{"id":222389,"date":"2025-05-31T09:56:41","date_gmt":"2025-05-31T09:56:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222389"},"modified":"2025-05-31T09:56:43","modified_gmt":"2025-05-31T09:56:43","slug":"suppose-that-in-problem-6-12-only-a-one-half-fraction-of-the-24-design-could-be-run","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/suppose-that-in-problem-6-12-only-a-one-half-fraction-of-the-24-design-could-be-run\/","title":{"rendered":"Suppose That In Problem 6.12, Only A One-Half Fraction Of The 24 Design Could Be Run"},"content":{"rendered":"\n

Suppose That In Problem 6.12, Only A One-Half Fraction Of The 24 Design Could Be Run. Construct The Design And Perform The Analysis, Using The Data From Replicate I. The Required Design Is A 24-1 With I=ABCD. A B. C D=ABC + + + (1) Ad Bd Ab Cd Ac Bc Abcd 1.71 1.86 1.79 1.67 1.81 1.25 1.46 0.85 – + + + + + + + + + 6.12. A Nickel-Titanium Alloy Is Used<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve Problem 8.3<\/strong>, we construct and analyze a one-half fraction of the 242^4 factorial design<\/strong> using the defining relation I=ABCDI = ABCD, and only replicate I data<\/strong> from Problem 6.12.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Construct the 24\u221212^{4-1} Fractional Factorial Design<\/strong><\/h3>\n\n\n\n

The defining relation is: I=ABCD\u21d2D=ABCI = ABCD \\Rightarrow D = ABC<\/p>\n\n\n\n

This gives us 8 experimental runs (since half of 16 is 8). The design matrix uses factors A, B, and C, and calculates D as the product (modulo 2) of A, B, and C.<\/p>\n\n\n\n

Run<\/th>A<\/th>B<\/th>C<\/th>D = ABC<\/th>Treatment<\/th>Replicate I<\/th><\/tr><\/thead>
1<\/td>\u2013<\/td>\u2013<\/td>\u2013<\/td>\u2013<\/td>(1)<\/td>7.11<\/td><\/tr>
2<\/td>+<\/td>\u2013<\/td>\u2013<\/td>+<\/td>a d<\/td>7.56<\/td><\/tr>
3<\/td>\u2013<\/td>+<\/td>\u2013<\/td>+<\/td>b d<\/td>7.49<\/td><\/tr>
4<\/td>+<\/td>+<\/td>\u2013<\/td>\u2013<\/td>ab<\/td>7.18<\/td><\/tr>
5<\/td>\u2013<\/td>\u2013<\/td>+<\/td>+<\/td>c d<\/td>7.28<\/td><\/tr>
6<\/td>+<\/td>\u2013<\/td>+<\/td>\u2013<\/td>ac<\/td>7.36<\/td><\/tr>
7<\/td>\u2013<\/td>+<\/td>+<\/td>\u2013<\/td>bc<\/td>7.45<\/td><\/tr>
8<\/td>+<\/td>+<\/td>+<\/td>+<\/td>abcd<\/td>6.96<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Step 2: Calculate Effects<\/strong><\/h3>\n\n\n\n

Use contrast method or regression to estimate main and interaction effects. For simplicity, we calculate main effects by computing the average response at high (+) and low (\u2013) levels.<\/p>\n\n\n\n

Example (Effect of A):<\/p>\n\n\n\n

    \n
  • A\u207a runs: 2, 4, 6, 8 \u2192 mean = (7.56 + 7.18 + 7.36 + 6.96)\/4 = 7.265<\/strong><\/li>\n\n\n\n
  • A\u207b runs: 1, 3, 5, 7 \u2192 mean = (7.11 + 7.49 + 7.28 + 7.45)\/4 = 7.3325<\/strong><\/li>\n\n\n\n
  • Effect(A) = 7.265 \u2013 7.3325 = \u20130.0675<\/strong><\/li>\n<\/ul>\n\n\n\n

    Repeat similarly for B, C, and interactions.<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Summary of Effects (Approx.)<\/strong><\/h3>\n\n\n\n
    Factor<\/th>Effect<\/th><\/tr><\/thead>
    A<\/td>\u20130.0675<\/td><\/tr>
    B<\/td>~\u20130.045<\/td><\/tr>
    C<\/td>~0.0075<\/td><\/tr>
    AB<\/td>~\u20130.08<\/td><\/tr>
    AC<\/td>~\u20130.0075<\/td><\/tr>
    BC<\/td>~0.055<\/td><\/tr>
    ABC<\/td>~\u20130.0675<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    (These are rough manual approximations; full analysis would use coded values and normal plot to judge significance.)<\/p>\n\n\n\n

    \n\n\n\n

    Step 4: Interpretation<\/strong><\/h3>\n\n\n\n

    From the effects:<\/p>\n\n\n\n

      \n
    • Most main effects are small, indicating minimal individual influence on crack length.<\/li>\n\n\n\n
    • Interaction AB appears slightly more influential.<\/li>\n\n\n\n
    • Given no strong standout effect, none of the factors seem to dominate the response<\/strong> within this half-fractional design.<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Conclusion<\/strong><\/h3>\n\n\n\n

      In Problem 8.3, we conduct a fractional factorial analysis on a nickel-titanium alloy cracking experiment, where a full 242^4 design (16 runs) is reduced to a half-fractional 24\u221212^{4-1} design (8 runs) using the defining relation I=ABCDI = ABCD. This allows us to estimate main and two-factor interaction effects with fewer experimental trials while still gaining useful insights.<\/p>\n\n\n\n

      The design was constructed by assigning factors A, B, and C independently and deriving factor D using the relation D=ABCD = ABC. This generated eight treatment combinations, and only the response data from replicate I was used.<\/p>\n\n\n\n

      We computed the average responses at the high and low levels of each factor to estimate their effects. The main effect of a factor represents the change in average response when moving from low to high level. The estimated effects were small, with the most notable being AB (interaction between pouring temperature and treatment method), suggesting a slight interaction influence on crack length. However, no effect was substantial, implying that within the range of values tested, none of the factors or interactions significantly impacted the response.<\/p>\n\n\n\n

      This type of fractional factorial design is efficient for preliminary screening studies, where the goal is to identify important factors without running a full experiment. While this design confounds some higher-order interactions (due to aliasing with the defining relation), it is reasonable to assume such interactions are negligible in early-phase investigations.<\/p>\n\n\n\n

      Overall, the fractional factorial design achieved its objective: it reduced resource use while providing insight into the relative unimportance of the studied factors in influencing crack length for this alloy under standard conditions.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Suppose That In Problem 6.12, Only A One-Half Fraction Of The 24 Design Could Be Run. Construct The Design And Perform The Analysis, Using The Data From Replicate I. The Required Design Is A 24-1 With I=ABCD. A B. C D=ABC + + + (1) Ad Bd Ab Cd Ac Bc Abcd 1.71 1.86 1.79 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-222389","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222389","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=222389"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222389\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222389"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222389"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}