Chapter 6 : Inference for categorical data - Presentation

Debabrata Kabiraj
04/03/2019

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.

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

No, the sample is NOT representative of the population.  
- This is an optional online survey and the sample is taken only from the students who took SAT. Not all high schoolers take the SAT; some take the ACT instead.
- It would be a good sample representation of all US high school seniors who took the SAT.
- Respondents only include those who took the SAT between April 25 and April 30 and were willing to respond. Those who were not willing to respond were excluded.
- Those who took the exam during another period of time (students who take SAT in April tend to be Juniors, so the senior population should be relatively small) would not be part of the sample

n <- 1509 # sample size
p <- .55 # percent successful participation
df <- data.frame("Success" = n*p, "Failure" = n*(1-p))
df

  Success Failure
1  829.95  679.05

Conditions for sampling distribution :
(1) Sample observations are independent and random
(2) We expect to see least 10 successes and 10 failures in our sample.

By these conditions, we can say that sampling distribution conditions are met. 
However this is an optional online survey, only the high school seniors who were interested in the online survey between April 25 and April 30, 2007 completed it.

(b) 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.

Standard Error = \[ \sqrt{((p*(1-p))/n)} \]

SE <-  sqrt((p*(1-p))/n)

ci <- 0.90 # 90% confidence interval
z <- qnorm(ci + (1-ci)/2) # 90% confidence interval

cip <- data.frame("Lower" = (p-z*SE), "Upper" = (p+z*SE))
cip

      Lower     Upper
1 0.5289346 0.5710654

We are 90% confident that 53% to 57% of high school seniors who took the SAT are fairly certain that they
will participate in a study abroad program in college.

(c) What does "90% confidence" mean?

By definition, a confidence interval is a plausible range of values for the population parameter. For a confidence interval, a 90% confidence means we are 90% confident that the confidence interval captured the true parameter.

When we randomly select a sample from the true population of all high school senior in the US, 90% of the population should have a proportion (percentage of high school seniors who are fairly certain that they will participate in a study abroad program in college) within this range (the confidence interval).

plot of chunk unnamed-chunk-8

As can be seen from above, that we are 90% confident that 53%-57% of high school seniors are most likely to participate in a study abroad program. We can say that 90% if random samples would give us 90% CI that includes the true proportion (.55).

(d) 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?

Based on the confidence interval results, it is fair to assume that majority of high school seniors will participate in a study abroad since the confidence interval is over 50% (majority).