{"id":2550,"date":"2019-08-27T17:37:00","date_gmt":"2019-08-27T17:37:00","guid":{"rendered":"https:\/\/analystprep.com\/cfa-level-1-exam\/?p=2550"},"modified":"2026-04-02T09:28:44","modified_gmt":"2026-04-02T09:28:44","slug":"p-value","status":"publish","type":"post","link":"https:\/\/analystprep.com\/cfa-level-1-exam\/quantitative-methods\/p-value\/","title":{"rendered":"The p-value in Hypothesis Testing"},"content":{"rendered":"\n<script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"VideoObject\",\n  \"name\": \"Hypothesis Testing (2025 Level I CFA\u00ae Exam \u2013 Quantitative Methods \u2013 Module 6)\",\n  \"description\": \"This CFA Level 1 session on Quantitative Methods simplifies Hypothesis Testing, covering key topics like null and alternative hypotheses, test statistics, errors, significance levels, and parametric vs. non-parametric tests, with practical examples and applications for financial analysis and exam success.\",\n  \"uploadDate\": \"2021-11-18T00:00:00+00:00\",\n  \"thumbnailUrl\": \"https:\/\/img.youtube.com\/vi\/twTTWMuEGx8\/maxresdefault.jpg\",\n  \"contentUrl\": \"https:\/\/youtu.be\/twTTWMuEGx8\",\n  \"embedUrl\": \"https:\/\/www.youtube.com\/embed\/twTTWMuEGx8\",\n  \"duration\": \"PT58M47S\"\n}\n<\/script>\n\n<script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"QAPage\",\n  \"mainEntity\": {\n    \"@type\": \"Question\",\n    \"name\": \"A CFA candidate conducts a statistical test about the mean value of a random variable X. The hypotheses are H\u2080: \u03bc = \u03bc\u2080 vs H\u2081: \u03bc \u2260 \u03bc\u2080. She obtains a test statistic of 2.2. Given a 5% significance level, what is the p-value?\",\n    \"text\": \"A CFA candidate conducts a statistical test about the mean value of a random variable X. The hypotheses are H\u2080: \u03bc = \u03bc\u2080 vs H\u2081: \u03bc \u2260 \u03bc\u2080. She obtains a test statistic of 2.2. Given a 5% significance level, determine the p-value.\",\n    \"answerCount\": 3,\n    \"acceptedAnswer\": {\n      \"@type\": \"Answer\",\n      \"text\": \"The correct answer is C. The p-value is 2.78%. Since this is a two-tailed test, the p-value is calculated as 2 \u00d7 P(Z > 2.2). From the standard normal distribution, P(Z > 2.2) \u2248 1.39%, so the p-value equals 2 \u00d7 1.39% = 2.78%.\"\n    }\n  }\n}\n<\/script>\n\n\n\n<iframe loading=\"lazy\" width=\"560\" height=\"315\" src=\"https:\/\/www.youtube.com\/embed\/twTTWMuEGx8?si=0AFswcIf_fpzO0Tr\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n\n\n\n<p>The <em>p<\/em>-value is the lowest level of significance at which we can reject a null hypothesis. It is the probability of coming up with a test statistic that would justify our rejection of a null hypothesis, assuming that the null hypothesis is indeed true.<\/p>\n\n\n\n<!--more-->\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Breaking Down the <em>p<\/em>-value<\/strong><\/h2>\n\n\n\n<p>When carrying out a statistical test with a fixed value of the significance level (\u03b1), we merely compare the observed test statistic with some critical value. For example, we might \u201creject a H<sub>0<\/sub> using a 5% test\u201d or \u201creject a H<sub>0<\/sub> at 1% significance level.\u201d The problem with this \u2018classical\u2019 approach is that it does not give us the details about the <strong>strength of the evidence<\/strong> against the null hypothesis.<\/p>\n\n\n\n<p>Determination of the <em>p<\/em>-value gives statisticians a more informative approach to hypothesis testing. The <em>p<\/em>-value is actually the lowest level at which we can reject a H<sub>0<\/sub>. This means that the strength of the evidence against a H<sub>0<\/sub> increases as the <em>p<\/em>-value becomes smaller.<\/p>\n\n\n\n<p>For one-tailed tests, the <em>p<\/em>-value is given by the probability that lies below the calculated test statistic for left-tailed tests. In the case of right-tailed tests, the probability that lies above the test statistic gives the <em>p<\/em>-value<em>. <\/em>However, if the test is two-tailed, this value is given by the sum of the probabilities in the two tails. We start by determining the probability lying below the negative value of the test statistic. After this, we add this to the probability lying above the positive value of the test statistic.<\/p>\n\n\n\n<div style=\"margin:20px 0\">\n<a href=\"https:\/\/analystprep.com\/free-trial\/\" target=\"_blank\"\nstyle=\"display:block;width:100%;text-align:center;padding:10px;border:2px solid #2f5bea;border-radius:40px;font-size:16px;color:#2f5bea;text-decoration:none\">\nPractice p-value questions with our free trial.\n<\/a>\n<\/div>\n\n\n<h3><strong>Example: <em>p<\/em>-value<\/strong><\/h3>\n<p>\u03b8 represents the probability of obtaining a head when a coin is tossed. Assume that we tossed a coin 200 times and the head came up in 85 out of of the 200 trials. Test the following hypothesis at a 5% level of significance.<\/p>\n<p>H<sub>0<\/sub>: \u03b8 = 0.5<\/p>\n<p>H<sub>1<\/sub>: \u03b8 &lt; 0.5<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, note that repeatedly tossing a coin follows a binomial distribution.<\/p>\n<p>Our p<em>-value<\/em> will be given by P(X &lt; 85) where X follows a binomial (200,0.5), assuming the H<sub>0<\/sub> is true.<\/p>\n<p>$$ \\begin{align*}<br \/>&amp; = P \\left[ Z &lt; \\frac {(85.5 \u2013 100)}{\\sqrt {50}} \\right] \\\\<br \/>&amp; = P(Z &lt; -2.05) = 1 \u2013 0.97982 = 0.02018 \\\\<br \/>\\end{align*} $$<\/p>\n<p><em>(We have applied the Central Limit Theorem by taking the binomial distribution as approximately normal.)<\/em><\/p>\n<p>Since the probability is less than 0.05, the H<sub>0 <\/sub>is extremely unlikely, and we actually have strong evidence against a H<sub>0<\/sub> that favors H<sub>1<\/sub>. Therefore, clearly expressing this result, we could say:<\/p>\n<p>\u201cThere is very strong evidence against the hypothesis that the coin is fair. We, therefore, conclude that the coin is biased against heads.\u201d<\/p>\n<p>Remember, failure to reject a H<sub>0<\/sub> does not mean it is true. It means there is insufficient evidence to justify the rejection of the H<sub>0<\/sub> given a certain level of significance.<\/p>\n<blockquote>\n<h2><strong>Question<\/strong><\/h2>\n<p>A CFA candidate conducts a statistical test about the mean value of a random variable X.<\/p>\n<p>H<sub>0<\/sub>: \u03bc = \u03bc<sub>0<\/sub> vs H<sub>1<\/sub>: \u03bc <em>\u2260<\/em> \u03bc<sub>0<\/sub><\/p>\n<p>She obtains a test statistic of 2.2. Given a 5% significance level, determine the <em>p<\/em>-value<em>.<\/em><\/p>\n<p>A. 1.39%.<\/p>\n<p>B. 2.78.<\/p>\n<p>C. 2.78%.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>The correct answer is <strong>C<\/strong>.<\/p>\n<p>$$ \\text P-\\text{value}= P(Z &gt; 2.2) = 1 \u2013 P(Z &lt; 2.2) = 1.39\\% \u00d7 2 = 2.78\\% $$<\/p>\n<p><em>(We have multiplied by two since this is a two-tailed test.)<\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-17065\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178.jpg\" alt=\"p-value\" width=\"1463\" height=\"929\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178.jpg 1463w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-300x190.jpg 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-768x488.jpg 768w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-1024x650.jpg 1024w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-400x254.jpg 400w\" sizes=\"auto, (max-width: 1463px) 100vw, 1463px\" \/><\/p>\n<p><strong><em>Interpretation:<\/em> <\/strong>The p<em>-value<\/em> (2.78%) is less than the level of significance (5%). Therefore, we have sufficient evidence to reject the H<sub>0<\/sub>. In fact, the evidence is so strong that we would also reject the H<sub>0 <\/sub>at significance levels of 4% and 3%. However, at significance levels of 2% or 1%, we would not reject the H<sub>0<\/sub> since the <em>p<\/em>-value surpasses these values.<\/p>\n<\/blockquote>\n<div class=\"notes_inv\">\n<hr \/><\/div>\n\n\n<p><iframe loading=\"lazy\" src=\"\/\/www.youtube.com\/embed\/twTTWMuEGx8\" width=\"611\" height=\"343\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The <em>p<\/em>-value is the lowest level of significance at which we can reject a null hypothesis. It is the probability of coming up with a test statistic that would justify our rejection of a null hypothesis, assuming that the null hypothesis is indeed true.<\/p>\n<p><!--more--><\/p>\n<h2><strong>Breaking Down the <em>p<\/em>-value <\/strong><\/h2>\n<p>When carrying out a statistical test with a fixed value of the significance level (\u03b1), we merely compare the observed test statistic with some critical value. For example, we might \u201creject a H<sub>0<\/sub> using a 5% test\u201d or \u201creject a H<sub>0<\/sub> at 1% significance level.\u201d The problem with this \u2018classical\u2019 approach is that it does not give us the details about the <strong>strength of the evidence<\/strong> against the null hypothesis.<\/p>\n<p>Determination of the <em>p<\/em>-value gives statisticians a more informative approach to hypothesis testing. The <em>p<\/em>-value is actually the lowest level at which we can reject a H<sub>0<\/sub>. This means that the strength of the evidence against a H<sub>0<\/sub> increases as the <em>p<\/em>-value becomes smaller.<\/p>\n<p>For one-tailed tests, the <em>p<\/em>-value is given by the probability that lies below the calculated test statistic for left-tailed tests. In the case of right-tailed tests, the probability that lies above the test statistic gives the <em>p<\/em>-value<em>. <\/em>However, if the test is two-tailed, this value is given by the sum of the probabilities in the two tails. We start by determining the probability lying below the negative value of the test statistic. After this, we add this to the probability lying above the positive value of the test statistic.<\/p>\n<h3><strong>Example: <em>p<\/em>-value<\/strong><\/h3>\n<p>\u03b8 represents the probability of obtaining a head when a coin is tossed. Assume that we tossed a coin 200 times and the head came up in 85 out of of the 200 trials. Test the following hypothesis at a 5% level of significance.<\/p>\n<p>H<sub>0<\/sub>: \u03b8 = 0.5<\/p>\n<p>H<sub>1<\/sub>: \u03b8 &lt; 0.5<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, note that repeatedly tossing a coin follows a binomial distribution.<\/p>\n<p>Our p<em>-value<\/em> will be given by P(X &lt; 85) where X follows a binomial (200,0.5), assuming the H<sub>0<\/sub> is true.<\/p>\n<p>$$ \\begin{align*}<br>&amp; = P \\left[ Z &lt; \\frac {(85.5 \u2013 100)}{\\sqrt {50}} \\right] \\\\<br>&amp; = P(Z &lt; -2.05) = 1 \u2013 0.97982 = 0.02018 \\\\<br>\\end{align*} $$<\/p>\n<p><em>(We have applied the Central Limit Theorem by taking the binomial distribution as approximately normal.)<\/em><\/p>\n<p>Since the probability is less than 0.05, the H<sub>0 <\/sub>is extremely unlikely, and we actually have strong evidence against a H<sub>0<\/sub> that favors H<sub>1<\/sub>. Therefore, clearly expressing this result, we could say:<\/p>\n<p>\u201cThere is very strong evidence against the hypothesis that the coin is fair. We, therefore, conclude that the coin is biased against heads.\u201d<\/p>\n<p>Remember, failure to reject a H<sub>0<\/sub> does not mean it is true. It means there is insufficient evidence to justify the rejection of the H<sub>0<\/sub> given a certain level of significance.<\/p>\n<blockquote>\n<h2><strong>Question<\/strong><\/h2>\n<p>A CFA candidate conducts a statistical test about the mean value of a random variable X.<\/p>\n<p>H<sub>0<\/sub>: \u03bc = \u03bc<sub>0<\/sub> vs H<sub>1<\/sub>: \u03bc <em>\u2260<\/em> \u03bc<sub>0<\/sub><\/p>\n<p>She obtains a test statistic of 2.2. Given a 5% significance level, determine the <em>p<\/em>-value<em>.<\/em><\/p>\n<p>A. 1.39%.<\/p>\n<p>B. 2.78.<\/p>\n<p>C. 2.78%.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>The correct answer is <strong>C<\/strong>.<\/p>\n<p>$$ \\text P-\\text{value}= P(Z &gt; 2.2) = 1 \u2013 P(Z &lt; 2.2) = 1.39\\% \u00d7 2 = 2.78\\% $$<\/p>\n<p><em>(We have multiplied by two since this is a two-tailed test.)<\/em><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-17065\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178.jpg\" alt=\"p-value\" width=\"1463\" height=\"929\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178.jpg 1463w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-300x190.jpg 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-768x488.jpg 768w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-1024x650.jpg 1024w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-178-400x254.jpg 400w\" sizes=\"auto, (max-width: 1463px) 100vw, 1463px\" \/><\/p>\n<p><strong><em>Interpretation:<\/em> <\/strong>The p<em>-value<\/em> (2.78%) is less than the level of significance (5%). Therefore, we have sufficient evidence to reject the H<sub>0<\/sub>. In fact, the evidence is so strong that we would also reject the H<sub>0 <\/sub>at significance levels of 4% and 3%. However, at significance levels of 2% or 1%, we would not reject the H<sub>0<\/sub> since the <em>p<\/em>-value surpasses these values.<\/p>\n<\/blockquote>\n<div class=\"notes_inv\"><hr><\/div>\n\n\n\n<div style=\"text-align:center;margin:40px 0\">\n<a href=\"https:\/\/analystprep.com\/free-trial\/\" target=\"_blank\"\nstyle=\"display:inline-block;padding:10px 26px;background:#3f78d7;color:#fff;border-radius:40px;font-size:16px;text-decoration:none\">\nStart Free Trial \u2192\n<\/a>\n<p style=\"margin-top:10px;max-width:600px;margin-left:auto;margin-right:auto;font-size:14px\">\nSolve CFA-style questions on p-values, hypothesis testing, and statistical significance.\n<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>The p-value is the lowest level of significance at which we can reject a null hypothesis. It is the probability of coming up with a test statistic that would justify our rejection of a null hypothesis, assuming that the null&#8230;<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[2],"tags":[],"class_list":["post-2550","post","type-post","status-publish","format-standard","hentry","category-quantitative-methods","blog-post","no-post-thumbnail","animate"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.9 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>P-value Question Example | CFA Level 1 - AnalystPrep<\/title>\n<meta name=\"description\" content=\"Learn about p-value in hypothesis testing through practical examples and how to interpret right-tailed test p-values.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/analystprep.com\/cfa-level-1-exam\/quantitative-methods\/p-value\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"P-value Question Example | CFA Level 1 - 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