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.
Ans:
False. The statements above given that we are 100% sure 46% of Americans in this sample support the decision of the US Supreme Court on the 2010 healthcare law.
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. We are 95% confident that between 43% and 49% of all Americans (i.e. the population) support the decision.
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:
False. The true population proportion is unknown. We are only 95% confident that the true population proportion is between 43% and 49%, but we cannot be sure about other sample proportions.
The margin of error at a 90% confidence level would be higher than 3%.
Ans:
When the confidence level decreases (from 95% to 90%), the margin of error should also decrease because MOE = z-score * standard error of the mean, where z-score will decrease if confidence level decreases.
(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.
Ans:
It is a sample statistic. There were only 1,259 US residents being asked.
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.
Ans:
The 95% confidence interval is about (0.4524, 0.5076). We are 95% confident that the true population proportion of all US residents who think the use of marijuana should be made legal falls within 45.24% and 50.76%.
## [1] 0.4524028 0.5075972A 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.
Ans:
True.
The sample size is 1,259, which is smaller enough comparing to the population of US residents. We can assume the sample observations are independent.
The statements np>10 and n(1-p)>10 is true with n=1259 and p=0.48. There are at least 10 cases in each of the responses.
As the two conditions are met, the sampling distribution follows a normal distribution. Therefore, the critic’s statement is true.
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?
Ans:
The 95% confidence interval of the true population parameter is between 45.24% and 50.76%, which includes 50% to 50.76%. However, the true population parameter could be between 45.24% and 49.99%, and we have 5% chance that the true population may fall off the range. Therefore, the news piece’s statement cannot be justified.
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 ?
Ans:
The result I got below by calculation is 2397.158. Therefore, We need to survey at least 2398 Americans to meet the requirements.
## [1] 2397.158According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insuffient 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:
We have the formula: \(SE_{\hat{p_{1}} - \hat{p_{2}}} = \sqrt{\frac{p_{1}(1-p_{1})}{n} + \frac{p_{2}(1-p_{2})}{n}}\)
# Let p_1 = 8.0% from California, and p_2 = 8.8% from Oregon
p1 <- 0.08
p2 <- 0.088
n1 <- 11545
n2 <- 4691
PE <- p1-p2
PE## [1] -0.008## [1] 0.004845984## [1] 0.009498128## [1] -0.017498128 0.001498128The difference in sample proportions of California to Oregon is -0.008.
The standard error of the difference in proportions is 0.004845984.
The margin of error of the difference in proportions at 95% confidence interval is 0.009498.
Therefore, the 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived is (-0.01750, 0.00145).
Since the confidence interval includes 0, there is no statistically difference between the proportions of Californians and Oregonians who are sleep deprived.
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.
Ans:
H0: The barking deer does not prefer to forage in certain habitats over others.
HA: The barking deer prefer to forage in certain habitats over others.
What type of test can we use to answer this research question?
Ans:
Chi-Squared test can be used to answer this research question.
Check if the assumptions and conditions required for this test are satisfied.
Ans:
We assume each case is independent of each other.
A condition requires that each cell-count must have at least 5 expected cases. By using the percentage and total, we have the expected cases list as below. The smallest count is “Woods” with 20 sites.
table_types <- c("Woods", "Cultivated Grass", "Deciduous Forests", "Others", "Total")
table_ratio <- c(0.048, 0.147, 0.396, 1-(0.048+0.147+0.396), 1)
table_expected <- c(round(426*0.048), round(426*0.147), round(426*0.396), 426-(round(426*0.048)+round(426*0.147)+round(426*0.396)), 426)
table_used <- c(4,16,61, 426-(4+16+61), 426)
table1 <- array(data=c(table_ratio, table_expected, table_used),
dim = c(5,3),
dimnames = list(table_types, c("Ratio", "Expected", "Used")))
table1## Ratio Expected Used
## Woods 0.048 20 4
## Cultivated Grass 0.147 63 16
## Deciduous Forests 0.396 169 61
## Others 0.409 174 345
## Total 1.000 426 426Do 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:
If p<0.05, we reject the null hypothesis. Therefore, we reject H0 and accept HA in this case. The barking deer does prefer to forage in certain habitats over others.
##
## Chi-squared test for given probabilities
##
## data: table_used[1:4]
## X-squared = 284.06, df = 3, p-value < 2.2e-16Researchers 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.
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What type of test is appropriate for evaluating if there is an association between coffee intake and depression?
Ans:
Chi-squared test 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).
Ans:
H0: There is no association between coffee intake and depression in women.
HA: There is association between coffee intake and depression in women.
Calculate the overall proportion of women who do and do not suffer from depression.
Ans:
The overall proportion of women who suffer from depression = 2607/50739 = 5.138%
The overall proportion of women who do not suffer from depression = 48132/50739 = 94.862%
coffee_consumption <- c("<=1 cup/week", "2-6 cups/week", "1 cup/day",
"2-3 cups/day", ">=4 cups/day", "TOTAL")
depression_yes <- c(670, 373, 905, 564, 95, 2607)
depression_no <- c(11545, 6244, 16329, 11726, 2288, 48132)
depression_total <- depression_yes + depression_no
table_coffee <- array(data=c(depression_yes , depression_no , depression_total),
dim=c(6,3),
dimnames=list(coffee_consumption,
c("Depression_Yes","Depression_No", "Total")))
table_coffee <- t(table_coffee)
table_coffee## <=1 cup/week 2-6 cups/week 1 cup/day 2-3 cups/day
## Depression_Yes 670 373 905 564
## Depression_No 11545 6244 16329 11726
## Total 12215 6617 17234 12290
## >=4 cups/day TOTAL
## Depression_Yes 95 2607
## Depression_No 2288 48132
## Total 2383 50739Identify the expected count for the highlighted cell, and calculate the contribution of this cell to the test statistic, i.e. \((Observed - Expected)^2 / Expected\).
Ans:
The expected count for the highlighted cell “373” is 339.98.
The contribution of this cell to the test statistic is 3.206.
## [1] 339.9815## [1] 3.206716The test statistic is \(\chi^2=20.93\). What is the p-value?
Ans:
The p-value is 0.000327.
## [1] 0.0003269507What is the conclusion of the hypothesis test?
Ans:
As the p<0.05, we reject the null hypothesis and accept the alternative hypothesis.
We find that there is association between coffee intake and depression in women.
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.
Ans:
The study were only done by observation but not experiments. There are still a lot of other factors may affect the stress and depression on the surveyed women. It is also not clear that the 50739 women is generally proportional to the population in different age, income group, or job field. Therefore, we cannot conclude that high coffee consumption can reduce rate of depression.