Inference for a Single Proportion

Medical Consultants

• 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

Hypotheses:

• \(H_0: p = 0.1\)
• \(H_A: p < 0.1\)

Data:

• consult data set is available on our Moodle page
• 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 off 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.

Hypotheses:

• \(H_0: p = 0.5\)
• \(H_A: p > 0.5\)

Data:

• payday data set is available on our Moodle page
• 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

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, \(N(0,1)\)

• 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

Normal model, \(N(0,1)\). P-value is area to the right of Z = 0.765 (shaded region).

• We are unable to reject the null hypothesis (p-value = 0.222)
• 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 100 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 100 simulated samples

• We can do the simulations using the Jamovi module
• Dotplot of 100 simulated proportions assuming \(H_0\)

Approximate null distribution with observed proportion of surgeries with complications (0.048) inticated by dashed vertical line.

• 9 out of 100 simulated proportions were less than or equal to the observed value of \(\hat{p}=0.048\)
• The approximate p-value is 9/100 = 0.09
• A p-value based on more simulations will be more accurate
• Let’s do 1,000 simulations this time

• With a large number of simulations, it is better to visualize the simulated proportions using a histogram

Approximate null distribution with observed proportion of surgeries with complications (0.048) inticated by dashed vertical line.

• 111 out of 1,000 simulated proportions were less than or equal to the observed value of \(\hat{p}=0.048\)
• The approximate p-value is 111/1,000 = 0.111
• With a p-value of 0.111 we are unable to reject the null hypothesis at the \(\alpha=0.05\) level
• It is plausible that the consultant has the same complication rate as the national average of 10%

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%)

• 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% bootstrap percentile confidence interval for the complication rate for the medical consultant
• This time we take repeated samples (with replacement) from our original sample to approximate the variability of the sampling distribution
• This is different from the parametric bootstrap simulation we used for hypothesis testing, because it does not assume \(H_0\) is true

• We will start with 100 bootstraps, because it is easier to visualize

100 bootstrap proportions.

• The 95% bootstrap percentile confidence interval is obtained by calculating the 2.5% and 97.5% percentiles for the bootstrapped statistics
• This gives us an approximate 95% confidence interval for the long-run complication proportion of 0.00766 to 0.105
• We can get a better estimate by increasing the number of bootstraps to 1,000

1,000 bootstrap proportions.

• This gives us an approximate 95% confidence interval for the long-run complication proportion of 0 to 0.113
• We are 95% confident that the consultant’s long-run complication rate is between 0 and 0.113
• Note that the national rate (0.1) is within the confidence interval, indicating that it is plausible that the consultant has the same complication rate as the national average