2010 Healthcare Law. (6.48, p. 248) 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.
- 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: This statement is FALSE because the confidence interval gives the range for population parameter but this statement mentioned sample.
- 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: This statement is TRUE because that is the correct interpretation for confidence interval.
- 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: This statement is FALSE because the margin error = Z-score * standard error. At 95% confidence interval, the Z-score is 1.96 while at 90% confidence interval, the Z-score is 1.645. Therefore, at a 90% confidence level, the margin error will be LOWER than 3%.
Legalization of marijuana, Part I. (6.10, p. 216) 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.
- Is 48% a sample statistic or a population parameter? Explain.
Answer: Since 48% is a sample proportion; therefore, it is a sample statistic.
- 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: For 95% confidence level, the z-score is 1.96. A 95% confidence interval for the proportion will be \(0.48\pm 1.96*sqrt[0.48*(1-0.48)/1259] = 0.48\pm 0.028 = (0.452,0.508) = (45.2%, 50.8%)\). So, we are confident that the true proportion of US residents who think marijuana should be made legal lies between 45.2% and 50.8%.
- 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: This is TRUE. 95% confidence interval is only accurate if the statistic follows a normal distribution. We can check this by doing np and n(1-p) whether both are greater than 10. np = 12590.48 = 604.32 and n(1-p) = 1259(1-0.48) = 654.48. Since both are greater than 10, the normal model is a good approximation for these data.
- 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: From part (b), a 95% confidence interval for the proportion of US residents who think marijuana should be made legal lies between 45.2% and 50.8%. Since the lower limit of confidence interval is less than 50% population who think that marijuana should be made legal. Therefore, the statement “Majority of Americans think marijuana should be legalized” is not justified.
Legalize Marijuana, Part II. (6.16, p. 216) As discussed in Exercise above, 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: Based on the calculation below, the 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived is (-0.0175,0.0015).
- California : sample size, n1 = 11545 and sample proportion, \(\hat { p } = 0.08\).
- Oregon : sample size, n2 = 4691 and sample proportion, \(\hat { p } = 0.088\).
- 95% confidence interval : \(\alpha = 1-0.95 = 0.05\). From the normal distribution table, the required Z(0.05) value for 95% confidence level is 1.96.
- The 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived is \((0.08-0.088) \pm 1.96 * SQRT[(0.08(1.0.08)/11545) + (0.088(1-0.088)/4691)] = -0.008 \pm 1.96 * SQRT(0.00002348) = -0.008 \pm (1.96*0.0048) = -0.008 \pm 0.0095 = (-0.008-0.0095, -0.008+0.0095) = (-0.0175,0.0015)\).
Barking deer. (6.34, p. 239) 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.
- Write the hypotheses for testing if barking deer prefer to forage in certain habitats over others.
Answer: The null and alternative hyptheses are as following:
- Null Hypothesis, \({ H }_{ o }\): The sites where barking deer forage were distributed according to the proportion of land for each habitat. p1 = 0.048, p2 = 0.147, p3 = 0.396, p4 = 0.409.
- Alternative Hypothesis, \({ H }_{ a }\) : The sites where barking deer forage is not evenly distributed across the habitiats of the region. In other words, at least one proportion is different from its expected value.
- What type of test can we use to answer this research question?
Answer: We can use chi-square goodness of fit test.
- Check if the assumptions and conditions required for this test are satisfied.
Answer: Th following conditions have been satisfied to perform a chi-square test.
- The data are obtained from a random sample.
- The varibale under study is categorical.
- The expected value of the number of sample observations in each level of the variable is at least 5. Expected value: Woods = 0.048426 = 20.5, Cultivated Grassplot = 0.147426 = 62.6, deciduous forest = 0.396426 = 168.7, Other = 0.409426 = 174.2.
- 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.
Answer: Since the test statistic value is greater than the critical value, we can reject the Null hypothesis. At 5% significance level, there is significant evidence that at least one proportion is different than its expected value. Therefore, we can conclude that these data provide convincing evidence that barking deer prefer to forage in certain habitats over others.
- Test Statistic = (4-20.5)^2/20.5 + (16-62.6)^2/62.6 + (61-168.7)^2/168.7 + (345-174.2)^2/174.2 = 13.23 + 34.71 + 68.75 + 167.37 = 284.06.
- Significance level = 0.05, degrees of freedom = 4-1=3, so from Chi-square table the critical value = 7.815.
- To get the p-value, see below R computation.
chi <- 284.06
pValue <- pchisq(chi,df=3,lower.tail = FALSE)
pValue
## [1] 2.801049e-61
Coffee and Depression. (6.50, p. 248) 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.
{
}
- What type of test is appropriate for evaluating if there is an association between coffee intake and depression?
Answer: The Chi-square test for association is appropriate for evaluating if there is an association between coffee intake and depression.
- Write the hypotheses for the test you identified in part (a).
Answer: The null and alternative hyptheses are as following:
- Null Hypothesis, \({ H }_{ o }\):Caffeinated coffee consumption and depression in women are not associated.
- Alternative Hypothesis, \({ H }_{ a }\) : Caffeinated coffee consumption and depression in women are associated.
- Calculate the overall proportion of women who do and do not suffer from depression.
Answer: Proportion of women who do suffer from depression = 2607/50739 = 0.05138. Proportion of women who do NOT suffer from depression = 48132/50739 = 0.9486.
- Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. (\(Observed - Expected)^2 / Expected\)).
Answer: The expected count for the highlighted cell = (2607*6617)/50739 = 340. The test statistic for the highlighted cell is (373-340)^2/340 = 3.2.
- The test statistic is \(\chi^2=20.93\). What is the p-value?
Answer: Based on the calculatiom below, the p-value is 0.000327.
- Degree of freedom = (no. of rows - 1) * (no. of columns - 1) = (2-1)*(5-1) = 4.
chi <- 20.93
pValue <- pchisq(chi,df=4,lower.tail = FALSE)
pValue
## [1] 0.0003269507
- What is the conclusion of the hypothesis test?
Answer: Since the p-value is less than 0.05, we can rejec the NULL Hypothesis. We can conclude that caffeinated coffee consumption and depression in women are associated.
- 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 with this statement as this is an observational study. So, we cannot infer that increase caffeinated coffee consumption will cause to decrease risk of depression in women. There could be other factors that cause to decrease depression in women who drink more coffee.