{"id":160366,"date":"2025-03-27T20:29:48","date_gmt":"2025-03-27T20:29:48","guid":{"rendered":"https:\/\/peraltafinancing.com\/agriculture\/completely-randomized-design-analysis-along-with-lsd-test\/"},"modified":"2025-03-27T20:29:48","modified_gmt":"2025-03-27T20:29:48","slug":"completely-randomized-design-analysis-along-with-lsd-test","status":"publish","type":"post","link":"https:\/\/fivemor.com\/?p=160366","title":{"rendered":"Completely Randomized Design Analysis along with LSD test"},"content":{"rendered":"


\n<\/p>\n

\n

Problem Statement<\/b><\/p>\n

In order to find out the yielding abilities of 5 varities of sesamum an experiment was conducted in the green house using a CRD with 4 pots per variety. The data for Yield (in g) is given below.<\/p>\n

\u00a0<\/p>\n\n\n\n\n\n\n\n\n
\n

Varieties<\/span><\/b><\/p>\n<\/td>\n<\/tr>\n

\n

V1<\/span><\/b><\/p>\n<\/td>\n

\n

V2<\/span><\/b><\/p>\n<\/td>\n

\n

V3<\/span><\/b><\/p>\n<\/td>\n

\n

V4<\/span><\/b><\/p>\n<\/td>\n

\n

V5<\/span><\/b><\/p>\n<\/td>\n<\/tr>\n

\n

10<\/span><\/p>\n<\/td>\n

\n

12<\/span><\/p>\n<\/td>\n

\n

20<\/span><\/p>\n<\/td>\n

\n

30<\/span><\/p>\n<\/td>\n

\n

17<\/span><\/p>\n<\/td>\n<\/tr>\n

\n

8<\/span><\/p>\n<\/td>\n

\n

15<\/span><\/p>\n<\/td>\n

\n

21<\/span><\/p>\n<\/td>\n

\n

32<\/span><\/p>\n<\/td>\n

\n

19<\/span><\/p>\n<\/td>\n<\/tr>\n

\n

12<\/span><\/p>\n<\/td>\n

\n

14<\/span><\/p>\n<\/td>\n

\n

25<\/span><\/p>\n<\/td>\n

\n

34<\/span><\/p>\n<\/td>\n

\n

18<\/span><\/p>\n<\/td>\n<\/tr>\n

\n

11<\/span><\/p>\n<\/td>\n

\n

13<\/span><\/p>\n<\/td>\n

\n

24<\/span><\/p>\n<\/td>\n

\n

29<\/span><\/p>\n<\/td>\n

\n

20<\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u00a0<\/p>\n

F value (0.05,4,15) = 3.05 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 t value (0.05,15) = 2.13<\/p>\n

\u00a0F value (0.01,4,15) = 4.89 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 t value (0.01,15) = 2.94<\/p>\n

\u00a0Work out the CRD analysis and perform LSD test if required. Interpret the results.<\/p>\n

\u00a0<\/p>\n

Step 1: Calculate Correction Factor<\/b><\/p>\n

Grand\u00a0Total\u00a0=\u00a0Sum\u00a0of all\u00a0observation\u00a0=\u00a010 +\u00a08\u00a0+12 +\u00a0. . .\u00a0+\u00a018 +\u00a020\u00a0=\u00a0384<\/p>\n

\ud835\udc36\ud835\udc39 =<\/span>\u00a0 \u00a0\ud835\udc3a\ud835\udc472<\/sup><\/span>\/r*t<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0= \u00a0\u00a0<\/span>3842<\/sup><\/span>\/4 \u2217 5<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0= \u00a0 7372.8<\/span><\/p>\n

Step 2: Calculate Total SS<\/b><\/p>\n

Total\u00a0SS =\u00a0Sum of Square\u00a0of all observations\u00a0\u2013 CF<\/p>\n

=\u00a0102<\/sup> + 82<\/sup> + 122<\/sup>+\u00a0.\u00a0.\u00a0.\u00a0+\u00a0182<\/sup> + 202<\/sup> \u2212 7372.8 \u00a0 = \u00a0 \u00a0 1127.20<\/span><\/p>\n

Step 3: Calculate Treatment SS<\/b><\/p>\n

\u00a0<\/span><\/p>\n\n\n\n
\n

V1 =\u00a041<\/span><\/p>\n<\/td>\n

\n

V2 =\u00a054<\/span><\/p>\n<\/td>\n

\n

V3\u00a0=\u00a090<\/span><\/p>\n<\/td>\n

\n

V4\u00a0=\u00a0125<\/span><\/p>\n<\/td>\n

\n

V5 =\u00a074<\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Treatment\u00a0SS\u00a0= (1\/r)* [\u00a0\ud835\udc472<\/sup> + \ud835\udc472<\/sup> + \ud835\udc472<\/sup> + \ud835\udc472<\/sup> + \ud835\udc472<\/sup>] \u2212 \ud835\udc36\ud835\udc39<\/span><\/p>\n

= (1\/4) * [\u00a0412<\/sup> + 542<\/sup> + 902<\/sup> + 1252<\/sup> + 742<\/sup> ] \u2212 7372.8<\/span><\/p>\n

=\u00a01076.7<\/span><\/p>\n

Step 4: Calculate Error SS<\/b><\/p>\n

Error SS = Total SS \u2013 Treatment SS<\/span><\/p>\n

= \u00a01127.20 – 1076.7 \u00a0<\/span>= \u00a050.50
<\/span><\/p>\n

\u00a0\u00a0<\/span><\/p>\n

Step 5: Calculate degree of freedom<\/b><\/span><\/p>\n

Treatment DF \u00a0 = \u00a0 t \u2013 1 \u00a0= \u00a0 5 – 1 \u00a0= \u00a0 4<\/span><\/p>\n

Total DF \u00a0 = r*t \u2013 1 \u00a0 = 5*4 \u2013 1 \u00a0= 19<\/span><\/p>\n

Error DF \u00a0= Total DF \u2013 Treatment DF \u00a0= 19 \u2013 4 \u00a0 = 15<\/span><\/p>\n

\u00a0<\/span><\/p>\n

Step 6: Calculate Mean Square for all sources<\/b><\/span><\/p>\n

Treatment MS = \u00a0Treatment SS \/ Treatment DF<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = \u00a01076.7 \/ 4<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = \u00a0 269.175<\/p>\n

\u00a0<\/span><\/p>\n

Error MS \u00a0= \u00a0Error SS \/ Error DF<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= \u00a050.50 \/ 15<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= \u00a03.37<\/span><\/p>\n

Step 7: Calculate F value<\/b><\/span><\/p>\n

Calculated F = \u00a0Treatment MS \/ Error MS<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= \u00a0269.175 \/ 3.37<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= \u00a079.95<\/span><\/p>\n

Step 8: Prepare ANOVA<\/b><\/span><\/p>\n

\n\n\n\n\n\n\n
\n

Source of Variation<\/span><\/b><\/p>\n<\/td>\n

\n

Degrees of Freedom<\/span><\/b><\/p>\n<\/td>\n

\n

Sum of Squares<\/span><\/b><\/p>\n<\/td>\n

\n

Mean Square<\/span><\/b><\/p>\n<\/td>\n

\n

Calculated F value<\/span><\/b><\/p>\n<\/td>\n<\/tr>\n

\n

Treatment<\/span><\/p>\n<\/td>\n

\n

4<\/span><\/p>\n<\/td>\n

\n

1076.70<\/span><\/p>\n<\/td>\n

\n

269.18<\/span><\/p>\n<\/td>\n

\n

79.95<\/span><\/p>\n<\/td>\n<\/tr>\n

\n

Error<\/span><\/p>\n<\/td>\n

\n

15<\/span><\/p>\n<\/td>\n

\n

50.50<\/span><\/p>\n<\/td>\n

\n

3.37<\/span><\/p>\n<\/td>\n

\n

–<\/span><\/p>\n<\/td>\n<\/tr>\n

\n

Total<\/span><\/p>\n<\/td>\n

\n

19<\/span><\/p>\n<\/td>\n

\n

1127.20<\/span><\/p>\n<\/td>\n

\n

–<\/span><\/p>\n<\/td>\n

\n

–<\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0<\/span><\/p>\n

The null hypothesis is\u00a0<\/span>\u00a0 which mean that mean performance of all the treatments is same.<\/span><\/p>\n

In order to check this null hypothesis we are required to compare the Calculate F value (79.95) with Table F value (3.05). As the Calculated F value is greater than Table F value our results are significant at 5 % level of significance. Further we can test for 1 % level by comparing Calculate F value (79.95) with Table F value (4.89), so result is also significant at 1 % level of significance, and we reject our Null hypothesis.\u00a0<\/span><\/p>\n

This mean that mean performance of all the treatment is not same. This raises the question that which treatment gives a better Yield. In order to get this answer, we need to perform LSD test.<\/span><\/p>\n

LSD TEST<\/b><\/span> (FOR MEAN COMPARISON)<\/b><\/p>\n

Step 1: Calculate CD (critical difference) value<\/b><\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/b><\/span><\/p>\n

Step 2: Calculate the treatment means and arrange them is orde<\/b>r<\/span><\/p>\n

The treatment mean is obtained by dividing treatment total with number of replications<\/span><\/p>\n\n\n\n
\n

V4 = 31.25<\/span><\/p>\n<\/td>\n

\n

V3 = 22.50<\/span><\/p>\n<\/td>\n

\n

V5 = 18.50<\/span><\/p>\n<\/td>\n

\n

V2 = 13.50\u00a0<\/span><\/p>\n<\/td>\n

\n

V1 = 10.25<\/span><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u00a0<\/span><\/p>\n

Step 3: Allocate same letters to treatment that are at par<\/span><\/p>\n

V4 mean \u2013 CD = 31.25 \u2013 3.82 = 27.43. The means ranging from 31.25 to 27.43 are at par with V4 but none of the means are in this range. So, V4a<\/sup>.<\/span><\/p>\n

V3 mean \u2013 CD = 22.50 \u2013 3.82 = 18.68. The means ranging from 22.50 to 18.68 are at par with V3 and share letter b but none of the mean are in this range. So, V3b<\/sup>.<\/span><\/p>\n

V5 mean \u2013 CD = 18.50 \u2013 3.82 = 14.68. The means ranging from 18.50 to 14.68 are at par with V5 and share letter c but none of the mean are in this range. So, V5c<\/sup>.<\/span><\/p>\n

V2 mean \u2013 CD = 13.50 \u2013 3.82 = 9.68. The means ranging from 13.50 to 9.68 are at par with V2 and share letter d and V1 (10.25) is in this range. So, V2d<\/sup> and V1d\u00a0<\/sup>.<\/span><\/p>\n

Calculation of CV<\/span><\/b><\/p>\n

\u00a0<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span><\/p>\n

Conclusion<\/span><\/b><\/p>\n

The ANOVA results reveal that the treatment component is significant at 1 % level of significance. The LSD test reveals that highest yield was observed for treatment V4 and none of the treatments was at par with it.\u00a0<\/span><\/p>\n

Similar Posts:<\/b><\/p>\n

Quiz on Completely Randomized Design<\/b><\/a><\/p>\n

Quiz on Randomized Complete Block Design<\/b><\/a><\/p>\n

R package for Analysis of Design of Experiments <\/a>
<\/b><\/p>\n<\/div>\n\n","protected":false},"excerpt":{"rendered":"

Problem Statement In order to find out the yielding abilities of 5 varities of sesamum an experiment was conducted in the green house using a CRD with 4 pots per variety. The data for Yield (in g) is given below. \u00a0 Varieties V1 V2 V3 V4 V5 10 12 20 30 17 8 15 21 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12026],"tags":[10990,21461,1185,65864,60392,13597],"dealstore":[],"offerexpiration":[],"class_list":["post-160366","post","type-post","status-publish","format-standard","hentry","category-agriculture","tag-analysis","tag-completely","tag-design","tag-lsd","tag-randomized","tag-test"],"yoast_head":"\nCompletely Randomized Design Analysis along with LSD test - Som2ny Network<\/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:\/\/fivemor.com\/?p=160366\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Completely Randomized Design Analysis along with LSD test - Som2ny Network\" \/>\n<meta property=\"og:description\" content=\"Problem Statement In order to find out the yielding abilities of 5 varities of sesamum an experiment was conducted in the green house using a CRD with 4 pots per variety. 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Som2ny Network","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/fivemor.com\/?p=160366","og_locale":"en_US","og_type":"article","og_title":"Completely Randomized Design Analysis along with LSD test - Som2ny Network","og_description":"Problem Statement In order to find out the yielding abilities of 5 varities of sesamum an experiment was conducted in the green house using a CRD with 4 pots per variety. The data for Yield (in g) is given below. \u00a0 Varieties V1 V2 V3 V4 V5 10 12 20 30 17 8 15 21 […]","og_url":"https:\/\/fivemor.com\/?p=160366","og_site_name":"Som2ny Network","article_published_time":"2025-03-27T20:29:48+00:00","author":"admin","twitter_card":"summary_large_image","twitter_misc":{"Written by":"admin","Est. reading time":"3 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/fivemor.com\/?p=160366#article","isPartOf":{"@id":"https:\/\/fivemor.com\/?p=160366"},"author":{"name":"admin","@id":"https:\/\/fivemor.com\/#\/schema\/person\/b85e3c3dc0e1daea076524dc8810c371"},"headline":"Completely Randomized Design Analysis along with LSD test","datePublished":"2025-03-27T20:29:48+00:00","mainEntityOfPage":{"@id":"https:\/\/fivemor.com\/?p=160366"},"wordCount":536,"commentCount":0,"publisher":{"@id":"https:\/\/fivemor.com\/#organization"},"keywords":["Analysis","Completely","Design","LSD","Randomized","Test"],"articleSection":["Agriculture"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/fivemor.com\/?p=160366#respond"]}]},{"@type":"WebPage","@id":"https:\/\/fivemor.com\/?p=160366","url":"https:\/\/fivemor.com\/?p=160366","name":"Completely Randomized Design Analysis along with LSD test - 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