--- title: "Inference for a Single Proportion" subtitle: | | IMS1 Ch. 16 | Math 219 author: "Yurk" format: revealjs: theme: beige editor: source --- ## Medical Consultants ```{r} #| include: false #| echo: false library(tidyverse) library(infer) library(openintro) ``` - Some organ donors work with a medical consultant who helps them throughout the process - The average complication rate for liver donor surgeries in the United States is about 10% - One consultant claims she has low rate of complications compared to national average. Is her claim supported? - Let $p$ be the consultant's long-run complication rate ------------------------------------------------------------------------ ```{r} #| include: false #| echo: false consult <- tibble(complication = c(rep("yes", 3), rep("no", 59))) ``` Hypotheses: - $H_0: p = 0.1$ - $H_A: p < 0.1$ Data: - `consult` dataset - She has served as a consultant for 62 liver donor surgeries - 3 (4.8%) resulted in complications - $\hat{p} = 0.048$ ## Payday Loan Regulations - Borrowers use payday loans to get a cash advance before their next payday - Borrower writes a check for loan amount + service fee - Lender holds check until borrower's payday - Very high APR equivalent (often over 300%) - Some borrowers take out second loan to pay of first, and so on ------------------------------------------------------------------------ - Michigan already has a law that limits the number of payday loans a borrower can hold (2) - Do most payday borrowers support additional regulation that would require payday lenders to do a credit check? - Let $p$ be the proportion of payday borrowers in MI that support additional regulation. ------------------------------------------------------------------------ ```{r} #| include: false #| echo: false payday <- tibble(support = c(rep("yes", 424), rep("no", 402))) ``` Hypotheses: - $H_0: p = 0.5$ - $H_A: p > 0.5$ Data: - `payday` dataset - Researchers selected a random sample of 826 payday borrowers - 424 (51.3%) said they would support a regulation - $\hat{p}=0.513$ ## Mathematical Model for a Proportion - We have learned that if certain conditions are met we can use a mathematical model to make inferences about a population - There is a version of the Central Limit Theorem for a single proportion ------------------------------------------------------------------------ ::: callout-note ## Sampling distribution of $\hat{p}$ The sampling distribution of $\hat{p}$ based on a sample of size $n$ from a poplation with true proportion $p$ will be approximately normal with mean $p$ and standard error $$SE=\sqrt{\frac{p(1-p)}{n}}$$\ if the following technical conditions are met: 1. independent observations (e.g., observations from SRS) 2. (**success-failure condition**) at least 10 expected successes and at least 10 expected failures. (i.e., $np\geq 10$ and $n(1-p)\geq 10$) ::: ## Checking Technical Conditions Consultant study - The success-failure condition is not met. Under the null hypothesis, we expect $62\times 0.1 = 6.2$ complications (less than 10) - Cannot model null distribution using a normal distribution - Use randomization instead (**parametric bootstrap simulation**) ------------------------------------------------------------------------ Payday study - The success-failure condition is met. Under the null hypothesis, we expect $0.5\times 826 = 413$ people to support the legislation and $(1-0.5)\times 826 = 413$ to not support the legislation. - It is appropriate to model the null distribution using a normal distribution ## Hypothesis Test Using Normal Approximation - Let $p_0$ be the proportion under the null hypothesis - We will use the **Z score** as the test statistic $$Z = \frac{\hat{p}-p_0}{SE(\hat{p})}=\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}$$ - Recall that if $\hat{p}$ is normally distributed, then $Z$ has a standard normal distribution, $N(0,1)$ ------------------------------------------------------------------------ ```{r} #| include: false #| echo: false p0 <- 0.5 phat <- 424/826 n <- 826 Z <- (phat - p0)/sqrt(p0*(1-p0)/n) ``` - For the payday study $p_0=0.5$, and $\hat{p}=0.513$, so $$Z = \frac{0.513 - 0.5}{\sqrt{0.5\cdot(1-0.5)/826}}=0.765$$ - The p-value is the probability that we would obtain a $Z$ score at least as large as 0.765 if the null hypothesis is true - We compute the p-value by finding the area under the density curve for $N(0,1)$ that is beyond 0.765 ```{r} #| include: false #| echo: false p0 <- 0.5 phat <- 424/826 n <- 826 Z <- (phat - p0)/sqrt(p0*(1-p0)/n) ``` ------------------------------------------------------------------------ ```{r} #| include: true #| echo: false #| fig-cap: "Normal model, $N(0,1)$. P-value is area of shaded region." normTail(m = 0, s = 1, U = Z) ``` ```{r} #| include: true #| echo: true 1 - pnorm(0.765, mean = 0, sd = 1) ``` ------------------------------------------------------------------------ - We are unable to reject the null hypothesis (p-value = 0.22) - Note that we cannot claim that 50% of payday borrowers support the new legislation (we cannot accept the null hypothesis) - However, 50% is a plausible value for the parameter ## Hypothesis Test Using Randomization - In the consultant study we cannot use a normal model for the null distribution - However, we can use **parametric bootstrap simulation** to approximate the null distribution - We simulate 1,000 random samples of 62 liver donors from a population in which the null hypothesis is true (10% complication rate) ------------------------------------------------------------------------ Parametric bootstrap simulation is equivalent to the following physical simulation: - For each donor simulate the outcome by spinning a spinner with 10% of the area representing "complication" and 90% representing "no complication" - For each sample, spin the spinner 62 times and record the proportion of complications in the sample - Repeat to obtain proportions for 1,000 simulated samples ------------------------------------------------------------------------ We can do the bootstrapping using the `infer` package ```{r} #| include: true #| echo: true set.seed(8675309) library(infer) consult_boot <- consult |> specify(response = complication, success = "yes") |> hypothesize(null = "point", p = 0.1) |> generate(reps = 1000, type = "draw") |> calculate(stat = "prop") ``` ------------------------------------------------------------------------ ```{r} #| include: true #| echo: false #| fig-cap: "Approximate null distribution with observed proportion of surgeries with complications (0.048) inticated by dashed vertical line." consult_boot |> ggplot(aes(stat)) + geom_histogram(bins = 15) + geom_vline(xintercept = 3/62, color = "red", linewidth = 2, linetype = "dashed") + labs(x = "bootstrapped proportion of surgeries with complications", title = "1,000 parametric bootstrapped proportions (p = 0.1)") + theme_minimal() ``` ------------------------------------------------------------------------ - The p-value is approximated by the proportion of bootstrapped proportions that are at least as extreme as the observed proportion ($\leq 0.048$) - With a p-value of 0.11 we are unable to reject the null hypothesis. - It is plausible that the consultant has the same complication rate as the national average of 10% ```{r} #| include: true #| echo: true consult_boot |> summarize(n_extreme = sum(stat <= 3/62), p_val = mean(stat <= 3/62)) ``` ## Confidence Interval - We can also use a normal distribution to find a confidence interval if technical conditions are met - Earlier we used $p_0$ as the mean and in the computation of SE, because we were trying to approximate the null distribution - A confidence interval estimates the value of the parameter - The best point-estimate we have is the sample proportion $\hat{p}$, so we use that as the mean and in the computation of SE ## Checking Conditions for CI - The success-failure condition is easier to check in this situation. - $n\hat{p}$ is the number of observed success, and $n(1-\hat{p})$ is the number of observed failures. - We just need to check if there were 10 successes and 10 failures in the sample. ------------------------------------------------------------------------ Consultant study - The success-failure condition is not met. There were 3 successes and 59 failures - Cannot use normal approximation to find a CI - Use randomization instead (**bootstrap** as in Chapter 12) ------------------------------------------------------------------------ Payday study - The success-failure condition is met. There were 424 successes and 402 failures - It is appropriate to use a normal approximation to find a CI ## Confidence Interval Using a Normal Approximation - If a normal approximation is appropriate, a confidence interval for a proportion can be written as $$\hat{p}\pm z^{\ast}\times SE$$ - SE is estimated using $$SE\approx\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ - $z^{\ast}$ is determined by the confidence level (e.g., 1.96 for 95%, 2.58 for 99%) ------------------------------------------------------------------------ ```{r} #| include: false #| echo: false n <- 826 phat <- 424/826 se <- sqrt(phat*(1-phat)/n) cil <- phat - 1.96*se ciu <- phat + 1.96*se ``` - The standard error for the proportion of borrowers that support the new regulation is $$SE \approx \sqrt{\frac{0.513\cdot(1-0.513)}{826}}=0.0174$$ - The 95% confidence interval is $$0.513\pm1.96\cdot0.0174 = 0.513\pm0.034$$ - We are 95% confident that the proportion of payday borrowers that support the new regulation is between 0.479 and 0.547 ## Confidence Interval Using a Bootstrapping - We use bootstrapping to find a 95% confidence interval for the complication rate for the medical consultant - This time we take repeated samples (with replacement) from our original sample ```{r} #| include: true #| echo: true consult_boot_ci <- consult |> specify(response = complication, success = "yes") |> generate(reps = 1000, type = "bootstrap") |> calculate(stat = "prop") ``` ------------------------------------------------------------------------ - The 95 % bootstrap confidence interval is obtained by calculating the 2.5% and 97.5% percentiles for the bootstrapped statistics. - We are 95% confident that the consultant's long-run complication rate is between 0 and 0.113 ```{r} #| include: true #| echo: true consult_boot_ci |> summarize(ci_lo = quantile(stat, 0.025), ci_hi = quantile(stat, 0.975)) ```