Young Americans, Part I. About 77% of young adults think they can achieve the American dream. Determine if the following statements are true or false, and explain your reasoning.
True. The values of the success-failure criteria are 15.4 and 3.6 so the mean will be to the right and tail will be to the left.
False. This is the criteria for sample means, not sample proportions.
Look at the Z score.
(.85-.77)/sqrt(.77*.23/60)## [1] 1.4725031.47. Less than 2 Standard Errors so not unusual.
(.85-.77)/sqrt(.77*.23/120) ## [1] 2.082434More than 2 Standard Errors, it looks unusual, so True.
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.
(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.
False. A CI is about a population, not a sample.
True. This is our boilerplace for how to report a confidence interval.
False. A CI is about the population proportion, not about a sample statistic.
False. The CI will be narrower and the so the margin of error will be smaller.
The Civil War. A national survey conducted in 2011 among a simple random sample of 1,507 adults shows that 56% of Americans think the Civil War is still relevant to American politics and political life.
\(H_0:p=.5\) \(H_A:p>.5\)
SF-Sample size is large, >10
Decision rule: reject null if p-value is less than 5%.
phat = 0.56
p0 = 0.5
n = 1507
z = (phat - p0)/sqrt(p0*(1-p0)/n)
1-pnorm(z,0,1)## [1] 1.593292e-06Z-score is 4.66. This is incredibly large.
p value is incredibly small. Less than %, reject the null.
Decision: The data provide convincing evidence that the majority of Americans still think the Civil War is relevant to politics.
The probabality of seeing 1,507 Americans or more thinking the Civil War is irrelevant in the distribution of the null is about zero.
phat = 0.56
n = 1507
U = phat + 1.65*sqrt(phat*(1-phat)/n)
L = phat - 1.65*sqrt(phat*(1-phat)/n)
c(L,U)## [1] 0.5389017 0.5810983With 90% confidence, between 53.9% and 58.1% of Americans believe that the Civil War is still relevant to US politics. 50% is still not in our interval.
Offshore drilling, Part I. A 2010 survey asked 827 randomly sampled registered voters in California “Do you support? Or do you oppose? Drilling for oil and natural gas off the Coast of California? Or do you not know enough to say?” Below is the distribution of responses, separated based on whether or not the respondent graduated from college.
pc = 104/438
pn = 131/389pc = .237
pn = .3367
(i)State Hypotheses
\(H_0: p_c = p_n\) no difference between the 2 proportions
\(H_A: p_c \ne p_n\)
Large independent sample, S-F criteria is met.
Decision rule: Reject null if p-value is less than 5%.
Test Stat and the p-val:
Z = (0.237-0.337)/sqrt(0.254*0.716/438+0.284*0.716/389)
2*pnorm(Z,0,1)## [1] 0.001093856Decision: reject the null since the p-value is less than 5%
There appears to be a significant difference in the proportion of college graduates that do not know enough about the topic.
Vegetarian college students. Suppose that 8% of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning.
False. This is the criteria for sample means, not sample proportions.
True. The SF criteria are 4 & 1 so the mean will be to the left and the tail will be to the right.
Z Score
(.12-.08)/sqrt(.08*.02/125)## [1] 11.1803411.18
Yes it would be very extremely highly unusual.
(.12-.08)/sqrt(.08*.02/250)## [1] 15.8113915.81
It would be even more ridiculously unusual.
?
Orange tabbies. Suppose that 90% of orange tabby cats are male. Determine if the following statements are true or false, and explain your reasoning.
This is not the type of data that is really skewed.
Using a sample size that is 4 times as large will reduce the standard error of the sample proportion by one-half.
The distribution of sample proportions of random samples of size 140 is approximately normal.
The distribution of sample proportions of random samples of size 280 is approximately normal.
Young Americans, Part II. About 25% of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statements are true or false, and explain your reasoning.
The distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump in random samples of size 12 is right skewed.
In order for the the distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump to be approximately normal, we need random samples where the sample size is at least 40.
A random sample of 50 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unusual.
A random sample of 150 young Americans where 20% have delayed starting a family due to the continued economic slump would be considered unusual.
Tripling the sample size will reduce the standard error of the sample proportion by one-third.
Elderly drivers. In January 2011, The Marist Poll published a report stating that 66% of adults nationally think licensed drivers should be required to retake their road test once they reach 65 years of age. It was also reported that interviews were conducted on 1,018 American adults, and that the margin of error was 3% using a 95% confidence level.
1.96*sqrt((.66*.44)/1018)## [1] 0.033104Yes it’s correct.
No, it provides convincing evidence tha 66% think so.
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.
48% is a sample statistic because no work has been done to apply this number to the population.
phat = .48
n = 1259
U = phat + 1.96*sqrt(phat*(1-phat)/n)
L = phat - 1.96*sqrt(phat*(1-phat)/n)
c(L,U)## [1] 0.4524028 0.5075972We are 95% confident that between 45% and 50.7% of Americans think that Marijuana should be made legal.
If the sample were repeated multiple times, it would eventually follow normal distribution.
No, on the upper end it is only just over half. I would say “almost half of Americans” to be more precise.
The Daily Show. A 2010 Pew Research foundation poll indicates that among 1,099 college graduates, 33% watch The Daily Show. Meanwhile, 22% of the 1,110 people with a high school degree but no college degree in the poll watch The Daily Show. A 95% confidence interval for (pcollege grad - pHSorless) where p is the proportion of those who watch The Daily Show, is (0.07, 0.15). Based on this information, determine if the following statements are true or false, and explain your reasoning if you identify the statement as false.
At the 5% significance level, the data provide convincing evidence of a difference between the proportions of college graduates and those with a high school degree or less who watch The Daily Show.
We are 95% confident that 7% less to 15% more college graduates watch The Daily Show than those with a high school degree or less.
95% of random samples of 1,099 college graduates and 1,110 people with a high school degree or less will yield differences in sample proportions between 7% and 15%.
A 90% confidence interval for pcollegegrad - pHSorless would be wider.
A 95% confidence interval for pcollegegrad - pHSorless is (-0.15,-0.07).
I need an office hour, or Math Lab, or some kind of help before I can do these!