-16If p is low (<0.05) the null must go, so in this case we accept the alternative hypothesis that there is a statistically significant difference in the proportion of where the deer forage.
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.
\(\chi^2\) test for two-way tables.
\(H_o\): There is no difference between the amount of coffee drank between the population of women without depression and the population of women with depression.
\(H_A\): There is a difference between the amount of coffee drank between the population of women without depression and the population of women with depression.
p_dep = 2607/50739
p_dep## [1] 0.05138059p_hth = 48132/50739
p_hth## [1] 0.9486194expt = 6617*p_dep
expt ## [1] 339.9854ts = (373-expt)^2/expt
ts## [1] 3.205914$df = (R-1)*(C-1) = $
# 2 rows, 5 col
df = (2-1)*(5-1)
df## [1] 4p = 1-pchisq(20.93,4)
p## [1] 0.0003269507Since \(p<0.05\) we reject the null hypothesis and accept the alternative hypothesis that there is a difference in the amount of coffee drank by women with depression compared to women without depression.
Yes I agree, the study only establishes statistical significance and not practical significance. There could be adverse effects to too much coffee that out weigh the benefits, or there could be a confounding factor that causes the two variables to be related.