inference-one-prop.qmd

# Inference for a single proportion {#sec-inference-one-prop}

```{r}
#| include: false
source("_common.R")
```

\vspace{-10mm}

::: {.chapterintro data-latex=""}
Focusing now on statistical inference for categorical data, we will revisit many of the foundational aspects of hypothesis testing from [Chapter -@sec-foundations-randomization].

The three data structures we detail are one binary variable, summarized using a single proportion; two binary variables, summarized using a difference of two proportions; and two categorical variables, summarized using a two-way table.
When appropriate, each of the data structures will be analyzed using the three methods from [Chapter -@sec-foundations-randomization], [Chapter -@sec-foundations-bootstrapping], and [Chapter -@sec-foundations-mathematical]: randomization test, bootstrapping, and mathematical models, respectively.

As we build on the inferential ideas, we will visit new foundational concepts in statistical inference.
For example, we will cover the conditions for when a normal model is appropriate; the two different error rates in hypothesis testing; and choosing the confidence level for a confidence interval.
:::

We encountered inference methods for a single proportion in [Chapter -@sec-foundations-bootstrapping], exploring point estimates and confidence intervals.
In this section, we'll do a review of these topics and how to choose an appropriate sample size when collecting data for single proportion contexts.

Note that there is only one variable being measured in a study which focuses on one proportion.
For each observational unit, the single variable is measured as either a success or failure (e.g., "surgical complication" vs. "no surgical complication").
Because the nature of the research question at hand focuses on only a single variable, there is not a way to randomize the variable across a different (explanatory) variable.
For this reason, we will not use randomization as an analysis tool when focusing on a single proportion.
Instead, we will apply bootstrapping techniques to test a given hypothesis, and we will also revisit the associated mathematical models.

\vspace{-3mm}

## Bootstrap test for a proportion {#sec-one-prop-null-boot}

The bootstrap simulation concept when $H_0$ is true is similar to the ideas used in the case studies presented in [Chapter -@sec-foundations-bootstrapping] where we bootstrapped without an assumption about $H_0.$ Because we will be testing a hypothesized value of $p$ (referred to as $p_0),$ the bootstrap simulation for hypothesis testing has a fantastic advantage that it can be used for any sample size (a huge benefit for small samples, a nice alternative for large samples).

We expand on the medical consultant example from @sec-case-study-med-consult, but instead of finding an interval estimate for the true complication rate, we test a specific claim.

### Observed data

Recall the set-up for the example:
People providing an organ for donation sometimes seek the help of a special "medical consultant".
These consultants assist the patient in all aspects of the surgery, with the goal of reducing the possibility of complications during the medical procedure and recovery.
Patients might choose a consultant based in part on the historical complication rate of the consultant's clients.
One consultant tried to attract patients by noting the average complication rate for liver donor surgeries in the US is about 10%, but her clients have only had 3 complications in the 62 liver donor surgeries she has facilitated.
She claims this is strong evidence that her work meaningfully contributes to reducing complications (and therefore she should be hired!).

::: {.workedexample data-latex=""}
Using the data, is it possible to assess the consultant's claim that her complication rate is less than 10%?

------------------------------------------------------------------------

No.
The claim is that there is a causal connection, but the data are observational.
Patients who hire this medical consultant may have lower complication rates for other reasons.

While it is not possible to assess this causal claim, it is still possible to test for an association using these data.
For this question we ask, could the low complication rate of $\hat{p} = 0.0484$ have simply occurred by chance, if her complication rate does not differ from the US standard rate?
:::

::: {.guidedpractice data-latex=""}
Write out hypotheses in both plain and statistical language to test for the association between the consultant's work and the true complication rate, $p,$ for the consultant's clients.[^16-inference-one-prop-1]
:::

[^16-inference-one-prop-1]: $H_0:$ There is no association between the consultant's contributions and the clients' complication rate.
    In statistical language, $p = 0.10.$ $H_A:$ Patients who work with the consultant tend to have a complication rate lower than 10%, i.e., $p < 0.10.$

Because, as it turns out, the conditions of working with the normal distribution are not met (see @sec-one-prop-norm), the uncertainty associated with the sample proportion should not be modeled using the normal distribution, as doing so would underestimate the uncertainty associated with the sample statistic.
However, we would still like to assess the hypotheses from the previous Guided Practice in absence of the normal framework.
To do so, we need to evaluate the possibility of a sample value $(\hat{p})$ as far below the null value, $p_0 = 0.10$ as what was observed.
The deviation of the sample value from the hypothesized parameter is usually quantified with a p-value.

The p-value is computed based on the null distribution, which is the distribution of the test statistic if the null hypothesis is true.
Supposing the null hypothesis is true, we can compute the p-value by identifying the probability of observing a test statistic that favors the alternative hypothesis at least as strongly as the observed test statistic.
Here we will use a bootstrap simulation to calculate the p-value.

### Variability of the statistic

We want to identify the sampling distribution of the test statistic $(\hat{p})$ if the null hypothesis was true.
In other words, we want to see the variability we can expect from sample proportions if the null hypothesis was true.
Then we plan to use this information to decide whether there is enough evidence to reject the null hypothesis.

Under the null hypothesis, 10% of liver donors have complications during or after surgery.
Suppose this rate was really no different for the consultant's clients (for *all* the consultant's clients, not just the 62 previously measured).
If this was the case, we could *simulate* 62 clients to get a sample proportion for the complication rate from the null distribution.
Simulating observations using a hypothesized null parameter value is often called a **parametric bootstrap simulation**\index{parametric bootstrap}, but we will refer to it descriptively as "simulating under the null hypothesis claim."

```{r}
#| include: false
terms_chp_16 <- c("parametric bootstrap")
```

Similar to the process described in [Chapter -@sec-foundations-bootstrapping], each client can be simulated using a bag of marbles with 10% red marbles and 90% white marbles.
Sampling a marble from the bag (with 10% red marbles) is one way of simulating whether a patient has a complication *if the true complication rate is 10%*.
If we select 62 marbles and then compute the proportion of patients with complications in the simulation, $\hat{p}_{sim1},$ then the resulting sample proportion is a sample from the null distribution.

There were 5 simulated cases with a complication and 57 simulated cases without a complication, i.e., $\hat{p}_{sim1} = 5/62 = 0.081.$

::: {.workedexample data-latex=""}
Is this one simulation enough to determine whether we should reject the null hypothesis?

------------------------------------------------------------------------

No.
To assess the hypotheses, we need to see a distribution of many values of $\hat{p}_{sim},$ not just a *single* draw from this sampling distribution.
:::

### Observed statistic vs. null statistics

One simulation isn't enough to get a sense of the null distribution; many simulation studies are needed.
Roughly 10,000 seems sufficient.
However, paying someone to simulate 10,000 studies by hand is a waste of time and money.
Instead, simulations are typically programmed into a computer, which is much more efficient.

```{r}
#| include: false
set.seed(334422)
medical_consultant_sim_dist <- tibble(stat = rbinom(10000, 62, 0.1)/62)

medical_consultant_n_sim <- medical_consultant_sim_dist |>
  filter(stat <= 0.0484) |>
  nrow()

medical_consultant_p_val <- round(medical_consultant_n_sim / 10000, 3)
```

@fig-nullDistForPHatIfLiverTransplantConsultantIsNotHelpful shows the results of 10,000 simulated studies.
The proportions that are equal to or less than $\hat{p} = 0.0484$ are shaded.
The shaded areas represent sample proportions under the null distribution that provide at least as much evidence as $\hat{p}$ favoring the alternative hypothesis.
There were `r medical_consultant_n_sim` simulated sample proportions with $\hat{p}_{sim} \leq 0.0484.$ We use these to construct the null distribution's left-tail area and find the p-value:

$$\text{left tail area} = \frac{\text{Number of observed simulations with }\hat{p}_{sim} \leq \text{ 0.0484}}{10000}$$

Of the 10,000 simulated $\hat{p}_{sim},$ `r medical_consultant_n_sim` were equal to or smaller than $\hat{p}.$ Since the hypothesis test is one-sided, the estimated p-value is equal to this tail area: `r medical_consultant_p_val`.

```{r}
#| label: fig-nullDistForPHatIfLiverTransplantConsultantIsNotHelpful
#| fig-cap: |
#|   The null distribution for $\hat{p},$ created from 10,000 simulated studies. 
#|   The left tail, representing the p-value for the hypothesis test is colored
#|   in blue.
#| fig-alt: |
#|   Histogram of 10,000 simulated sample proportions,
#|   from the null distribution, where the true proportion is 0.1. The left tail, representing the 
#|   p-value for the hypothesis test, is colored in blue.
#| fig-asp: 0.4
#| fig-width: 7
ggplot(medical_consultant_sim_dist, aes(x = stat)) +
  geom_histogram(binwidth = 0.0167) +
  gghighlight(stat <= 00.0484) +
  scale_x_continuous(breaks = seq(0, 0.25, 0.05), labels = label_number(accuracy = 0.01)) +
  labs(
    x = expression(hat(p)[sim]),
    y = "Number of\nsimulated scenarios"
  )
```

::: {.guidedpractice data-latex=""}
Because the estimated p-value is `r medical_consultant_p_val`, which is larger than the discernibility level 0.05, we cannot reject the null hypothesis.
Explain what this means in plain language in the context of the problem.[^16-inference-one-prop-2]
:::

[^16-inference-one-prop-2]: There is not enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
    We cannot conclude that there is evidence that the consultant's surgery complication rate is lower than the US standard rate of 10%.  That said, we also cannot conclude that there is evidence that the consultant's surgery complication rate is higher than the US standard rate of 10%.  When the p-value is larger than the discernibility level, we are unable to make conclusions about the research statement.

::: {.guidedpractice data-latex=""}
Does the conclusion in the previous Guided Practice imply the consultant is good at their job?
Explain.[^16-inference-one-prop-3]
:::

[^16-inference-one-prop-3]: Not necessarily.
    There is no evidence to make a claim in either direction, so we cannot make any claims about whether the consultant is good at their job.

::: {.important data-latex=""}
**Null distribution of** $\hat{p}$ **with bootstrap simulation.**

Regardless of the statistical method chosen, the p-value is always derived by analyzing the null distribution of the test statistic.
The normal model poorly approximates the null distribution for $\hat{p}$ (the sample proportion) when the success-failure condition is not satisfied.
As a substitute, we can generate the null distribution using simulated sample proportions and use this distribution to compute the tail area, i.e., the p-value.
:::

\vspace{-5mm}

In the previous Guided Practice, the p-value is *estimated*.
It is not exact because the simulated null distribution itself is only a close approximation of the sampling distribution of the sample statistic.
An exact p-value can be generated using the binomial distribution, but that method will not be covered in this text.

## Mathematical model for a proportion {#sec-one-prop-norm}

### Conditions

In @sec-normalDist, we introduced the normal distribution and showed how it can be used as a mathematical model to describe the variability of a statistic.
There are conditions under which a sample proportion $\hat{p}$ is well modeled with a normal distribution.
When the observations are independent and the sample size is sufficiently large, the normal model will describe the sampling distribution of the sample proportion quite well; when the observations violate the conditions, the normal model can be inaccurate.
Particularly, it can underestimate the variability of the sample proportion.

::: {.important data-latex=""}
**Sampling distribution of** $\hat{p}.$

The sampling distribution for $\hat{p}$ (the sample proportion) based on a sample of size $n$ from a population with a true proportion $p$ is nearly normal when:

1.  The sample's observations are independent, e.g., are from a simple random sample.
2.  We expected to see at least 10 successes and 10 failures in the sample, i.e., $np\geq10$ and $n(1-p)\geq10.$ This is called the **success-failure condition**.

When these conditions are met, then the sampling distribution of $\hat{p}$ is nearly normal with mean $p$ and standard error of $\hat{p}$ as $SE = \sqrt{\frac{\ \hat{p}(1-\hat{p})\ }{n}}.$
:::

Recall that the margin of error is defined by the standard error.
The margin of error for $\hat{p}$ can be directly obtained from $SE(\hat{p}).$

::: {.important data-latex=""}
**Margin of error for** $\hat{p}.$

The margin of error is $z^\star \times \sqrt{\frac{\ \hat{p}(1-\hat{p})\ }{n}}$ where $z^\star$ is calculated from a specified percentile on the normal distribution.
:::

\index{success-failure condition} \index{standard error (SE)!single proportion}\index{single proportion!standard error (SE)}

```{r}
#| include: false
terms_chp_16 <- c(terms_chp_16, "success-failure condition", "SE single proportion")
```

Typically we do not know the true proportion $p,$ so we substitute some value to check conditions and estimate the standard error.
For confidence intervals, the sample proportion $\hat{p}$ is used to check the success-failure condition and compute the standard error.
For hypothesis tests, typically the null value -- that is, the proportion claimed in the null hypothesis -- is used in place of $p.$

The independence condition is a more nuanced requirement.
When it isn't met, it is important to understand how and why it is violated.
For example, there exist no statistical methods available to truly correct the inherent biases of data from a convenience sample.
On the other hand, if we took a cluster sample (see @sec-samp-methods), the observations wouldn't be independent, but suitable statistical methods are available for analyzing the data (but they are beyond the scope of even most second or third courses in statistics).

::: {.workedexample data-latex=""}
In the examples based on large sample theory, we modeled $\hat{p}$ using the normal distribution.
Why is this not appropriate for the case study on the medical consultant?

------------------------------------------------------------------------

The independence assumption may be reasonable if each of the surgeries is from a different surgical team.
However, the success-failure condition is not satisfied.
Under the null hypothesis, we would anticipate seeing $62 \times 0.10 = 6.2$ complications, not the 10 required for the normal approximation.
:::

While this book is scoped to well-constrained statistical problems, do remember that this is just the first book in what is a large library of statistical methods that are suitable for a very wide range of data and contexts.

### Confidence interval for a proportion

\index{point estimate!single proportion}
\index{single proportion!point estimate}

A confidence interval provides a range of plausible values for the parameter $p,$ and when $\hat{p}$ can be modeled using a normal distribution, the confidence interval for $p$ takes the form $\hat{p} \pm z^{\star} \times SE.$ We have seen $\hat{p}$ to be the sample proportion.
The value $z^{\star}$ determines the confidence level (previously set to be 1.96) and will be discussed in detail in the examples following.
The value of the standard error, $SE,$ depends heavily on the sample size.

::: {.important data-latex=""}
**Standard error of one proportion,** $\hat{p}.$

When the conditions are met so that the distribution of $\widehat{p}$ (the sample proportion) is nearly normal, the **variability** of a single proportion, $\widehat{p}$ is well described by:

\vspace{-5mm}

$$SE(\hat{p}) = \sqrt{\frac{p(1-p)}{n}}$$

Note that we almost never know the true value of $p$ (the population probability or proportion). A more helpful formula to use is:

\vspace{-5mm}

$$
SE(\hat{p}) \approx \sqrt{\frac{(\mbox{best guess of }p)(1 - \mbox{best guess of }p)}{n}}
$$

For hypothesis testing, we use $p_0$ (the proportion specified in the null hypothesis) as the best guess of $p.$ For confidence intervals, we use $\widehat{p}$ as the best guess of $p.$
:::

::: {.guidedpractice data-latex=""}
Consider taking many polls of registered voters (i.e., random samples) of size 300 asking them if they support legalized marijuana.
It is suspected that about 2/3 of all voters support legalized marijuana.
To understand how the sample proportion $(\hat{p})$ would vary across the samples, calculate the standard error of $\hat{p}.$[^16-inference-one-prop-4]
:::

[^16-inference-one-prop-4]: Because the $p$ is unknown but expected to be around 2/3, we will use 2/3 in place of $p$ in the formula for the standard error.
    $SE = \sqrt{\frac{p(1-p)}{n}} \approx \sqrt{\frac{2/3 (1 - 2/3)} {300}} = 0.027.$

### Variability of the sample proportion

::: {.workedexample data-latex=""}
A simple random sample of 826 payday loan borrowers was surveyed to better understand their interests around regulation and costs.
70% of the responses supported new regulations on payday lenders.

1.  Is it reasonable to model the distribution of $\hat{p}$ using a normal distribution?

2.  Estimate the standard error of $\hat{p}.$

3.  Construct a 95% confidence interval for $p,$ the proportion of payday borrowers who support increased regulation for payday lenders.

------------------------------------------------------------------------

1.  The data are a random sample, so it is reasonable to assume that the observations are independent and representative of the population of interest. We also must check the success-failure condition, using $\hat{p}$ in place of $p$ when computing a confidence interval. Since both values are at least 10, we can use the normal distribution to model $\hat{p}.$

$$
\begin{aligned}
  \text{Support: }
      n p &
          \approx 826 \times 0.70
      = 578\\
  \text{Not: }
      n (1 - p) &
        \approx 826 \times (1 - 0.70)
      = 248
\end{aligned}
$$
2.  Because $p$ is unknown and the standard error is for a confidence interval, use $\hat{p}$ in place of $p$ in the formula.

$$SE = \sqrt{\frac{p(1-p)}{n}} \approx \sqrt{\frac{0.70 (1 - 0.70)} {826}} = 0.016.$$

3.  Using $\hat{p} = 0.70$, $z^{\star} = 1.96$ for a 95% confidence interval, and the standard error $SE = 0.016$ from the previous Guided Practice, the confidence interval is

$$ 
\begin{aligned}
\text{point estimate} \ &\pm \ z^{\star} \times \ SE \\
0.70 \ &\pm \ 1.96 \ \times \ 0.016 \\ 
(0.669 \ &, \ 0.731)
\end{aligned}
$$
We are 95% confident that the true proportion of payday borrowers who supported regulation at the time of the poll was between 0.669 and 0.731.
:::

::: {.important data-latex=""}
**Constructing a confidence interval for a single proportion.**

There are three steps to constructing a confidence interval for $p$ (the true population proportion or probability).

1.  Check if it seems reasonable to assume the observations are independent and check the success-failure condition using $\hat{p}$ (the sample proportion). If the conditions are met, the sampling distribution of $\hat{p}$ may be well-approximated by the normal model.
2.  Calculate the standard error using $\hat{p}$ instead of $p$.
3.  Apply the general confidence interval formula.
:::

For additional one-proportion confidence interval examples, see @sec-ConfidenceIntervals.

\clearpage

### Changing the confidence level

\index{confidence level}

Suppose we want to consider confidence intervals where the confidence level is somewhat higher than 95%: perhaps we would like a confidence level of 99%.
Think back to the analogy about trying to catch a fish: if we want to be more sure that we will catch the fish, we should use a wider net.
To create a 99% confidence level, we must also widen our 95% interval.
On the other hand, if we want an interval with lower confidence, such as 90%, we could make our original 95% interval slightly slimmer.

The 95% confidence interval structure provides guidance in how to make intervals with new confidence levels.
Below is a general 95% confidence interval for a point estimate that comes from a nearly normal distribution:

\vspace{-5mm}

$$
\text{point estimate} \ \pm \ 1.96 \ \times \ SE
$$

There are three components to this interval: the point estimate, "1.96", and the standard error.
The choice of $1.96 \times SE$ was based on capturing 95% of the data since the estimate is within 1.96 standard errors of the true value about 95% of the time.
1.96 corresponds to the 95% confidence level.

::: {.guidedpractice data-latex=""}
If $X$ is a normally distributed random variable, how often will $X$ be within 2.58 standard deviations of the mean?[^16-inference-one-prop-5]
:::

[^16-inference-one-prop-5]: This is equivalent to asking how often the $Z$ score will be larger than -2.58 but less than 2.58.
    (For a picture, see @fig-choosingZForCI.) To determine this probability, look up -2.58 and 2.58 in the normal probability table (0.0049 and 0.9951).
    Thus, there is a $0.9951-0.0049 \approx 0.99$ probability that the unobserved random variable $X$ will be within 2.58 standard deviations of the mean.

```{r}
#| label: fig-choosingZForCI
#| fig-cap: |
#|   The area between -$z^{\star}$ and $z^{\star}$ increases as $|z^{\star}|$ 
#|   becomes larger. If the confidence level is 99%, we choose $z^{\star}$ such 
#|   that 99% of the normal curve is between -$z^{\star}$ and $z^{\star},$ 
#|   which corresponds to 0.5% in the lower tail and 0.5% in the upper 
#|   tail: $z^{\star}=2.58.$
#| fig-alt: |
#|   A normal curve with 99% of the area between negative z star and positive z star.
#|   The area between negative z star and positive z star increases as $| z star
#|   becomes larger. If the confidence level is 99 percent, we choose z star such
#|   that 99 percent of the normal curve is between negative z star and positive z star,
#|   which corresponds to 0.5 percent  in the lower tail and 0.5 percent in the upper
#|   tail.  The z star value is 2.58.
#| fig-asp: 0.5
par(mar = c(3.3, 1, .5, 1), mgp = c(2.1, 0.6, 0))
X <- rev(seq(-4, 4, 0.025))
Y <- dt(X, 10) # makes better visual
plot(X, Y, type = "l", xlab = "Standard deviations from the mean", ylab = "", axes = FALSE, xlim = 3.3 * c(-1, 1), ylim = c(0, 0.59), col = IMSCOL["gray", "full"])
axis(1, at = -3:3)
abline(h = 0) 
yMax <- 0.41
X <- seq(-4, 4, 0.025)
Y <- dt(X, 10) # makes better visual
lines(X, Y, col = IMSCOL["gray", "full"])

these <- (X < 2.58 & X > -2.58)
x <- c(-2.58, X[these], 2.58)
y <- c(0, dt(X[these], 10), 0)
polygon(x, y, col = IMSCOL["blue", "f2"], border = "#00000000")
these <- (X < 1.96 & X > -1.96)
x <- c(-1.96, X[these], 1.96)
y <- c(0, dt(X[these], 10), 0)
polygon(x, y, col = IMSCOL["blue", "full"], border = "#00000000")

lines(1.96 * c(-1, 1), rep(yMax, 2), lwd = 2)
lines(rep(-1.96, 2), c(0, yMax), lty = 2, col = IMSCOL["gray", "full"])
lines(rep(1.96, 2), c(0, yMax), lty = 2, col = IMSCOL["gray", "full"])
text(0, yMax, "95%, extends -1.96 to 1.96", pos = 3)

yMax <- 0.53
lines(2.58 * c(-1, 1), rep(yMax, 2), lwd = 2)
lines(rep(-2.58, 2), c(0, yMax), lty = 2, col = "#00000055")
lines(rep(2.58, 2), c(0, yMax), lty = 2, col = "#00000055")
text(0, yMax, "99%, extends -2.58 to 2.58", pos = 3)
```

\index{confidence interval}\index{Z-CI!single proportion}

To create a 99% confidence interval, change 1.96 in the 95% confidence interval formula to be $2.58.$ The previous Guided Practice highlights that 99% of the time a normal random variable will be within 2.58 standard deviations of its mean.
This approach -- using the Z scores in the normal model to compute confidence levels -- is appropriate when the point estimate is associated with a normal distribution and we can properly compute the standard error.
Thus, the formula for a 99% confidence interval is:

$$
\text{point estimate} \ \pm \ 2.58 \ \times \ SE
$$

The normal approximation is crucial to the precision of the $z^\star$ confidence intervals (in contrast to the bootstrap percentile confidence intervals).
When the normal model is not a good fit, we will use alternative distributions that better characterize the sampling distribution or we will use bootstrapping procedures.

::: {.guidedpractice data-latex=""}
Create a 99% confidence interval for the impact of the stent on the risk of stroke using the data from @sec-case-study-stents-strokes.
The point estimate is 0.090, and the standard error is $SE = 0.028.$ It has been verified for you that the point estimate can reasonably be modeled by a normal distribution.[^16-inference-one-prop-6]
:::

[^16-inference-one-prop-6]: Since the necessary conditions for applying the normal model have already been checked for us, we can go straight to the construction of the confidence interval: $\text{point estimate} \pm 2.58 \times SE$ Which gives an interval of (0.018, 0.162).\$ We are 99% confident that implanting a stent in the brain of a patient who is at risk of stroke increases the risk of stroke within 30 days by a rate of 0.018 to 0.162 (assuming the patients are representative of the population).

::: {.important data-latex=""}
**Mathematical model confidence interval for any confidence level.**

If the point estimate follows the normal model with standard error $SE,$ then a confidence interval for the population parameter is

$$
\text{point estimate} \ \pm \ z^{\star} \ \times \ SE
$$

where $z^{\star}$ corresponds to the confidence level selected.
:::

@fig-choosingZForCI provides a picture of how to identify $z^{\star}$ based on a confidence level.
We select $z^{\star}$ so that the area between -$z^{\star}$ and $z^{\star}$ in the normal model corresponds to the confidence level.

::: {.guidedpractice data-latex=""}
Previously, we found that implanting a stent in the brain of a patient at risk for a stroke *increased* the risk of a stroke.
The study estimated a 9% increase in the number of patients who had a stroke, and the standard error of this estimate was about $SE = 2.8%.$ Compute a 90% confidence interval for the effect.[^16-inference-one-prop-7]
:::

[^16-inference-one-prop-7]: We must find $z^{\star}$ such that 90% of the distribution falls between -$z^{\star}$ and $z^{\star}$ in the standard normal model, $N(\mu=0, \sigma=1).$ We can look up -$z^{\star}$ in the normal probability table by looking for a lower tail of 5% (the other 5% is in the upper tail), thus $z^{\star} = 1.65.$ The 90% confidence interval can then be computed as $\text{point estimate} \pm 1.65 \times SE \to (4.4\%, 13.6\%).$ (Note: the conditions for normality had earlier been confirmed for us.) That is, we are 90% confident that implanting a stent in a stroke patient's brain increased the risk of stroke within 30 days by 4.4% to 13.6%.\
    Note, the problem was set up as 90% to indicate that there was not a need for a high level of confidence (such as 95% or 99%).
    A lower degree of confidence increases potential for error, but it also produces a more narrow interval.

### Hypothesis test for a proportion

One possible regulation for payday lenders is that they would be required to do a credit check and evaluate debt payments against the borrower's finances.
We would like to know: would borrowers support this form of regulation?

::: {.guidedpractice data-latex=""}
Set up hypotheses to evaluate whether borrowers have a majority support for this type of regulation.[^16-inference-one-prop-8]
:::

[^16-inference-one-prop-8]: $H_0:$ there is not support for the regulation; $H_0:$ $p \leq 0.50.$ $H_A:$ the majority of borrowers support the regulation; $H_A:$ $p > 0.50.$

To apply the normal distribution framework in the context of a hypothesis test for a proportion, the independence and success-failure conditions must be satisfied.
In a hypothesis test, the success-failure condition is checked using the null proportion: we verify $np_0$ and $n(1-p_0)$ are at least 10, where $p_0$ is the null value.

::: {.important data-latex=""}
**The test statistic for assessing a single proportion is a Z.**

The **Z score**\index{Z score}\index{Z score!single proportion}\index{single proportion!Z score} is a ratio of how the sample proportion differs from the hypothesized proportion $(p_0)$ as compared to the expected variability of the $\hat{p}$ (sample proportion) values.

$$
Z = \frac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}}
$$

When the null hypothesis is true and the conditions are met, Z has a standard normal distribution.

Conditions:

-   independent observations\
-   large samples $(n p_0 \geq 10$ and $n (1-p_0) \geq 10)$\
:::

```{r}
#| include: false
terms_chp_16 <- c(terms_chp_16, "Z score")
```

::: {.guidedpractice data-latex=""}
Do payday loan borrowers support a regulation that would require lenders to pull their credit report and evaluate their debt payments?
From a random sample of 826 borrowers, 51% said they would support such a regulation.
Is it reasonable to use a normal distribution to model $\hat{p}$ for a hypothesis test here?[^16-inference-one-prop-9]
:::

[^16-inference-one-prop-9]: Independence holds since the poll is based on a random sample.
    The success-failure condition also holds, which is checked using the null value $(p_0 = 0.5)$ from $H_0:$ $np_0 = 826 \times 0.5 = 413,$ $n(1 - p_0) = 826 \times 0.5 = 413.$ Recall that here, the best guess for $p$ is $p_0$ which comes from the null hypothesis (because we assume the null hypothesis is true when performing the testing procedure steps).
    $H_0:$ there is not support for the regulation; $H_0:$ $p \leq 0.50.$ $H_A:$ the majority of borrowers support the regulation; $H_A:$ $p > 0.50.$

::: {.important data-latex=""}
**Mathematical model hypothesis test for a proportion.**

Set up hypotheses and verify the conditions using the null value, $p_0,$ to ensure $\hat{p}$ (the sample proportion) is nearly normal under $H_0.$ If the conditions hold, calculate the standard error, again using $p_0,$ and show the p-value in a drawing.
Lastly, compute the p-value and evaluate the hypotheses.
:::

For additional one-proportion hypothesis test examples, see @sec-HypothesisTesting.

::: {.workedexample data-latex=""}
Using the hypotheses and data from the previous Guided Practices, evaluate whether the poll on lending regulations provides convincing evidence that a majority of payday loan borrowers support a new regulation that would require lenders to pull credit reports and evaluate debt payments.

------------------------------------------------------------------------

With hypotheses already set up and conditions checked, we can move onto calculations.
The standard error in the context of a one-proportion hypothesis test is computed using the null value, $p_0:$

\vspace{-5mm}

$$
SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.5 (1 - 0.5)}{826}} = 0.017
$$

A picture of the normal model is shown below with the p-value represented by the shaded region.

```{r}
#| label: normTail-51
#| fig-alt: |
#|   A normal curve centered at 0.5 with a standard deviation of 0.017. 
#|   The area to the right of 0.51 is shaded.
#| fig-asp: 0.4
#| out-width: 60%
#| fig-align: center
par(mar = c(2, 0, 0, 0))
normTail(0.5, 0.017, U = 0.51, col = IMSCOL["blue", "full"])  
```

Based on the normal model, the test statistic can be computed as the Z score of the point estimate:

\vspace{-5mm}

$$
Z = \frac{\text{point estimate} - \text{null value}}{SE} = \frac{0.51 - 0.50}{0.017} = 0.59
$$

The single tail area which represents the p-value is 0.2776.
Because the p-value is larger than 0.05, we do not reject $H_0.$ The poll does not provide convincing evidence that a majority of payday loan borrowers support regulations around credit checks and evaluation of debt payments.

In @sec-two-prop-errors we discuss two-sided hypothesis tests of which the payday example may have been better structured.
That is, we might have wanted to ask whether the borrowers **support or oppose** the regulations (to study opinion in either direction away from the 50% benchmark).
In that case, the p-value would have been doubled to 0.5552 (again, we would not reject $H_0).$ In the two-sided hypothesis setting, the appropriate conclusion would be to claim that the poll does not provide convincing evidence that a majority of payday loan borrowers support or oppose regulations around credit checks and evaluation of debt payments.

In both the one-sided or two-sided setting, the conclusion is somewhat unsatisfactory because there is no conclusion.
That is, there is no resolution one way or the other about public opinion.
We cannot claim that exactly 50% of people support the regulation, but we cannot claim a majority in either direction.
:::

### Violating conditions

We've spent a lot of time discussing conditions for when $\hat{p}$ can be reasonably modeled by a normal distribution.
What happens when the success-failure condition fails?
What about when the independence condition fails?
In either case, the general ideas of confidence intervals and hypothesis tests remain the same, but the strategy or technique used to generate the interval or p-value change.

When the success-failure condition isn't met for a hypothesis test, we can simulate the null distribution of $\hat{p}$ using the null value, $p_0,$ as seen in @sec-one-prop-null-boot.
Unfortunately, methods for dealing with observations which are not independent (e.g., repeated measurements on subjects where measurements are taken pre and post study) are outside the scope of this book.

\clearpage

## Chapter review {#sec-chp16-review}

### Summary

Building on the foundational ideas from the previous few ideas, this chapter focused exclusively on the single population proportion as the parameter of interest.
Note that it is not possible to do a randomization test with only one variable, so to do computational hypothesis testing, we applied a bootstrapping framework.
The bootstrap confidence interval and the mathematical framework for both hypothesis testing and confidence intervals are similar to those applied to other data structures and parameters.
When using the mathematical model, keep in mind the success-failure conditions.
Additionally, know that bootstrapping is always more accurate with larger samples.

### Terms

The terms introduced in this chapter are presented in @tbl-terms-chp-16.
If you're not sure what some of these terms mean, we recommend you go back in the text and review their definitions.
You should be able to easily spot them as **bolded text**.

```{r}
#| label: tbl-terms-chp-16
#| tbl-cap: Terms introduced in this chapter.
#| tbl-pos: H
make_terms_table(terms_chp_16)
```

\clearpage

## Exercises {#sec-chp16-exercises}

Answers to odd-numbered exercises can be found in [Appendix -@sec-exercise-solutions-16].

::: {.exercises data-latex=""}
{{< include exercises/_16-ex-inference-one-prop.qmd >}}
:::