The following problems are copied from the chapter 17 exercises from Introduction to Modern Statistics First Edition by Mine Çetinkaya-Rundel and Johanna Hardin (https://openintro-ims.netlify.app/inference-one-prop.html)

  1. National Health Plan.

A Kaiser Family Foundation poll for US adults in 2019 found that 79% of Democrats, 55% of Independents, and 24% of Republicans supported a generic “National Health Plan.” There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. 79% of 347 Democrats and 55% of 617 Independents support a National Health Plan. (K. F. Foundation 2019)

  1. Is it appropriate to use the formulas discussed in class to calculate a confidence interval for the difference in proportions? Explain your answer.

ANSWER: Yes, all three groups have large enough numbers to be above 10 for both success’s and failure’s and there is no reason to not believe that the samples were not taken independently from one another. Both samples are also above 20 and are sufficiently large. Therefore, a formula can be used to calculate

  1. Calculate a 99% confidence interval for the difference between the proportion of Democrats and Independents who support a National Health Plan (\(p_D - p_I\))

\(\hat p_D\) = .79 \(\hat p_I\) = .55 z = 2.576

prop.test(x=c(274, 347), n=c (339, 617), conf.level= .9, correct=FALSE)
## 
##  2-sample test for equality of proportions without continuity
##  correction
## 
## data:  c(274, 347) out of c(339, 617)
## X-squared = 58.101, df = 1, p-value = 2.49e-14
## alternative hypothesis: two.sided
## 90 percent confidence interval:
##  0.1977357 0.2939861
## sample estimates:
##    prop 1    prop 2 
## 0.8082596 0.5623987
  1. Interpret it in this context. We have already checked conditions for you.

ANSWER: We are 99% confident that the true difference in proportions of US Adult Democrats and Independents lies within the interval (.198, .294)

  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 National Health Plan than the Independent.

ANSWER: True

  1. Sleep deprivation, CA vs. OR, confidence interval. (This is slightly modified from the book)

According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient 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.

  1. Calculate a 90% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived using the bootstrap method to calculate the standard error. List all the values you used in your calculations. (\(p_C - p_O\))

difference = -.008 \(z^* =\) 1.645 SE \(\approx\) .0047 interval \(\approx\) (.0127, .0686)

  1. Calculate a 90% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived using the formulas we discussed in class.
prop.test(x=c (413, 4691), n=c(924, 11545), conf.level= .9, correct=FALSE)
## 
##  2-sample test for equality of proportions without continuity
##  correction
## 
## data:  c(413, 4691) out of c(924, 11545)
## X-squared = 5.8461, df = 1, p-value = 0.01561
## alternative hypothesis: two.sided
## 90 percent confidence interval:
##  0.01271249 0.06858074
## sample estimates:
##    prop 1    prop 2 
## 0.4469697 0.4063231
  1. Interpret the interval from part b in context.

ANSWER: We are 90% Confident that the true difference in proportions of Californians and Oregonians who are sleep deprived lies within the interval (.0127, .0686)

  1. Heart transplant success.

The Stanford University Heart Transplant Study was conducted to determine whether an experimental heart transplant program increased lifespan. Each patient entering the program was officially designated a heart transplant candidate, meaning that he was gravely ill and might benefit from a new heart. Patients were randomly assigned into treatment and control groups. Patients in the treatment group received a transplant, and those in the control group did not. The visualization below displays how many patients survived and died in each group.175 (Turnbull, Brown, and Hu 1974)

Suppose we are interested in estimating the difference in survival rate between the control and treatment groups using a confidence interval. Explain why we cannot construct such an interval using the normal approximation (formulas).

ANSWERS: There is less than 10 survivors in the control group. Therefore we cannot find the difference in survival rate between the control and treatment groups using formulas.