Chapter 6 - Inference for Categorical Data

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.

Answer: False. From the point estimate we know that 46% of Americans in the sample support the decision.

2. 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.

Answer: True. Out of 100 sample drawn 95 sample Americans decision will lie between 43% and 49% confidence interval.

3. 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%.

Answer: True. 95% sample decision will be between 43% and 49%.

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

Answer: False. For same standard error 90% confidence interval will have a lower margin of error. # 6.12 Legalization of marijuana, Part I. 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.

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

Answer: It’s a sample statistic because it’s collected from a subset of the population.

2. 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.

Answer:

n <- 1259
p <- 0.48
se <- sqrt((p*(1-p))/n)

t <- qt(0.975,n-1)
me <- t * se
p-me

## [1] 0.4523767

p + me

## [1] 0.5076233

95% confidence interval is (0.4523767, 0.5076233), out of 100 sample surveys 95 will have the survey result will be between the interval.

3. 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.

Answer: It’s true for the data. Two things must need to be satisfied for the data to be normally distributed. The sample is independent since collected from a survey.For success failure condition is also satisfied since \(\hat{p}\) = 48% of 1259 is greater than 10.

4. 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?

Answer: It’s not justified since 95% confidence interval is below 51% which is almost half of the total respondent..

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 ?

Answer:

p <- 0.48
se <- 0.02 / 1.96

n <- (p*(1-p))/(se^2)
ceiling(n)

## [1] 2398

The sample size should be 2398.

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 difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.53

Answer:

se <- sqrt((0.08*0.92/11545)+(0.088*0.0912/4691))
margin_eror <- 1.96*se

round(0.008 - margin_eror,4)

## [1] 0.0024

round(0.008 + margin_eror,4)

## [1] 0.0136

95% confidence interval is (0.0024, 0.0136)

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.

Woods Cultivated grassplot Deciduous forests Other Total 4 16 61 345426 (a) Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.

Answer: Null Hypothesis: Barking deer has no preference for foraging in certain habitats over Others Alternative Hypothesis: Barking deer has preference for foraging in certain habitats over others.

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

Answer: We can chi squared test to answer this research question.

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

Answer: Independence of observations is satisfied since each area is counted once .

Expected cell counts need to be greater or equal to 5. For woods habitat expected cell count is 0.048*426 = 20.49. Hence the second assumption is also satisfied.

4. Do these data provide convincing evidence that barking deer pre- fer to forage in certain habitats over others? Conduct an appro- priate hypothesis test to answer this research question.

Answer:

obsTotal <- c(4/426, 116/426, 67/426, 345 / 426)

popTotal <- c(0.048,0.147,0.396, (1-0.048+ 0.147+0.396))
chisq.test(obsTotal, popTotal)

## Warning in chisq.test(obsTotal, popTotal): Chi-squared approximation may be
## incorrect

## 
##  Pearson's Chi-squared test
## 
## data:  obsTotal and popTotal
## X-squared = 12, df = 9, p-value = 0.2133

Since the p value of the test 0.2133 > 0.05 at 5% level of significance we don’t reject the null hypothesis. We conclude that there is no evidence that barking deer prefer one type of grazing land over another.

6.48 Coffee and Depression. 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.63

Caffeinated coffee consumption ≤1 2-6 1 2-3 ≥4 cups/day 95 2,288 2,383 cup/week Clinical Yes 670 depression No 11,545 Total 12,215 cups/week 6,244 6,617 cup/day 905 16,329 17,234 cups/day 564 11,726 12,290 Total 2,607 48,132 50,739

1. What type of test is appropriate for evaluating if there is an association between coffee intake and depression? Answer: A chi-squared test will be appropriate here.

2. Write the hypotheses for the test you identified in part (a).

Answer: Null Hypothesis: Coffee consumption and clinical depression are independent.

Alternative Hypothesis: Coffe consumption and clinical ddpression are dependent.

3. Calculate the overall proportion of women who do and do not suffer from depression.

Answer:

round(2607/50739,2)

## [1] 0.05

round(1-2607/50739,2)

## [1] 0.95

5% suffer from depression and 95% don’t suffer from depression.

4. 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.

expected <-round( 2607*6617/50739)
expected

## [1] 340

(373-expected)^2/expected

## [1] 3.202941

Expected count for the cell is 340 and the ammount it contributes is 3.202941.

5. The test statistic is χ2 = 20.93. What is the p-value?

Answer:

pchisq(20.93,4,lower.tail = FALSE)

## [1] 0.0003269507

The p value is 0.0003269507

6. What is the conclusion of the hypothesis test?

Answer: Sine the p value is less than 0.05 we reject the null hypothesis and conclude that coffee consumption and depression are related.

7. 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.

Answer: Yes. I agree . The studies statistical significance doesn’t necessarily implies clinical significance difference. More resarch needs to be done before recommending coffee.