Load necessary libraries -
library(ggplot2)
library('DATA606')##
## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics
## This package is designed to support this course. The text book used
## is OpenIntro Statistics, 3rd Edition. You can read this by typing
## vignette('os3') or visit www.OpenIntro.org.
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## The getLabs() function will return a list of the labs available.
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## The demo(package='DATA606') will list the demos that are available.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.
(a)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.
Ans: FALSE. A confidence interval is constructed to estimate the population proportion, not the sample proportion.
(b) 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.
Ans: TRUE. A confidence interval is constructed to estimate the population propotion which is 46% ± 3%, or 43% to 49%.
(c) 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%.
Ans: TRUE. By the definition of the confidence level.
(d) The margin of error at a 90% confidence level would be higher than 3%.
Ans: FALSE. At a 90% confidence level, the margin of error will be less than 3% and confidence interval will be more narrow since we need to be less confident that the population probability is within the interval.
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.
Ans: The sample was selected simple random process and it repsents less than 10% of the population. We have at least 10 successes and failures for both states, so the distribution can be approximated using the normal model.
p_ca <- 0.08
p_or <- 0.088
p <- p_ca - p_or
n_ca <- 11545
n_or <- 4691
SE2_ca <- (p_ca * (1-p_ca)) / n_ca
SE2_or <- (p_or * (1-p_or)) / n_or
SE <- sqrt(SE2_ca + SE2_or)
( ME <- 1.96 * SE )## [1] 0.009498128The confidence interval is \((-0.0174981,0.0014981)\).
We are 95% confident that the difference between the proportion of Californians and Oregonians who are sleep deprived is between -0.0174981 and 0.0014981.
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.
observed <- c(4, 16, 61, 345, 426)
expected_prop <- c(0.048, 0.147, 0.396, 1-0.048-0.147-0.396, 1)
expected <- expected_prop * 426
deer <- rbind(observed, expected)
colnames(deer) <- c("woods", "grassplot", "forests", "other", "total")
deer## woods grassplot forests other total
## observed 4.000 16.000 61.000 345.000 426
## expected 20.448 62.622 168.696 174.234 426(a) Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.
Ans: *
\({ H }_{ 0 }:\) Barking deer has no preference of certain habitats for foraging.
\({ H }_{ A }:\) Barking deer prefers some habitats over others for foraging.
(b) What type of test can we use to answer this research question?
Ans: We can use chi-square goodness of fit test to this hypothesis.
(c) Check if the assumptions and conditions required for this test are satisfied.
Ans: Although it is possible that something in the behavior of barking deer makes cases dependent on each other, it is more likely that the cases are independent. Each expected value is above 5.
(d) Do these data provide convincing evidence that barking deer prefer to forage in certain habitats over others? Conduct an appropriate hypothesis test to answer this research question.
Ans:
k <- 4
df <- k-1
chi2 <- sum(((deer[1,] - deer[2,])^2)/deer[2,])
( p_value <- 1 - pchisq(chi2, df) )## [1] 0The p-value is practically 0. Even at 99% confidence level, this value is below the significance level, so we reject the null hypothesis. Barking deer prefers to forage in some habitats over others.