25:["$","title",null,{"children":"Political polls | Data Viz Insights | Observable"}] 49:Tdd7,# Political polls ## How sure can we be? Note: The question is *not* "How can we be sure?". The reality is that we can never be 100% sure that the results of a political poll are accurate to within some desired precision. The real question is > How sure can we be? Let's try to make this more precise. Suppose there are some number of candidates running for one particular office. We'll call them Candidate 1, Candidate 2, Candidate 3, etc. Suppose we survey ${tex`N`} randomly selected voters and ask them who they intend to vote for. For concreteness sake, maybe ${tex`N=1000`}. Perhaps, we find that - A proportion ${tex`\hat{p}_1`} intend to vote for Candidate 1, - A proportion ${tex`\hat{p}_2`} intend to vote for Candidate 2, - A proportion ${tex`\hat{p}_3`} intend to vote for Candidate 3, - etc. Again, those represent measured proportions. If 450 liked Candidate 1, 400 liked Candidate 2, and 150 liked Candidate 3, then - ${tex`\hat{p}_1 = 0.45`}, - ${tex`\hat{p}_2 = 0.4`}, and - ${tex`\hat{p}_3 = 0.15`}. Here are two questions that arise from this scenario: - If we like Candidate 1, how sure can we that Candidate 1 is really ahead of Candidate 2? - If we wanted to be more sure, how many people should we survey? ## Polling and randomness Let's turn things around a bit. Referring to the same candidates Candidate 1, Candidate 2, Candidate 3, etc. We'll also suppose that there are well defined proportions ${tex`p_1,`} ${tex`p_2,`} ${tex`p_3`}, etc. that represent the actual proportions of voters who intend to vote for those respective candidates. Those proportions are unknown; the purpose of the poll is to approximate them *and* to provide some sort of estimate of how accurate those approximations might be. The numbers ${tex`\hat{p}_1,`} ${tex`\hat{p}_2,`} ${tex`\hat{p}_3`}, etc. that arise from the poll are *approximations* to the actual values. The real question is just how close are those estimates to the actual values? In trying to wrap our heads around this it's important to understand that we've got a *random process*. That is, if we go and select a different random sample, we're going to get somewhat different numbers. If we do that a bunch of times though, it turns out the results are distributed in a predictable way. If we understand that distribution, then we can make probabilistic assessments of our accuracy. That is, we might be able to say things like > The data imply that Candidate 1 is 95% likely to defeat Candidate 2. ## Simulation We can illustrate the process to improve our intuition with a little simulation. For concreteness sake, we might suppose that - ${tex`p_1 = 0.48`}, - ${tex`p_2 = 0.45`}, and - ${tex`n=1000`}. That is, the actual percentage of folks who prefer Candidate 1 is 48%, the actual percentage of folks who prefer Candidate 2 is 45%, and we'll approximate these numbers using a sample with 1000 individuals. Evidently, there are some other candidates that are preferred by the other 7% of the voters. We are interested in comparing the likely winners and, as it turns out, we can do so using only the numbers ${tex`\hat{p}_1`} and ${tex`\hat{p}_2`}, which we will assume represent the proportions of voters who prefer the top two candidates, together with the sample size ${tex`n`}. To evaluate the simulation process, we'll actually need one more parameter, namely ${tex`N`}, which represents the number of simulations. For illustrative purposes, we'll take ${tex`N=2000`}. In computer parlance, we can define and store these parameters like so:4a:T4d4,This run came up with the mathematical approximations - ${tex`p_1 \approx \hat{p}_1 = ${one_simulation.pp1}`} and - ${tex`p_2 \approx \hat{p}_2 = ${one_simulation.pp2}`}. If we run the simulation again, though, those approximations will almost certainly change. In general, they should be kinda close to - ${tex`p_1 = ${default_parameters.p1}`} and - ${tex`p_2 = ${default_parameters.p2}`}. Note that the `simulate` function also returns something called `diff`. This represents the *difference* `pp1-pp2`. Recall that we ultimately want to understand what the likelihood that ${tex.block`p_1 > p_2 \text{ given that } \hat{p}_1 > \hat{p}_2.`} Well, that can be rephrased as the likelihood that ${tex.block`p_1 - p_2 > 0 \text{ given that } \hat{p}_1 - \hat{p}_2 > 0.`} The advantage is that this boils down to the single random expression ${tex`\hat{p}_1 - \hat{p}_2`}, which is represented by `diff` in the simulation. So... here's what we're going to do: We're going to run the `simulation` function a *bunch* of times (say `N=2000`). We're then going to plot a histogram of the resulting `diff`s. Hopefully, that will give us a clue how about how `diff` is distributed. Here are the results of one such experimental run:4b:T4a2,{ let bins = d3 .bin() .value((o) => o.diff) .thresholds(10)(data); let areas = bins.map((d) => d.length / default_parameters.N); let M = d3.max( bins.map((d) => d.length / (default_parameters.N * (d.x1 - d.x0))) ); let m = d3.mean(data, (o) => o.diff); let plot = Plot.plot({ width: 987, height: 610, x: { domain: [-0.08, 0.14] }, marks: [ Plot.rectY(bins, { x1: "x0", x2: "x1", inset: 1, y: (d) => d.length / (default_parameters.N * (d.x1 - d.x0)), fill: (d) => (d.x0 < 0 ? "red" : "steelblue") }), Plot.dot(data, { x: "diff", y: -0.1, fill: "black", opacity: 0.1 }), Plot.line( [ [m, -0.1 * M], [m, 1.1 * M] ], { strokeDasharray: [5, 8], strokeWidth: 3, stroke: "black" } ), Plot.ruleY([0]), Plot.axisX({ y: 0 }), Plot.frame({ anchor: "left" }) // show_curve // ? Plot.line( // d3.range(-0.1, 0.2, 0.001).map((x) => [x, pdf(x)]), // { stroke: "black", strokeWidth: 3 } // ) // : null ] }); return plot; }4c:T733,Generally, a *histogram* is a way to understand the distribution of numeric data. The dots along the horizontal axis in the figure correspond to the actual numeric values of `diff`. The heights of the rectangles are proportional to the number of points in the interval under each rectangle. Thus, the figure paints a vivid picture of how the simulation *most often* produces points near the mean and tapers off as we move away from the mean. The red rectangles correspond to those points where ${tex.block`\hat{p}_1 - \hat{p}_2 < 0`} Those are exactly the points where the poll got the prediction wrong! We can assess the probability that this process gets the result wrong by counting those dots and dividing by the number of times we ran the simulation. Well it turns out that - there are ${data.filter((o) => o.diff < 0).length} dots in the red and - we ran the simulation ${default_parameters.N} times so that - the odds of getting it wrong are about ${tex`\displaystyle \frac{${data.filter((o) => o.diff < 0).length}}{${default_parameters.N}} = ${data.filter((o) => o.diff < 0).length / data.length}`}. To be clear, though, that refers only to this specific difference and sample size. It suggests that if the difference is ${tex`p_1-p_2=0.03`} and the sample size of ${tex`n=1000`}, then we can be ${d3.format("0.2%")(1 - data.filter((o) => o.diff < 0).length / data.length)} confident that our candidate is ahead. It doesn't address other sample sizes or confidence levels. It also doesn't address maybe the most important question: > If we want to have a certain degree of confidence (say 95%), then how many people should we survey? We can get a grip on this by experimenting with the sample size. This was ${tex`n=1000`} in our parameters above but you can view the effect of changing sample size using the radio buttons below:4d:T709,{ return make_histogram(make_parameters(n)); function make_parameters(n) { return { p1: 0.48, // The proportion of voters who lke Candidate 1 p2: 0.45, // The proportion of voters who lke Candidate 2 n: n, // The sample size N: 2000 // Number of times we'll simulate the process. }; } function make_histogram(parameters) { let m0 = parameters.p1 - parameters.p2; let s = Math.sqrt((parameters.p1 + parameters.p2 - m0 ** 2) / parameters.n); let pdf = (x) => Math.exp(-((x - m0) ** 2 / (2 * s ** 2))) / (Math.sqrt(2 * Math.PI) * s); let data = d3.range(parameters.N).map(() => simulate(parameters)); let bins = d3 .bin() .value((o) => o.diff) .thresholds(10)(data); let areas = bins.map((d) => d.length / parameters.N); let M = d3.max(bins.map((d) => d.length / (parameters.N * (d.x1 - d.x0)))); let m = d3.mean(data, (o) => o.diff); let plot = Plot.plot({ width: 987, height: 610, marks: [ Plot.rectY(bins, { x1: "x0", x2: "x1", inset: 1, y: (d) => d.length / (parameters.N * (d.x1 - d.x0)), fill: (d) => (d.x0 < 0 ? "red" : "steelblue") }), Plot.dot(data, { x: "diff", y: -0.1, fill: "black", opacity: 0.2 }), Plot.line( [ [m, -0.1 * M], [m, 1.1 * M] ], { strokeDasharray: [5, 8], strokeWidth: 3, stroke: "black" } ), Plot.ruleY([0]), Plot.axisX({ y: 0 }), Plot.frame({ anchor: "left" }), Plot.line( d3.range(-0.15, 0.25, 0.001).map((x) => [x, pdf(x)]), { stroke: "black", strokeWidth: 3 } ) ] }); return plot; } }4e:Tf98,normal_histogram = { // Name of the "Do it again" button steely_dan_says; // SVG set up let w = 800; let h = 500; let svg = d3 .create("svg") .attr("width", "100%") .attr("viewBox", [0, 0, w, h]) .style("border", "solid 1px black"); let [xmin, xmax, ymin, ymax] = [-3.5, 3.5, -0.2, 1.4]; let pad = 20; let x_scale = d3 .scaleLinear() .domain([xmin, xmax]) .range([pad, w - pad]); let y_scale = d3 .scaleLinear() .domain([ymin, ymax]) .range([h - pad, pad]); let path = d3 .line() .x((d) => x_scale(d[0])) .y((d) => y_scale(d[1])); // Generate some random data let n = 1000; let m = d3.randomUniform(-2, 2)(); let s = d3.randomUniform(0.3, 1.3)(); let pts = d3.range(n).map(d3.randomNormal(m, s)); // Bin the data. // Uncomment the return call to see what I mean let bins = d3.bin().thresholds(20)(pts); // return bins; // The width of each bin and a normalizer // to get the areas of the bins to be 1. let dx = bins[1].x1 - bins[1].x0; let normalizer = 1 / (n * dx); // Set up the rectangles let rect_group = svg .append("g") .attr("fill", "steelblue") .attr("stroke", "white") .attr("stroke-width", 3); rect_group .selectAll("rect") .data(bins) .join("rect") .attr("x", (b) => x_scale(b.x0)) .attr("width", (b) => x_scale(b.x1) - x_scale(b.x0)) .attr("y", y_scale(0)) // Note the rects are initial right on the x-axis .attr("height", 0) // And they are very short .on("pointerenter", function (evt, d) { let rect = d3.select(this); rect.attr("stroke", "black").raise(); svg .append("text") .attr("class", "tip") .attr("x", parseFloat(rect.attr("x")) + (x_scale(dx) - x_scale(0)) / 4) .attr("y", parseFloat(rect.attr("y")) - 10) .text(d.length); }) .on("pointerleave", function () { d3.select(this).attr("stroke", "white"); svg.selectAll(".tip").remove(); }); // Graph the normal curve with the same mean and standard deviation as the data. let f = (x) => Math.exp(-((x - m) ** 2 / (2 * s ** 2))) / Math.sqrt(2 * Math.PI * s ** 2); let graph = svg .append("path") .attr("d", path(d3.range(-4, 4, 0.01).map((x) => [x, f(x)]))) .attr("fill", "none") .attr("stroke", "black") .attr("stroke-width", 3); // Draw the axes svg .append("g") .attr("transform", `translate(0, ${y_scale(0)})`) .call( d3 .axisBottom(x_scale) .tickValues([-3, -2, -1, 1, 2, 3]) .tickFormat(d3.format("d")) .tickSizeOuter(0) ); svg .append("g") .attr("transform", `translate(${x_scale(0)})`) .call(d3.axisLeft(y_scale).tickValues([0.5, 1, 1.5]).tickSizeOuter(0)); // Set up the point group let pt_group = svg.append("g").attr("fill", "black"); pt_group .selectAll("circle") .data(pts) .join("circle") .attr("cx", (pt) => x_scale(pt)) .attr("cy", y_scale(0)) .attr("r", 4) .attr("opacity", 0); // The points are initially invisible // Wait 300 milliseconds then transition the points to fade // in over 600 milliseconds. Promises.delay(300).then(() => pt_group .selectAll("circle") .transition() .duration(600) .attr("opacity", 0.02) ); // Another 1300-900 = 400 milliseconds later, // transition the heights of the rects so that // they're set after another half second. Promises.delay(1300).then(function () { rect_group .selectAll("rect") .transition() .duration(500) .attr("y", (b) => y_scale(normalizer * b.length)) .attr("height", (b) => y_scale(0) - y_scale(normalizer * b.length)); }); // Finally, draw the graph. let length = graph.node().getTotalLength(); graph.attr("stroke-dasharray", [0, length]); Promises.delay(1800).then(() => graph.transition().duration(800).attr("stroke-dasharray", [length, length]) ); return svg.node(); }4f:T428,plot = { if (adjustable_parameters.p1 + adjustable_parameters.p2 <= 1) { let this_pdf = pdf(adjustable_parameters); let xmin = this_pdf.m0 - 3 * this_pdf.s; let xmax = this_pdf.m0 + 3 * this_pdf.s; let plot = Plot.plot({ marks: [ Plot.areaY( d3.range(xmin, 0, Math.abs(xmin) / 400).map((x) => [x, this_pdf(x)]), { x: (d) => d[0], y: (d) => d[1], fill: "red" } ), Plot.areaY( d3.range(0, xmax, xmax / 400).map((x) => [x, this_pdf(x)]), { x: (d) => d[0], y: (d) => d[1], fill: "blue" } ), Plot.line( d3 .range(xmin, xmax, (xmax - xmin) / 800) .map((x) => [x, this_pdf(x)]), { strokeWidth: 3 } ), Plot.ruleX([0]), Plot.ruleY([0]), Plot.axisY({ x: 0 }) ] }); plot.pdf = this_pdf; return plot; } else { return Plot.plot({ marks: [Plot.ruleX([0]), Plot.ruleY([0]), Plot.axisY({ x: 0 })], x: { domain: [-1, 1] }, y: { domain: [0, 1] } }); } }50:[] 29:{"id":"465220d059bc6ecb","slug":null,"title":"Political polls","thumbnail":"ac429ac6831dbb29eec5275cc5dd116c460c4cc1805ac4103733031709be9a5d","access_level":"public","create_time":"$D2023-09-15T15:14:05.791Z","update_time":"$D2023-09-15T15:14:05.791Z","view_time":null,"first_public_time":"$D2023-09-15T15:14:05.791Z","has_unpublished_edits":null,"fork":true,"stars":0,"starred":false,"owner":{"type":"team","id":"e59531852f3a26f9","login":"data-viz-insights","name":"Data Viz Insights","avatar_url":"https://avatars-next.observableusercontent.com/307a2fe144554c3274dc52c733e9dcb2a90d30a16bdd26ad4e2317c6c1f99dab","tier":"education_2026","bio":"","followed_by_you":false,"follows_you":false},"version":180,"publish_version":null,"role":null,"authored":false,"authors":[{"type":"individual","id":"19d5361f4eebf591","login":"owen-vaziri","name":"Owen Vaziri","avatar_url":"https://avatars-next.observableusercontent.com/40bcd7a5c42f0daba336fda0c6bff8c4781340b73acedaa22d611d41f75f66c7","tier":null,"bio":"","followed_by_you":false,"follows_you":false}],"members":[{"type":"team","id":"e59531852f3a26f9","login":"data-viz-insights","name":"Data Viz Insights","avatar_url":"https://avatars-next.observableusercontent.com/307a2fe144554c3274dc52c733e9dcb2a90d30a16bdd26ad4e2317c6c1f99dab","tier":"education_2026","role":"editor"},{"type":"individual","id":"19d5361f4eebf591","login":"owen-vaziri","name":"Owen Vaziri","avatar_url":"https://avatars-next.observableusercontent.com/40bcd7a5c42f0daba336fda0c6bff8c4781340b73acedaa22d611d41f75f66c7","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"lapsed":null,"subscribed":null,"forks":1,"collection":null,"type":"notebook","body":{"titling":"auto","title":"Political polls","cells":[{"id":0,"value":"$49","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":11,"value":"default_parameters = ({\n p1: 0.48, // The proportion of voters who lke Candidate 1\n p2: 0.45, // The proportion of voters who lke Candidate 2\n n: 1000, // The sample size\n N: 2000 // Number of times we'll simulate the process.\n})","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":13,"value":"We can now define a function called `choose` which \"chooses\" either Candidate 1, Candidate 2, or neither depending on the probabilities ${tex`p_1`} and ${tex`p_2`}","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":15,"value":"function choose(parameters) {\n let p = Math.random();\n if (p < parameters.p1) {\n return 1;\n } else if (p < parameters.p1 + parameters.p2) {\n return 2;\n } else {\n return null;\n }\n}","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":17,"value":"We can use this function like so:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":19,"value":"choice = choose(default_parameters)","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":21,"value":"That time, the function chose ${choice ? `Candidate ${choice}` : `neither Candidate 1 nor Candidate 2`}.\n\nWe could run the `choose` function 20 times in a row like so:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":23,"value":"d3.range(20).map(() => choose(default_parameters))","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":26,"value":"Generally, that will return mostly `1`s and `2`s and maybe 1 or 2 or 3 `null`s. If we ran `choose` a *bunch* of times (say `n=1000`) times, we could then compute the proportions of `1`s and `2`s to compute *estimates* of `p1` and `p2`.\n\nHere's a function, called `simulate` that runs one such simulation:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":29,"value":"function simulate(parameters) {\n let d = d3.range(parameters.n).map(() => choose(parameters));\n let pp1 = d.filter((x) => x == 1).length / parameters.n;\n let pp2 = d.filter((x) => x == 2).length / parameters.n;\n\n return { pp1, pp2, diff: pp1 - pp2 };\n}","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":31,"value":"Note that we have\n- computer code symbols `pp1` and `pp2`, which are approximations to `p1` and `p2`, kinda like\n- the mathematical symbols ${tex`\\hat{p}_1`} and ${tex`\\hat{p}_1`}, which are approximations to ${tex`p_1`} and ${tex`p_2`}.\n\nNow, here's how to run the `simulate` function:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":33,"value":"one_simulation = simulate(default_parameters)","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":35,"value":"$4a","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":37,"value":"viewof data = {\n let data = d3\n .range(default_parameters.N)\n .map(() => simulate(default_parameters));\n return Inputs.table(data);\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":40,"value":"And here's a *histogram* of the `diff` column.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":44,"value":"$4b","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":50,"value":"$4c","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":91,"value":"viewof n = Inputs.radio([500, 1000, 2000, 4000], {\n value: 1000,\n label: \"Sample size:\"\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":94,"value":"$4d","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":108,"value":"It certainly looks like our results improve as sample size increases, which makes only all the sense in the world.\n\n## The normal distribution\n\nThe picture also includes the graph of a specific function That's the graph of the so-called *normal distribution* and it's chosen so that its mean and standard deviation agree with the data. The normal distribution *models* the data in a concrete sense and we can use it compute probabilities from the data very efficiently. We can also use it to understand and prove statements like the following:\n\n- The standard deviation of the data decreases with sample size ${tex`n`}.\n - In fact, it decreases like ${tex`1/\\sqrt{n}`}.\n- If a well designed survey of 3000 likely voters indicates that our candidate is 3 percentage points ahead of their closest competitor, then we can be 95% confident that they'll win the election.\n\nThe normal distribution is a workhorse in data analysis since it arises in a natural way from the averaging process. Here's another look at it:","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":115,"value":"viewof steely_dan_says = Inputs.button(\"Do it again\")","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":112,"value":"$4e","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":134,"value":"## A tool to check probabilities of interest\n\nThe tool below allows you to compute the types of probabilities described in this notebook for given values of \n\n- ${tex`\\hat{p}_1`}, the proportion of people who prefer Candidate 1 to Candidate 2,\n- ${tex`\\hat{p}_2`}, the proportion of people who prefer Candidate 1 to Candidate 2, and\n- ${tex`n`}, the overall sample size.\n\nRecall that we must have ${tex`\\hat{p}_1 + \\hat{p}_2 \\leq 1`} and the inequality could be strict if there are other candidates. Generally, we might need to consider all pairs of candidates to predict the winner. In most US elections, though, there are two overwhelming favorites so that we really need to only consider those two, particularly when the winner is determined by plurality.\n\nTo use the tool, simply enter ${tex`\\hat{p}_1`}, ${tex`\\hat{p}_2`}, and ${tex`n`} in the boxes below (or user the sliders). The estimates should automagically appear.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":136,"value":"viewof adjustable_parameters = Inputs.form({\n p1: Inputs.range([0.01, 1], {\n step: 0.0001,\n label: tex`\\hat{p}_1:`,\n value: 0.48\n }),\n p2: Inputs.range([0.01, 1], {\n step: 0.0001,\n label: tex`\\hat{p}_2:`,\n value: 0.48\n }),\n n: Inputs.range([100, 10000], {\n step: 1,\n label: \"Sample size :\",\n value: 1000\n })\n})","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":180,"value":"","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":139,"value":"{\n if (adjustable_parameters.p1 + adjustable_parameters.p2 <= 1) {\n return md`\n- Probability that Candidate 1 defeats Candidate 2: ${d3.format(\n \"0.4\"\n )(d3.max([0, integrate(plot.pdf, 0, plot.pdf.m0 + 4 * plot.pdf.s)]))}\n- Probability that Candidate 2 defeats Candidate 1: ${d3.format(\n \"0.4\"\n )(d3.max([0, integrate(plot.pdf, plot.pdf.m0 - 4 * plot.pdf.s, 0)]))}\n`;\n } else {\n return md`The parameters need to satisfy ${tex`p_1+p_2<1/2`}.`;\n }\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":141,"value":"$4f","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":176,"value":"617.0,0.5130942,0.3512664","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":144,"value":"The tool works by using a numerical integrator to compute the area under the normal curve\n${tex.block`f(x) = \\frac{1}{\\sqrt{2\\pi}\\sigma} e^{-(x-\\mu)^2/(2\\sigma^2)},`}\nwith mean ${tex`\\mu`} and standard deviatin ${tex`\\sigma`} determined by the sample size ${tex`n`} and the computed proportions ${tex`\\hat{p}_1`} and ${tex`\\hat{p}_2`} via \n- ${tex`\\mu = \\hat{p}_1-\\hat{p}_2`} and\n- ${tex`\\sigma = \\sqrt{(\\hat{p}_1+\\hat{p}_2 + (\\hat{p}_1-\\hat{p}_2)^2))/n}`}\nas described in [this paper](https://abcnews.go.com/images/PollingUnit/MOEFranklin.pdf).","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":166,"value":"## Code","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"","database":"$undefined"},{"id":132,"value":"function pdf(parameters) {\n let m0 = parameters.p1 - parameters.p2;\n let s = Math.sqrt((parameters.p1 + parameters.p2 - m0 ** 2) / parameters.n);\n let pdf = (x) =>\n Math.exp(-((x - m0) ** 2 / (2 * s ** 2))) / (Math.sqrt(2 * Math.PI) * s);\n\n pdf.m0 = m0;\n pdf.s = s;\n\n return pdf;\n}","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":130,"value":"function integrate(ff, a, b) {\n return adaptiveSimpson(ff, a, b, {\n tolerance: 1e-11,\n maxDepth: 8,\n minDepth: 7\n });\n}","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"},{"id":123,"value":"import { adaptiveSimpson } from \"@rreusser/integration@3056\"","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","tool":"$undefined","providerMetadata":"$undefined","attachments":"$undefined","output":"$undefined","database":"$undefined"}],"files":"$W50","stdlib":"1"}} 24:null