{"id":1390,"date":"2024-10-15T13:39:54","date_gmt":"2024-10-15T13:39:54","guid":{"rendered":"https:\/\/www.allendowney.com\/blog\/?p=1390"},"modified":"2024-10-15T13:39:54","modified_gmt":"2024-10-15T13:39:54","slug":"bootstrapping-a-proportion","status":"publish","type":"post","link":"https:\/\/www.allendowney.com\/blog\/2024\/10\/15\/bootstrapping-a-proportion\/","title":{"rendered":"Bootstrapping a Proportion"},"content":{"rendered":"\n
\n

It’s another installment in Data Q&A: Answering the real questions with Python<\/em>. Previous installments are available from the Data Q&A<\/em> landing page<\/a>.<\/p>\n<\/blockquote>\n\n\n\n

Here\u2019s a question from the Reddit statistics forum<\/a>.<\/p>\n\n\n\n

\n

How do I use bootstrapping to generate confidence intervals for a proportion\/ratio? The situation is this:<\/p>\n\n\n\n

I obtain samples of text with differing numbers of lines. From several tens to over a million. I have no control over how many lines there are in any given sample. Each line of each sample may or may not contain a string S. Counting lines according to S presence or S absence generates a ratio of S to S\u2019 for that sample. I want to use bootstrapping to calculate confidence intervals for the found ratio (which of course will vary with sample size).<\/p>\n\n\n\n

To do this I could either:<\/p>\n\n\n\n

A. Literally resample (10,000 times) of size (say) 1,000 from the original sample (with replacement) then categorise S (and S\u2019), and then calculate the ratio for each resample, and finally identify highest and lowest 2.5% (for 95% CI), or<\/p>\n\n\n\n

B. Generate 10,000 samples of 1,000 random numbers between 0 and 1, scoring each stochastically as above or below original sample ratio (equivalent to S or S\u2019). Then calculate CI as in A.<\/p>\n\n\n\n

Programmatically A is slow and B is very fast. Is there anything wrong with doing B? The confidence intervals generated by each are almost identical.<\/p>\n<\/blockquote>\n\n\n\n

The answer to the immediate question is that A and B are equivalent, so there\u2019s nothing wrong with B. But in follow-up responses, a few related questions were raised:<\/p>\n\n\n\n

    \n
  1. Is resampling a good choice for this problem?<\/li>\n\n\n\n
  2. What size should the resamplings be?<\/li>\n\n\n\n
  3. How many resamplings do we need?<\/li>\n<\/ol>\n\n\n\n

    I don\u2019t think resampling is really necessary here, and I\u2019ll show some alternatives. And I\u2019ll answer the other questions along the way.<\/p>\n\n\n\n

    Click here to run this notebook on Colab<\/a>.<\/p>\n\n\n\n

    I\u2019ll download a utilities module with some of my frequently-used functions, and then import the usual libraries.<\/p>\n\n\n\n

    Pallor and Probability<\/h2>\n\n\n\n

    As an example, let\u2019s use one of the exercises from Think Python<\/em>:<\/p>\n\n\n\n

    \n

    The Count of Monte Cristo<\/em> is a novel by Alexandre Dumas that is considered a classic. Nevertheless, in the introduction of an English translation of the book, the writer Umberto Eco confesses that he found the book to be \u201cone of the most badly written novels of all time\u201d.<\/p>\n\n\n\n

    In particular, he says it is \u201cshameless in its repetition of the same adjective,\u201d and mentions in particular the number of times \u201cits characters either shudder or turn pale.\u201d<\/p>\n\n\n\n

    To see whether his objection is valid, let\u2019s count the number number of lines that contain the word pale in any form, including pale, pales, paled, and paleness, as well as the related word pallor. Use a single regular expression that matches all of these words and no others.<\/p>\n<\/blockquote>\n\n\n\n

    The following cell downloads the text of the book from Project Gutenberg.<\/p>\n\n\n\n

    download('https:\/\/www.gutenberg.org\/cache\/epub\/1184\/pg1184.txt');\n<\/pre>\n\n\n\n
    We\u2019ll use the following functions to remove the additional material that appears before and after the text of the book.<\/p>\n\n\n\n
    def is_special_line(line):\n    return line.startswith('*** ')\n<\/pre>\n\n\n\n
    def clean_file(input_file, output_file):\n    reader = open(input_file)\n    writer = open(output_file, 'w')\n\n    for line in reader:\n        if is_special_line(line):\n            break\n\n    for line in reader:\n        if is_special_line(line):\n            break\n        writer.write(line)\n        \n    reader.close()\n    writer.close()\n\nclean_file('pg1184.txt', 'pg1184_cleaned.txt')\n<\/pre>\n\n\n\n
    And we\u2019ll use the following function to count the number of lines that contain a particular pattern of characters.<\/p>\n\n\n\n
    import re\n\ndef count_matches(lines, pattern):\n    count = 0\n    for line in lines:\n        result = re.search(pattern, line)\n        if result:\n            count += 1\n    return count\n<\/pre>\n\n\n\n
    readlines<\/code> reads the file and creates a list of strings, one for each line.<\/p>\n\n\n\n
    lines = open('pg1184_cleaned.txt').readlines()\nn = len(lines)\nn\n<\/pre>\n\n\n\n
    61310\n<\/pre>\n\n\n\n
    There are about 61,000 lines in the file.<\/p>\n\n\n\n
    The following pattern matches \u201cpale\u201d and several related words.<\/p>\n\n\n\n
    pattern = r'\\b(pale|pales|paled|paleness|pallor)\\b'\nk = count_matches(lines, pattern)\nk\n<\/pre>\n\n\n\n
    223\n<\/pre>\n\n\n\n
    These words appear in 223 lines of the file.<\/p>\n\n\n\n
    p_est = k \/ n\np_est\n<\/pre>\n\n\n\n
    0.0036372533028869677\n<\/pre>\n\n\n\n
    So the estimated proportion is about 0.0036. To quantify the precision of that estimate, we\u2019ll compute a confidence interval.<\/p>\n\n\n\n
    Resampling<\/h2>\n\n\n\n
    First we\u2019ll use the method OP called A \u2013 literally resampling the lines of the file. The following function takes a list of lines and selects a sample, with replacement, that has the same size.<\/p>\n\n\n\n
    def resample(lines):\n    return np.random.choice(lines, len(lines), replace=True)\n<\/pre>\n\n\n\n
    In a resampled list, the same line can appear more than once, and some lines might not appear at all. So in any resampling, the forbidden words might appear more times than in the original text, or fewer. Here\u2019s an example.<\/p>\n\n\n\n
    np.random.seed(1)\ncount_matches(resample(lines), pattern)\n<\/pre>\n\n\n\n
    201\n<\/pre>\n\n\n\n
    In this resampling, the words appear in 201 lines, fewer than in the original (223).<\/p>\n\n\n\n
    If we repeat this process many times, we can compute a sample of possible values of k<\/code>. Because this method is slow, we\u2019ll only repeat it 101 times.<\/p>\n\n\n\n
    ks_resampling = [count_matches(resample(lines), pattern) for i in range(101)]\n<\/pre>\n\n\n\n
    With these different values of k<\/code>, we can divide by n<\/code> to get the corresponding values of p<\/code>.<\/p>\n\n\n\n
    ps_resampling = np.array(ks_resampling) \/ n\n<\/pre>\n\n\n\n
    To see what the distribution of those values looks like, we\u2019ll plot the CDF.<\/p>\n\n\n\n
    from empiricaldist import Cdf\n\nCdf.from_seq(ps_resampling).plot(label='resampling')\ndecorate(xlabel='Resampled proportion', ylabel='Density')\n<\/pre>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    So that\u2019s the slow way to compute the sampling distribution of the proportion. The method OP calls B is to simulate a Bernoulli trial with size n<\/code> and probability of success p_est<\/code>. One way to do that is to draw random numbers from 0 to 1 and count how many are less than p_est<\/code>.<\/p>\n\n\n\n
    (np.random.random(n) < p_est).sum()\n<\/pre>\n\n\n\n
    229\n<\/pre>\n\n\n\n
    Equivalently, we can draw a sample from a Bernoulli distribution and add it up.<\/p>\n\n\n\n
    from scipy.stats import bernoulli\n\nbernoulli(p_est).rvs(n).sum()\n<\/pre>\n\n\n\n
    232\n<\/pre>\n\n\n\n
    These values follow a binomial distribution with parameters n<\/code> and p_est<\/code>. So we can simulate a large number of trials quickly by drawing values from a binomial distribution.<\/p>\n\n\n\n
    from scipy.stats import binom\n\nks_binom = binom(n, p_est).rvs(10001)\n<\/pre>\n\n\n\n
    Dividing by n<\/code>, we can compute the corresponding sample of proportions.<\/p>\n\n\n\n
    ps_binom = np.array(ks_binom) \/ n\n<\/pre>\n\n\n\n
    Because this method is so much faster, we can generate a large number of values, which means we get a more precise picture of the sampling distribution.<\/p>\n\n\n\n
    The following figure compares the CDFs of the values we got by resampling and the values we got from the binomial distribution.<\/p>\n\n\n\n
    Cdf.from_seq(ps_resampling).plot(label='resampling')\nCdf.from_seq(ps_binom).plot(label='binomial')\ndecorate(xlabel='Resampled proportion', ylabel='CDF')\n<\/pre>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    If we run the resampling method longer, these CDFs converge, so the two methods are equivalent.<\/p>\n\n\n\n
    To compute a 90% confidence interval, we can use the values we sampled from the binomial distribution.<\/p>\n\n\n\n
    np.percentile(ps_binom, [5, 95])\n<\/pre>\n\n\n\n
    array([0.0032458 , 0.00404502])\n<\/pre>\n\n\n\n
    Or we can use the inverse CDF of the binomial distribution, which is even faster than drawing a sample. And it\u2019s deterministic \u2013 that is, we get the same result every time, with no randomness.<\/p>\n\n\n\n
    binom(n, p_est).ppf([0.05, 0.95]) \/ n\n<\/pre>\n\n\n\n
    array([0.0032458 , 0.00404502])\n<\/pre>\n\n\n\n
    Using the inverse CDF of the binomial distribution is a good way to compute confidence intervals. But before we get to that, let\u2019s see how resampling behaves as we increase the sample size and the number of iterations.<\/p>\n\n\n\n
    Sample Size<\/h2>\n\n\n\n
    In the example, the sample size is more than 60,000, so the CI is very small. The following figure shows what it looks like for more moderate sample sizes, using p=0.1<\/code> as an example.<\/p>\n\n\n\n
    p = 0.1\nns = [50, 500, 5000]\nci_df = pd.DataFrame(index=ns, columns=['low', 'high'])\n\nfor n in ns:\n    ks = binom(n, p).rvs(10001)\n    ps = ks \/ n\n    Cdf.from_seq(ps).plot(label=f\"n = {n}\")\n    ci_df.loc[n] = np.percentile(ps, [5, 95])\n    \ndecorate(xlabel='Proportion', ylabel='CDF')\n<\/pre>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    As the sample size increases, the spread of the sampling distribution gets smaller, and so does the width of the confidence interval.<\/p>\n\n\n\n
    ci_df['width'] = ci_df['high'] - ci_df['low']\nci_df\n<\/pre>\n\n\n\n
    <\/th>low<\/th>high<\/th>width<\/th><\/tr><\/thead>
    50<\/th>0.04<\/td>0.18<\/td>0.14<\/td><\/tr>
    500<\/th>0.078<\/td>0.122<\/td>0.044<\/td><\/tr>
    5000<\/th>0.0932<\/td>0.1072<\/td>0.014<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
    With resampling methods, it is important to draw samples with the same size as the original dataset \u2013 otherwise the result is wrong.<\/p>\n\n\n\n
    But the number of iterations doesn\u2019t matter as much. The following figure shows the sampling distribution if we run the sampling process 101, 1001, and 10,001 times.<\/p>\n\n\n\n
    p = 0.1\nn = 100\niter_seq = [101, 1001, 100001]\n\nfor iters in iter_seq:\n    ks = binom(n, p).rvs(iters)\n    ps = ks \/ n\n    Cdf.from_seq(ps).plot(label=f\"iters = {iters}\")\n    \ndecorate()\n<\/pre>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    The sampling distribution is the same, regardless of how many iterations we run. But with more iterations, we get a better picture of the distribution and a more precise estimate of the confidence interval. For most problems, 1001 iterations is enough, but if you can generate larger samples fast enough, more is better.<\/p>\n\n\n\n
    However, for this problem, resampling isn\u2019t really necessary. As we\u2019ve seen, we can use the binomial distribution to compute a CI without drawing a random sample at all. And for this problem, there are approximations that are even easier to compute \u2013 although they come with some caveats.<\/p>\n\n\n\n
    Approximations<\/h2>\n\n\n\n
    If n<\/code> is large and p<\/code> is not too close to 0 or 1, the sampling distribution of a proportion is well modeled by a normal distribution, and we can approximate a confidence interval with just a few calculations.<\/p>\n\n\n\n
    For a given confidence level, we can use the inverse CDF of the normal distribution to compute a<\/p>\n\n\n\n
    score, which is the number of standard deviations the CI should span \u2013 above and below the observed value of p<\/code> \u2013 in order to include the given confidence.<\/p>\n\n\n\n
    from scipy.stats import norm\n\nconfidence = 0.9\nz = norm.ppf(1 - (1 - confidence) \/ 2)\nz\n<\/pre>\n\n\n\n
    1.6448536269514722\n<\/pre>\n\n\n\n
    A 90% confidence interval spans about 1.64 standard deviations.<\/p>\n\n\n\n
    Now we can use the following function, which uses p<\/code>, n<\/code>, and this z<\/code> score to compute the confidence interval.<\/p>\n\n\n\n
    def confidence_interval_normal_approx(k, n, z):\n    p = k \/ n\n    margin_of_error = z * np.sqrt(p * (1 - p) \/ n)\n    \n    lower_bound = p - margin_of_error\n    upper_bound = p + margin_of_error\n    return lower_bound, upper_bound\n<\/pre>\n\n\n\n
    To test it, we\u2019ll compute n<\/code> and k<\/code> for the example again.<\/p>\n\n\n\n
    n = len(lines)\nk = count_matches(lines, pattern)\nn, k\n<\/pre>\n\n\n\n
    (61310, 223)\n<\/pre>\n\n\n\n
    Here\u2019s the confidence interval based on the normal approximation.<\/p>\n\n\n\n
    ci_normal = confidence_interval_normal_approx(k, n, z)\nci_normal\n<\/pre>\n\n\n\n
    (0.003237348046298746, 0.00403715855947519)\n<\/pre>\n\n\n\n
    In the example, n<\/code> is large, which is good for the normal approximation, but p<\/code> is small, which is bad. So it\u2019s not obvious whether we can trust the approximation.<\/p>\n\n\n\n
    An alternative that\u2019s more robust is the Wilson score interval, which is reliable for values of p<\/code> close to 0 and 1, and sample sizes bigger than about 5.<\/p>\n\n\n\n
    def confidence_interval_wilson_score(k, n, z):    \n    p = k \/ n\n    factor = z**2 \/ n\n    denominator = 1 + factor\n    center = p + factor \/ 2\n    half_width = z * np.sqrt((p * (1 - p) + factor \/ 4) \/ n)\n    \n    lower_bound = (center - half_width) \/ denominator\n    upper_bound = (center + half_width) \/ denominator\n    \n    return lower_bound, upper_bound\n<\/pre>\n\n\n\n
    Here\u2019s the 90% CI based on Wilson scores.<\/p>\n\n\n\n
    ci_wilson = confidence_interval_wilson_score(k, n, z)\nci_wilson\n<\/pre>\n\n\n\n
    (0.003258660468175958, 0.00405965209814987)\n<\/pre>\n\n\n\n
    Another option is the Clopper-Pearson interval, which is what we computed earlier with the inverse CDF of the binomial distribution. Here\u2019s a function that computes it.<\/p>\n\n\n\n
    from scipy.stats import binom\n\ndef confidence_interval_exact_binomial(k, n, confidence=0.9):\n    alpha = 1 - confidence\n    p = k \/ n\n\n    lower_bound = binom.ppf(alpha \/ 2, n, p) \/ n if k > 0 else 0\n    upper_bound = binom.ppf(1 - alpha \/ 2, n, p) \/ n if k < n else 1\n    \n    return lower_bound, upper_bound\n<\/pre>\n\n\n\n
    And here\u2019s the interval we get.<\/p>\n\n\n\n
    ci_binomial = confidence_interval_exact_binomial(k, n)\nci_binomial\n<\/pre>\n\n\n\n
    (0.003245800032621106, 0.0040450171260805745)\n<\/pre>\n\n\n\n
    A final alternative is the Jeffreys interval, which is derived from Bayes\u2019s Theorem. If we start with a Jeffreys prior and observe k<\/code> successes out of n<\/code> attempts, the posterior distribution of p<\/code> is a beta distribution with parameters a = k + 1\/2<\/code> and b = n - k + 1\/2<\/code>. So we can use the inverse CDF of the beta distribution to compute a CI.<\/p>\n\n\n\n
    from scipy.stats import beta\n\ndef bayesian_confidence_interval_beta(k, n, confidence=0.9):\n    alpha = 1 - confidence    \n    a, b = k + 1\/2, n - k + 1\/2\n    \n    lower_bound = beta.ppf(alpha \/ 2, a, b)\n    upper_bound = beta.ppf(1 - alpha \/ 2, a, b)\n    \n    return lower_bound, upper_bound\n<\/pre>\n\n\n\n
    And here\u2019s the interval we get.<\/p>\n\n\n\n
    ci_beta = bayesian_confidence_interval_beta(k, n)\nci_beta\n<\/pre>\n\n\n\n
    (0.003254420914221609, 0.004054683138668112)\n<\/pre>\n\n\n\n
    The following figure shows the four intervals we just computed graphically.<\/p>\n\n\n\n
    intervals = {\n    'Normal Approximation': ci_normal,\n    'Wilson Score': ci_wilson,\n    'Clopper-Pearson': ci_binomial,\n    'Jeffreys': ci_beta\n}\ny_pos = np.arange(len(intervals))\n\nfor i, (label, (lower, upper)) in enumerate(intervals.items()):\n    middle = (lower + upper) \/ 2\n    xerr = [[(middle - lower)], [(upper - middle)]]\n    plt.errorbar(x=middle, y=i-0.2, xerr=xerr, fmt='o', capsize=5)\n    plt.text(middle, i, label, ha='center', va='top')\n    \ndecorate(xlabel='Proportion', ylim=[3.5, -0.8], yticks=[])\n<\/pre>\n\n\n\n
    \"\"<\/figure>\n\n\n\n
    In this example, because n<\/code> is so large, the intervals are all similar \u2013 the differences are too small to matter in practice. For smaller values of n<\/code>, the normal approximation becomes unreliable, and for very small values, none of them are reliable.<\/p>\n\n\n\n
    The normal approximation and Wilson score interval are easy and fast to compute. On my old laptop, they take 1-2 microseconds.<\/p>\n\n\n\n
    %timeit confidence_interval_normal_approx(k, n, z)\n<\/pre>\n\n\n\n
    1.04 \u00b5s \u00b1 4.04 ns per loop (mean \u00b1 std. dev. of 7 runs, 1,000,000 loops each)\n<\/pre>\n\n\n\n
    %timeit confidence_interval_wilson_score(k, n, z)\n<\/pre>\n\n\n\n
    1.64 \u00b5s \u00b1 28.6 ns per loop (mean \u00b1 std. dev. of 7 runs, 1,000,000 loops each)\n<\/pre>\n\n\n\n
    Evaluating the inverse CDF of the binomial and beta distributions are more complex computations \u2013 they take about 100 times longer.<\/p>\n\n\n\n
    %timeit confidence_interval_exact_binomial(k, n)\n<\/pre>\n\n\n\n
    195 \u00b5s \u00b1 7.53 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1,000 loops each)\n<\/pre>\n\n\n\n
    %timeit bayesian_confidence_interval_beta(k, n)\n<\/pre>\n\n\n\n
    269 \u00b5s \u00b1 4.6 \u00b5s per loop (mean \u00b1 std. dev. of 7 runs, 1,000 loops each)\n<\/pre>\n\n\n\n
    But they still take less than 300 microseconds, so unless you need to compute millions of confidence intervals per second, the difference in computation time doesn\u2019t matter.<\/p>\n\n\n\n
    Discussion<\/h2>\n\n\n\n
    If you took a statistics class and learned one of these methods, you probably learned the normal approximation. That\u2019s because it is easy to explain and, because it is based on a form of the Central Limit Theorem, it helps to justify time spent learning about the CLT. But in my opinion it should never be used in practice because it is dominated by the Wilson score interval \u2013 that is, it is worse than Wilson in at least one way and better in none.<\/p>\n\n\n\n
    I think the Clopper-Pearson interval is equally easy to explain, but when n<\/code> is small, there are few possible values of k<\/code>, and therefore few possible values of p<\/code> \u2013 and the interval can be wider than it needs to be.<\/p>\n\n\n\n
    The Jeffreys interval is based on Bayesian statistics, so it takes a little more explaining, but it behaves well for all values of n<\/code> and p<\/code>. And when n<\/code> is small, it can be extended to take advantage of background information about likely values of p<\/code>.<\/p>\n\n\n\n
    For these reasons, the Jeffreys interval is my usual choice, but in a computational environment that doesn\u2019t provide the inverse CDF of the beta distribution, I would use a Wilson score interval.<\/p>\n\n\n\n
    OP is working in LiveCode, which doesn\u2019t provide a lot of math and statistics libraries, so Wilson might be a good choice. Here\u2019s a LiveCode implementation generated by ChatGPT.<\/p>\n\n\n\n
    -- Function to calculate the z-score for a 95% confidence level (z \u2248 1.96)\nfunction zScore\n    return 1.96\nend zScore\n\n-- Function to calculate the Wilson Score Interval with distinct bounds\nfunction wilsonScoreInterval k n\n    -- Calculate proportion of successes\n    put k \/ n into p\n    put zScore() into z\n    \n    -- Common term for the interval calculation\n    put (z^2 \/ n) into factor\n    put (p + factor \/ 2) \/ (1 + factor) into adjustedCenter\n    \n    -- Asymmetric bounds\n    put sqrt(p * (1 - p) \/ n + factor \/ 4) into sqrtTerm\n    \n    -- Lower bound calculation\n    put adjustedCenter - (z * sqrtTerm \/ (1 + factor)) into lowerBound\n    \n    -- Upper bound calculation\n    put adjustedCenter + (z * sqrtTerm \/ (1 + factor)) into upperBound\n    \n    return lowerBound & comma & upperBound\nend wilsonScoreInterval\n<\/pre>\n\n\n\n
    Data Q&A: Answering the real questions with Python<\/em><\/a><\/p>\n\n\n\n
    Copyright 2024 Allen B. Downey<\/a><\/p>\n\n\n\n
    <\/p>\n","protected":false},"excerpt":{"rendered":"
    It’s another installment in Data Q&A: Answering the real questions with Python. Previous installments are available from the Data Q&A landing page. Here\u2019s a question from the Reddit statistics forum. How do I use bootstrapping to generate confidence intervals for a proportion\/ratio? The situation is this: I obtain samples of text with differing numbers of lines. From several tens to over a million. I have no control over how many lines there are in any given sample. Each line of…<\/p>\n
     Read More  Read More<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[1],"tags":[],"class_list":["post-1390","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"yoast_head":"\nBootstrapping a Proportion - Probably Overthinking It<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.allendowney.com\/blog\/2024\/10\/15\/bootstrapping-a-proportion\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Bootstrapping a Proportion - Probably Overthinking It\" \/>\n<meta property=\"og:description\" content=\"It’s another installment in Data Q&A: Answering the real questions with Python. Previous installments are available from the Data Q&A landing page. Here\u2019s a question from the Reddit statistics forum. How do I use bootstrapping to generate confidence intervals for a proportion\/ratio? The situation is this: I obtain samples of text with differing numbers of lines. From several tens to over a million. I have no control over how many lines there are in any given sample. Each line of... 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To find out, you could use data from the General Social Survey (GSS), which asks questions like: Do you think there should be laws against marriages between Blacks\/African-Americans and whites?Should a man who admits[mfn]If you find the wording\u2026","rel":"","context":"In \"general social survey\"","block_context":{"text":"general social survey","link":"https:\/\/www.allendowney.com\/blog\/tag\/general-social-survey\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2021\/05\/fepol_vs_age_by_cohort10.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2021\/05\/fepol_vs_age_by_cohort10.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2021\/05\/fepol_vs_age_by_cohort10.jpg?resize=525%2C300&ssl=1 1.5x"},"classes":[]},{"id":169,"url":"https:\/\/www.allendowney.com\/blog\/2019\/02\/06\/the-marriage-strike-continues\/","url_meta":{"origin":1390,"position":5},"title":"The marriage strike continues","author":"AllenDowney","date":"February 6, 2019","format":false,"excerpt":"Last month The National Survey of Family Growth released new data from 5,554 respondents interviewed between 2015 and 2017. I've worked on several studies using data from the NSFG, so it's time to do some updates! In 2015 I gave a talk at SciPy called \"Will Millennials Ever Get Married\"\u2026","rel":"","context":"Similar post","block_context":{"text":"Similar post","link":""},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2019\/02\/Screenshot-at-2019-02-06-10-02-10.png?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2019\/02\/Screenshot-at-2019-02-06-10-02-10.png?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2019\/02\/Screenshot-at-2019-02-06-10-02-10.png?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2019\/02\/Screenshot-at-2019-02-06-10-02-10.png?resize=700%2C400&ssl=1 2x, https:\/\/i0.wp.com\/www.allendowney.com\/blog\/wp-content\/uploads\/2019\/02\/Screenshot-at-2019-02-06-10-02-10.png?resize=1050%2C600&ssl=1 3x"},"classes":[]}],"_links":{"self":[{"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/posts\/1390","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/comments?post=1390"}],"version-history":[{"count":1,"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/posts\/1390\/revisions"}],"predecessor-version":[{"id":1396,"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/posts\/1390\/revisions\/1396"}],"wp:attachment":[{"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/media?parent=1390"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/categories?post=1390"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.allendowney.com\/blog\/wp-json\/wp\/v2\/tags?post=1390"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}