CHAPTER 6 HOMEWORK - 6.6, 6.12, 6.20, 6.28, 6.44, 6.48

6.6

6.6 2010 Healthcare Law. 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.

  1. 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.

False - We are 95% confident that between 43% and 49% of the American population support the decision.

  1. 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.

True

  1. 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%.

True - This is the confidence interval

  1. The margin of error at a 90% confidence level would be higher than 3%.

False - It would be lower than 3% since we are lowering our confidence

6.12

Legalization of marijuana, Part I. 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.

  1. Is 48% a sample statistic or a population parameter? Explain.

It is a sample statistics becuase it was taken from a sample of the population

  1. 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.

We are 95% confident, this result is the proportion of US residents who think marijuana should be made legal

p <- 0.48
n <- 1259
se <- sqrt(p*0.52/n)
alpha <- (1-0.95)/2
z_score <- abs(qnorm(alpha))

c(p-z_score*se, p+z_score*se)
## [1] 0.4524033 0.5075967
  1. 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.

Yes, All required conditions have been met.The sample size is large enough and Samples are independent.

  1. 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?

No, this statement is incorrect - 48% does not represent the majority

6.20

6.20 Legalize Marijuana, Part II. 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 ?

much <- 0.48*0.52/(0.02/z_score)^2
ceiling(much)
## [1] 2398

2398 Americans needed for survey for Margin of Error of a 95% confidence interval to 2%_ is

6.28

Sleep deprivation, CA vs. OR, Part I. 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 di↵erence between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.

p <- abs(0.08-0.088)
se <- sqrt(0.08*0.92/11545 + 0.088*0.912/4691)

alpha <- (1-0.95)/2
z_score <- abs(qnorm(alpha))

c(p-z_score*se, p+z_score*se)
## [1] -0.001497954  0.017497954

We are 95% confident the difference in proportions between California and Oregon residents who are sleep deprived is in the resulting interval = -0.001497954 0.017497954

6.44

Barking deer. 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

  1. Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.

H0 - Barking Deer forage in various habitats according to the naturally occurring elements of said habitat.

HA - Barking Deer forage in various habitats according to a preference other than the naturally occurring elements.

  1. What type of test can we use to answer this research question?

We can use a chi-squared test to answer this research question

  1. Check if the assumptions and conditions required for this test are satisfied.

Assuming each deer is an independent thinker and considering the counts in each bin are equal or greater than 5, all assumptions and conditions required for this test have been satisfied.

  1. 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.
chisq <- (4-20.448)^2/20.448 + (16-62.622)^2/62.622 + (61-168.696)^2/168.696 +(345-174.234)^2/174.234

chisq # Chi Squared for Deer
## [1] 284.0609
pchisq(chisq, 3, lower.tail = F) # p value
## [1] 2.799724e-61

Since the P value is so low we must reject the NULL hypothesis __

6.48

Coffee and Depression. Researchers conducted a study investigating the relationship between ca↵einated co↵ee 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

  1. What type of test is appropriate for evaluating if there is an association between coffee intake and depression?

A two table chi squared test is appropriate for evaluating if there is an association between coffee intake and depression

  1. Write the hypotheses for the test you identified.

H0 - There is an independent relationship between depression and coffee consumption.

HA - There is a dependent relationship between depression and coffee consumption.

  1. Calculate the overall proportion of women who do and do not suffer from depression.
2607/50739  
## [1] 0.05138059

Suffer from depression

48132/50739  
## [1] 0.9486194

Do NOT suffer from depression

  1. 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.
6617 * 0.05
## [1] 330.85

expected count for highlighted cell

(373 - 330.85)^2 / 330.85
## [1] 5.369873

test statistic calculation

  1. The test statistic is #2 = 20.93. What is the p-value?
(5-1) * (2-1)
## [1] 4

degree of freedom

According to the table, a chi squared value of 20.93 with 4 degrees of freedom has a p value less than 0.005 = we must reject the null hypothesis

  1. What is the conclusion of the hypothesis test?

We reject the NULL hypothesis, therefore after proving there is a dependent relationship between depression and coffee consumption

  1. 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.64 Do you agree with this statement? Explain your reasoning.

I agree it is oo early to recommend that women load up on extra coffee. further tests would need to be conducted before we can explicitly state that coffee effectively treats depression in women.