{"id":2553,"date":"2019-08-27T17:37:00","date_gmt":"2019-08-27T17:37:00","guid":{"rendered":"https:\/\/analystprep.com\/cfa-level-1-exam\/?p=2553"},"modified":"2026-03-31T11:03:48","modified_gmt":"2026-03-31T11:03:48","slug":"t-test-z-test","status":"publish","type":"post","link":"https:\/\/analystprep.com\/cfa-level-1-exam\/quantitative-methods\/t-test-z-test\/","title":{"rendered":"Uses of the t-test and the z-test"},"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\": \"Master Hypothesis Testing for the 2025 CFA\u00ae Level I Quantitative Methods curriculum. This comprehensive lesson walks through the full hypothesis testing framework, including null and alternative hypotheses, one-tailed vs. two-tailed tests, Type I and Type II errors, significance levels, power of a test, p-values, and decision rules. You\u2019ll learn how to select and interpret the appropriate test statistics for means, variances, correlations, and contingency tables, covering both parametric and nonparametric tests. Designed to mirror CFA exam style, this session helps you determine statistical and economic significance with confidence.\",\n  \"uploadDate\": \"2021-11-18\",\n  \"thumbnailUrl\": \"https:\/\/img.youtube.com\/vi\/twTTWMuEGx8\/hqdefault.jpg\",\n  \"contentUrl\": \"https:\/\/www.youtube.com\/watch?v=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\": \"ImageObject\",\n  \"@id\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/images\/p-value-quantitative-methods\",\n  \"contentUrl\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value.png\",\n  \"url\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value.png\",\n  \"caption\": \"p-Value Illustration (t-test \/ z-test)\",\n  \"width\": 534,\n  \"height\": 224,\n  \"copyrightNotice\": \"\u00a9 2024 AnalystPrep\",\n  \"acquireLicensePage\": \"https:\/\/analystprep.com\/license-info\",\n  \"creditText\": \"AnalystPrep Design Team\",\n  \"creator\": {\n    \"@type\": \"Organization\",\n    \"name\": \"AnalystPrep\"\n  },\n  \"isPartOf\": {\n    \"@type\": \"WebPage\",\n    \"@id\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/quantitative-methods\/t-test-z-test\/\"\n  }\n}\n<\/script>\n<script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"QAPage\",\n  \"mainEntity\": {\n    \"@type\": \"Question\",\n    \"name\": \"What is the value of t when the level of significance is reduced to 0.5%?\",\n    \"text\": \"What is the value of t in the example above if the level of significance is reduced from 5% to 0.5%, and does this change the decision rule?\\n\\nA. 2.02; it does not change the decision rule.\\nB. 3.25; it does not change the decision rule.\\nC. 3.25; it does change the decision rule.\",\n    \"answerCount\": 1,\n    \"acceptedAnswer\": {\n      \"@type\": \"Answer\",\n      \"text\": \"B. 3.25; it does not change the decision rule. At a significance level of 0.5%, the critical value from the t-distribution is 3.25, and this higher critical value does not change the original decision rule.\"\n    }\n  }\n}\n<\/script>\n\n\n\n<iframe loading=\"lazy\" width=\"560\" height=\"315\" src=\"https:\/\/www.youtube.com\/embed\/twTTWMuEGx8?si=2VUQ7P_PXBNmU9Vy\" 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<h2 class=\"wp-block-heading\"><strong>The z-test<\/strong><\/h2>\n\n\n\n<p>The z-test is the ideal hypothesis test to conduct in a <strong>normal distribution<\/strong> of a random variable. In addition, the variance of the population must be <strong>known<\/strong>. The z-statistic refers to the test statistic computed for hypothesis testing.<\/p>\n\n\n\n<!--more-->\n\n\n\n<p><strong>Testing H<sub>0<\/sub>: \u03bc = \u03bc<sub>0<\/sub> using the z-test<\/strong><\/p>\n\n\n\n<p>Given a&nbsp;random sample of size <em>n<\/em> from a normally distributed population with mean \u03bc and variance \u03c3<sup>2<sub>, <\/sub><\/sup>and a sample mean X\u0304, we can compute the z-statistic as:<\/p>\n\n\n\n<p>$$ \\text z-\\text{statistic} =\\cfrac {(\\bar{X} &#8211; \\mu_0)}{\\left(\\frac {\\sigma}{\\sqrt n} \\right)} $$<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<p>\\(\\bar{X}\\) is the sample mean.<\/p>\n\n\n\n<p>\\(\u03bc_0\\) is the hypothesized mean of the population.<\/p>\n\n\n\n<p>\\(\u03c3\\) is the standard deviation of the population.<\/p>\n\n\n\n<p>\\(n\\) is the sample size.<\/p>\n\n\n\n<p>Once computed, the z-statistic is compared to the critical value that corresponds to the level of significance of the test. For example, if the significance level is 5%, the z-statistic is screened against the upper or lower 95% point of the normal distribution (\u00b11.96). The decision rule is to reject the H<sub>0 <\/sub>if the z-statistic falls within the critical or rejection region.<\/p>\n\n\n\n<div style=\"text-align: center; margin: 22px 0;\">\n  <div style=\"max-width: 680px; margin: 0 auto;\">\n    <a href=\"https:\/\/analystprep.com\/free-trial\/\" target=\"_blank\" rel=\"noopener noreferrer\"\n       style=\"display: flex; align-items: center; justify-content: center;\n       width: 100%; padding: 10px 18px;\n       border: 2px solid #1e5bd8; color: #1e5bd8;\n       border-radius: 9999px; text-decoration: none; font-weight: 600;\">\n      Practice t-tests and z-tests with our free trial\n    <\/a>\n  <\/div>\n<\/div>\n\n\n<h3><strong>Example: z-test<\/strong><\/h3>\n<p>Academics carried out a study on 50 former United States presidents and found an average IQ of 135. You are required to carry out a 5% statistical test to determine whether the average IQ of presidents is greater than 130. (IQs are distributed normally, and previous studies indicate that \u03c3 = 25.)<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, you have to state the hypothesis:<\/p>\n<p>H<sub>0<\/sub>: \u03bc = 130<\/p>\n<p>H<sub>1<\/sub>: \u03bc &gt; 130<\/p>\n<p>Assuming the H<sub>0<\/sub> is true, \\(\\frac {(\\bar{X} &#8211; 130)}{\\left(\\frac {\\sigma}{\\sqrt n} \\right)} \\sim N(0,1)\\).<\/p>\n<p>The z-statistic is \\(\\cfrac {(135 \u2013 130)}{\\left(\\frac {25}{\\sqrt {50}} \\right)} = 1.414\\).<\/p>\n<p>This is a right-tailed test. Therefore, we compare our test statistic to the upper 95% point of the standard normal distribution (1.6449). Since 1.414 is less than 1.6449, we <strong>do not have<\/strong> sufficient evidence to reject the H<sub>0<\/sub>. As such, it would be <strong>reasonable <\/strong>to conclude that the average IQ of U.S. presidents is not more than 130.<\/p>\n<h2><strong>The t-test<\/strong><\/h2>\n<p>The t-test is based on the t-distribution. The test is appropriate for testing the value of a population mean when:<\/p>\n<ul>\n<li>\u03c3 is unknown; and<\/li>\n<li>The sample size is large (n \u2265 30), and if n &lt; 30, the distribution must either be normal or approximately normal.<\/li>\n<\/ul>\n<h3><strong>Testing H<sub>0<\/sub>: \u03bc = \u03bc<sub>0<\/sub> Using the t-test<\/strong><\/h3>\n<p>We compute a t-statistic with n \u2013 1 degrees of freedom as:<\/p>\n<p>$$ t_{n-1} = \\cfrac {(\\bar{X} &#8211; \\mu_0)}{\\left(\\frac {s}{\\sqrt n} \\right)} $$<\/p>\n<p>Where:<\/p>\n<p>\\(\\bar{X}\\) is the sample mean.<\/p>\n<p>\\(\u03bc_0\\) is the hypothesized mean of the population.<\/p>\n<p>\\(s\\) is the standard deviation of the sample.<\/p>\n<p>\\(n\\) is the sample size.<\/p>\n<h3><strong>Example: t-test<\/strong><\/h3>\n<p>The annual rate of rainfall (cm) in a certain equatorial country over the last 10 years is given below:<\/p>\n<p>{ 25\u00a0\u00a0 26\u00a0\u00a0 25\u00a0\u00a0\u00a0 27\u00a0\u00a0 28\u00a0\u00a0 29\u00a0\u00a0 28\u00a0\u00a0 27\u00a0\u00a0 26\u00a0\u00a0 25 }<\/p>\n<p>Financial analysts in the country wish to determine whether the average rate of rainfall has increased from its former value of 23. Carry out a statistical test at the 5% level.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>As always, you should begin by stating the hypothesis:<\/p>\n<p>H<sub>0<\/sub>: \u03bc\u00a0\u2264 23<\/p>\n<p>H<sub>1<\/sub>: \u03bc &gt; 23<\/p>\n<p>If we assume that the annual rainfall quantities are distributed normally and recorded independently, then:<\/p>\n<p>$$ \\cfrac {(\\bar{X} &#8211; 23)}{\\left(\\frac {S}{\\sqrt n} \\right)} \\sim t_{n- 1} $$<\/p>\n<p>Please, confirm that X\u0304 = 26.6 and S = 1.43.<\/p>\n<p>Therefore, our t-statistic = \\(\\cfrac {(26.6 &#8211; 23)}{\\left(\\frac {1.43}{\\sqrt {10}}\\right)} = 7.96\\).<\/p>\n<p>Our test statistic (7.96) is greater than the upper 95% point of the t<sub>0.05,9<\/sub> distribution (1.833).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-38626\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value.png\" alt=\"\" width=\"534\" height=\"224\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value.png 534w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value-300x126.png 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value-400x168.png 400w\" sizes=\"auto, (max-width: 534px) 100vw, 534px\" \/><\/p>\n<p>Therefore, we have <strong>sufficient evidence<\/strong> to reject the H<sub>0<\/sub>. As such, it is <strong>reasonable<\/strong> to conclude that the average annual rainfall has increased from its former long-term average of 23.<\/p>\n<blockquote>\n<h2><strong>Question<\/strong><\/h2>\n<p>What is the value of <em>t<\/em> in the example above if the level of significance is reduced from 5% to 0.5%, and does this change the decision rule?<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li data-tadv-p=\"keep\">2.02; it does not change the decision rule.<\/li>\n<li data-tadv-p=\"keep\">3.25; it does not change the decision rule.<\/li>\n<li data-tadv-p=\"keep\">3.25; it does change the decision rule.<\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<p>The correct answer is <strong>B<\/strong>.<\/p>\n<p>A quick glance at the t<sub>0.005,9<\/sub> distribution when \u03b1 = 0.5% gives a value of 3.25.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-38627\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value-2.png\" alt=\"\" width=\"534\" height=\"224\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value-2.png 534w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value-2-300x126.png 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-p-value-2-400x168.png 400w\" sizes=\"auto, (max-width: 534px) 100vw, 534px\" \/><\/p>\n<p>However, the evidence against the H<sub>0<\/sub> is too strong since our test statistic (7.96) is still greater than 3.25. As such, the conclusion would remain unchanged.<\/p>\n<\/blockquote>\n<p><em>(<\/em><strong>Note to candidates:<\/strong><em> You might also work out the solutions to the above examples and questions using p-values instead of critical values. The decision rules would remain unchanged provided you work out the p-values correctly.)<\/em><\/p>\n<div class=\"notes_inv\">\n<hr \/><\/div>\n\n\n<div style=\"text-align: center; margin: 30px 0;\">\n  <a style=\"display: inline-flex; align-items: center; justify-content: center; padding: 12px 26px; border-radius: 9999px; background: #1e5bd8; color: #ffffff; font-weight: bold; text-decoration: none;\" href=\"https:\/\/analystprep.com\/free-trial\/\" target=\"_blank\" rel=\"noopener noreferrer\">\n    Start Free Trial \u2192\n  <\/a>\n  <p style=\"margin-top: 12px; font-size: 16px; line-height: 1.5;\">\n    Practice hypothesis testing, test selection, and statistical inference with CFA Level I exam-style questions.\n  <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>The z-test The z-test is the ideal hypothesis test to conduct in a normal distribution of a random variable. In addition, the variance of the population must be known. The z-statistic refers to the test statistic computed for hypothesis testing.<\/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-2553","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>t-Test vs. z-Test | CFA Level 1 - AnalystPrep<\/title>\n<meta name=\"description\" content=\"The z-test is used when the population variance is known, while the t-test is preferred when dealing with small samples and an unknown variance.\" \/>\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\/t-test-z-test\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"t-Test vs. z-Test | CFA Level 1 - 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