6.5 Prop 19 in California.

In a 2010 Survey USA poll, 70% of the 119 respondents between the ages of 18 and 34 said they would vote in the 2010 general election for Prop 19, which would change California law to legalize marijuana and allow it to be regulated and taxed. At a 95% confidence level, this sample has an 8% 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 62% and 78% of the California voters in this sample support Prop 19.

Ans: False. confidence interval is calculated to estimate the population, not te sample.

  1. We are 95% confident that between 62% and 78% of all California voters between the ages of 18 and 34 support Prop 19.

Ans : True

confidence interval = sample mean +- margin_of_err

margin_of_err = .08 samplemean = .70

lower <- .70 - .08
lower
## [1] 0.62
upper <- .70 + .08
upper
## [1] 0.78
  1. If we considered many random samples of 119 California voters between the ages of 18 and 34, and we calculated 95% confidence intervals for each, 95% of them will include the true population proportion of 18-34 year old Californians who support Prop 19.

Ans : True. Based on the above calculation and if the condition to construct the confident interval 1. Observations should be independent 2. Sucess- Failure rate should be > 10.

  1. In order to decrease the margin of error to 4%, we would need to quadruple (multiply by 4) the sample size.

Margin of error in inversely proportional to sqrt(n). So this is true.

  1. Based on this confidence interval, there is sufficient evidence to conclude that a majority of California voters between the ages of 18 and 34 support Prop 19.

Ans: True

6.11 Study abroad.

A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey between April 25 and April 30, 2007 shows that 55% of high school seniors are fairly certain that they will participate in a study abroad program in college.43

  1. Is this sample a representative sample from the population of all high school seniors in the US? Explain your reasoning.

Ans : Yes. conditions to be checked.

Independence : True sucess-failue : True

  1. Let’s suppose the conditions for inference are met. Even if your answer to part (a) indicated that this approach would not be reliable, this analysis may still be interesting to carry out (though not report). Construct a 90% confidence interval for the proportion of high school seniors (of those who took the SAT) who are fairly certain they will participate in a study abroad program in college, and interpret this interval in context.

COnfidence interval - (0.53 , 0.57)

sample_mean <- .55
n <- 1509
p <- .55
SE <- sqrt(p*(1-p)/n)
zval <- qnorm(0.90)

lower <- sample_mean - zval * SE
lower
## [1] 0.5335873
upper <- sample_mean + zval * SE
upper
## [1] 0.5664127
  1. What does “90% confidence” mean?

90% confidence means if we take more samples, we are confident that 95% of samples exhibit 53% - 57% of people are fairly certain that they will particilate in a abroad study program.

  1. Based on this interval, would it be appropriate to claim that the majority of high school seniors are fairly certain that they will participate in a study abroad program in college?

Ans : True. Because confidence interval is greater than 50. we can assume that majority of high school seniors are fairly certain that they will participate in a study abroad program in college.

6.27 Public Option, Part III.

Exercise 6.13 presents the results of a poll evaluating support for the health care public option plan in 2009. 70% of 819 Democrats and 42% of 783 Independents support the public option. (a) Calculate a 95% confidence interval for the di???erence between (pD ??? pI ) and interpret it in this context. We have already checked conditions for you.

The confident interval is calculated as (0.23 ,0.33 )

PE <- 0.70-0.42
SE <- sqrt((0.70*(1-0.70)/819) + (0.42*(1-0.42)/783))
z <- 1.96
CI <- z*SE
PE - CI
## [1] 0.2333075
PE + CI
## [1] 0.3266925
  1. True or false: If we had picked a random Democrat and a random Independent at the time of this poll, it is more likely that the Democrat would support the public option than the Independent.

Since the confidence (lower, upper) is greater that than 0(positive), democrat would support the public opinion.