{"id":2565,"date":"2019-08-27T17:37:00","date_gmt":"2019-08-27T17:37:00","guid":{"rendered":"https:\/\/analystprep.com\/cfa-level-1-exam\/?p=2565"},"modified":"2024-12-03T05:40:06","modified_gmt":"2024-12-03T05:40:06","slug":"chi-square-test-example","status":"publish","type":"post","link":"https:\/\/analystprep.com\/cfa-level-1-exam\/quantitative-methods\/chi-square-test-example\/","title":{"rendered":"Chi-square Test of a Single Population Variance and F-test of the Ratio of Two Population Variances"},"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 video covers hypothesis testing in quantitative methods. It explains key concepts, including null and alternative hypotheses, one- and two-tailed tests, Type I\/II errors, p-values, and parametric\/non-parametric tests. Using examples, it simplifies statistical techniques, decision rules, and applications for interpreting financial data, preparing candidates for exam questions.\",\n  \"uploadDate\": \"2018-11-18T00:00:00+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\": \"ImageObject\",\n  \"url\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-189.jpg\",\n  \"caption\": \"Image of the two-tailed chi-squared test (5% significance).\",\n  \"width\": 1373,\n  \"height\": 860,\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}\n<\/script>\n\n<script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"ImageObject\",\n  \"url\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-chi-square-table-1.png\",\n  \"caption\": \"Image of the chi-square table.\",\n  \"width\": 444,\n  \"height\": 427,\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}\n<\/script>\n\n<script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"ImageObject\",\n  \"url\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-190.jpg\",\n  \"caption\": \"Image of the two-tailed chi-squared test (5% significance).\",\n  \"width\": 1373,\n  \"height\": 976,\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}\n<\/script>\n\n<script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"ImageObject\",\n  \"url\": \"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-F-distribution.png\",\n  \"caption\": \"Image of the F-distribution.\",\n  \"width\": 742,\n  \"height\": 453,\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}\n<\/script>\n\n\n<p><iframe loading=\"lazy\" src=\"\/\/www.youtube.com\/embed\/twTTWMuEGx8\" width=\"611\" height=\"343\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><strong>Testing the Variances of a Normally Distributed Population using the Chi-square Test<\/strong><\/h2>\n<p>A chi-square test is used to establish whether a hypothesized value of variance is equal to, less than, or greater than the true population variance. Unlike most distributions covered in the CFA\u00ae Program curriculum, the chi-square distribution is asymmetrical. However, the distribution approaches the \u201cnormal\u201d one as the degrees of freedom increase.<\/p>\n<p><!--more--><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-17069 size-full\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-189.jpg\" alt=\"Image of the two-tailed chi-squared test (5% significance)\" width=\"1373\" height=\"860\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-189.jpg 1373w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-189-300x188.jpg 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-189-768x481.jpg 768w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-189-1024x641.jpg 1024w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-189-400x251.jpg 400w\" sizes=\"auto, (max-width: 1373px) 100vw, 1373px\" \/><\/p>\n<p>As a natural consequence, the chi-square distribution has no negative values and is bounded by zero. A chi-square statistic with (n \u2013 1) degrees of freedom is computed as:<\/p>\n<p>$$ { \\chi }_{ n-1 }^{ 2 }=\\frac { \\left( n-1 \\right) { S }^{ 2 } }{ { \\sigma }_{ 0 }^{ 2 } } $$<\/p>\n<p>Where:<\/p>\n<p>\\(n\\) = Sample size.<\/p>\n<p>\\(S^2\\)&nbsp;= Sample variance.<\/p>\n<p>\\({ \\sigma }_{ 0 }^{ 2 }\\) = Hypothesized population variance.<\/p>\n<h3><strong>Example: Chi-square Test<br><\/strong><\/h3>\n<p>For the 15-year period between 1995 and 2010, ABC\u2019s monthly return had a standard deviation of 5%. John Matthew, CFA, wishes to establish whether the standard deviation witnessed during that period still adequately describes the long-term standard deviation of the company\u2019s return. To achieve this end, he collects data on the monthly returns recorded between 1<sup>st<\/sup> January, 2015 and 31<sup>st<\/sup> December, 2016 and computes a monthly standard deviation of 4%.<\/p>\n<p>Carry out a 5% test to determine if the standard deviation computed in the latter period is different from the 15-year value.<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>As is the norm, start by writing down the hypothesis<\/p>\n<p>\\(H_0: \\sigma_0^2= 0.0025\\)<\/p>\n<p>\\(H_1: \\sigma^2 \\neq 0.0025\\)<\/p>\n<p>Since the latter period has 24 months, <em>n<\/em> = 24. The test statistic is:<\/p>\n<p>$$ { \\chi }_{ 24-1 }^{ 2 }=\\frac { \\left( 24-1 \\right) { 0.0016 } }{ 0.0025 } =14.72 $$<\/p>\n<p>This is a two-tailed test. As such, we have to divide the significance level by two and screen our test statistic against the lower and upper 2.5% points of \\({ \\chi }_{ 23 }^{ 2 }\\).<\/p>\n<p>Consulting the chi-square table, the test statistic (14.72) lies between the lower (11.689) and the upper (38.076) 2.5% points of the chi-square distribution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-38688 size-full\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-chi-square-table-1.png\" alt=\"Image of the chi-square table\" width=\"444\" height=\"427\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-chi-square-table-1.png 444w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-chi-square-table-1-300x289.png 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-chi-square-table-1-400x385.png 400w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-chi-square-table-1-24x24.png 24w\" sizes=\"auto, (max-width: 444px) 100vw, 444px\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-17070 size-full\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-190.jpg\" alt=\"Image of the two-tailed chi-squared test (5% significance)\" width=\"1373\" height=\"976\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-190.jpg 1373w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-190-300x213.jpg 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-190-768x546.jpg 768w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-190-1024x728.jpg 1024w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/page-190-400x284.jpg 400w\" sizes=\"auto, (max-width: 1373px) 100vw, 1373px\" \/><\/p>\n<p>Therefore, we have insufficient evidence to reject the H<sub>0<\/sub>. It is, therefore, reasonable to conclude that the latter standard deviation value is close enough to the 15-year value.<\/p>\n<h2><strong>Testing the Equality of the Variances of Two Normally Distributed Populations based on Two Independent Random Samples using the F-distribution<\/strong><\/h2>\n<p>Assume that we have 2 independent random samples of sizes n<sub>1<\/sub> and n<sub>2<\/sub> from \\(N(\\mu_1, \\sigma_1^2)\\) and \\(N(\\mu_2, \\sigma_2^2)\\).<\/p>\n<p>Also, let us consider a scenario where we have the sample variances as \\(S_1^2\\) and \\(S_2^2\\). The basic situation usually involves the following hypothesis:<\/p>\n<p>\\(H_0: \\ \\sigma_1^2 =\\sigma_2^2\\)<\/p>\n<p>\\(H_1: \\ \\sigma_1^2 \\neq \\sigma_2^2\\)<\/p>\n<p>The test statistic is \\(\\frac {S_1^2}{S_2^2} \\sim F_{n1 \u2013 1, n2 \u2013 1 } \\text{ under } H_0\\).<\/p>\n<p>The decision rule is to reject the null hypothesis if the test statistic falls within the critical region of the F-distribution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-38692 size-full\" src=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-F-distribution.png\" alt=\"Image of the F-distribution\" width=\"742\" height=\"453\" srcset=\"https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-F-distribution.png 742w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-F-distribution-300x183.png 300w, https:\/\/analystprep.com\/cfa-level-1-exam\/wp-content\/uploads\/2019\/08\/cfa-level-1-F-distribution-400x244.png 400w\" sizes=\"auto, (max-width: 742px) 100vw, 742px\" \/><\/p>\n<h4><strong>Example: <\/strong><strong>F-test<\/strong><\/h4>\n<p>Use the following data to calculate the test statistic. Establish if the variances of the two populations \u201cA\u201d and \u201cB\u201d can be considered to be equal at the 95% confidence level.<\/p>\n<p>\\(\\sum A^2 =&nbsp; 56,430\\) and \\(\\sum B^2 = 59,520\\) and \\(\\bar{A}= 75\\) while \\(\\bar{B} = 77\\), \\(n_1 = 10\\) and \\(n_2 = 10\\)<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, state the hypothesis<\/p>\n<p>\\(H_0: \\ \\sigma_A^2 = \\sigma_B^2\\)<\/p>\n<p>\\(H_1: \\ \\sigma_A^2 \\neq \\sigma_B^2\\)<\/p>\n<p>Using the formula:<\/p>\n<p>$$Var(X)=s^2=\\frac{1}{n-1}\\left[\\sum{X^2}-n\\bar{X}^2\\right]$$<\/p>\n<p>We have:<\/p>\n<p>$$ \\begin{align*} S_A^2 &amp; =\\cfrac { \\left\\{ 56,430 \u2013 (10 \u00d7 75^2) \\right\\} }{9} = 20 \\\\ S_B^2 &amp; =\\cfrac { \\left\\{ 59,520 \u2013 (10 \u00d7 77^2) \\right\\} }{9} = 25.556 \\\\ \\end{align*} $$<\/p>\n<p>Now,<\/p>\n<p>$$ \\cfrac {S_A^2}{ S_B^2}=\\cfrac {20}{25.556} = 0.783 $$<\/p>\n<p>We should compare the test statistic with the upper and lower 2.5% points of the F-distribution. These points are 4.026 and 1\/4.026 = 0.2484. Since the test statistic lies between the two limits, we have <strong>insufficient<\/strong> evidence against the H<sub>0<\/sub>. Therefore, it would be reasonable to conclude that the two samples have equal variance.<\/p>","protected":false},"excerpt":{"rendered":"<p>Testing the Variances of a Normally Distributed Population using the Chi-square Test A chi-square test is used to establish whether a hypothesized value of variance is equal to, less than, or greater than the true population variance. Unlike most distributions&#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-2565","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 v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Chi-square test Question Example | CFA Level 1<\/title>\n<meta name=\"description\" content=\"Learn how to perform a chi-square test with solved examples. 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