Four kinds of fertilizer f1, f2, f3nd f4 are used to study the yield of beans. The soil is divi

step 1 Okay, this question says four kinds of fertilizer, F1, F2, F3 and F4. I used to study the yield

Four kinds of fertilizer f1, f2, f3nd f4 are used to study...

Key Concepts

Randomized Complete Block Design

This is an experimental design where subjects or experimental units are grouped into blocks based on a factor that is expected to affect the outcome, and then each treatment is randomly assigned within each block. This design helps control for variation among blocks, allowing a clearer assessment of treatment effects by reducing error variability attributed to block differences.

Analysis of Variance (ANOVA)

ANOVA is a statistical method used to compare means among multiple groups to determine if at least one differs significantly from the others. In the context of a randomized complete block design, ANOVA partitions the total variability into components due to treatments, blocks, and random error, enabling researchers to test the significance of the treatment effects while accounting for block variability.

Hypothesis Testing and Significance Levels

This concept involves formulating a null hypothesis, which typically states that there is no effect or difference among groups, and an alternative hypothesis that indicates the presence of an effect. Significance levels, such as 0.05 or 0.01, are used to decide whether to reject the null hypothesis based on the probability of observing the test statistic under that hypothesis, hence guiding the interpretation of experimental results.

Single-Degree-of-Freedom Contrasts

Contrasts are linear combinations of treatment means that allow for specific comparisons between groups in an experiment. A single-degree-of-freedom contrast tests a specific hypothesis about these means, providing a focused analysis that can yield more precise insights into differences between particular treatments. This approach is useful when comparing just two groupings or pairs of treatments within the overall experimental design.

Transcript

00:03 Okay, this question says four kinds of fertilizer, f1, f2, f3 and f4.

00:09 I used to study the yield of beans.

00:13 The soil is divided into three blocks, each containing four homogeneous plots.

00:19 The yields in kilograms per plot and the corresponding treatment are as follows.

00:24 Okay, so we have block one is f1, f3, f4 and f2.

00:30 Block two is f3, f1, f2 and f4.

00:34 Right, then in block three there, which is f4, f2, f1, and f3 there with the figures 5 .151, 56 .3, 51 .9, and 53 .5 for block three.

00:49 Then it says conduct analysis of variance at the 0 .05 level of significance using the randomized complete block model.

00:58 Then p says, use single degree of freedom, contract.

01:04 At 0 .01 level of significance to compare the fertilizers f1 f3 versus f2 f1 and draw conclusions there all right so how do we do this one so first of all let's establish our hypothesis we're saying our now hypothesis is right let me get a red pen there our now hypothesis you're saying alpha 1 is goes to alpha 2 which is goes to alpha 3 which is what to alpha 4 and there are all equal to zero there we are saying fertilizer effects are zero so the effects of fertilizer is zero that's this is our now hypothesis and our alternative hypothesis we are saying at least one of the alphas is not equal to zero so one of the alpha differs from the rest there okay so and then the level of significance is zero comma zero five so do the computations there so what we will do with the computations first of all we must get the cf which is the connection factor there after getting the connection factor we must get the s s t there right some totals there then after getting the sum totals you must get the ss a and then get the sp right then we will apply the anover there okay i just put it this way so that it can be easy for us to to flow in that way all right so our cf you square um you add all the values the figures that we have from block one block two and block three the totals you square them then divide by the how many are 12 or any talk so to get the cf there so when we did that right my cf my cf my cf was 2477 770 comma 2533 okay so with this one i'll have to use this cf to then calculate right the s s the s t there so my s t is you sum all right the y and you square them up okay it's the squares is the summation of the squares there right and then you divide by you divide by n divide by our n there there okay so that means you square everything inside there you square all the 12 data sets there and then you must you subtract your cf there so when you do that the figure that i quote is was 4 8 7 2 2 .6 7 which is the figure that you see here in the total thing this is the sst here so now after getting this we have to get the ssa a there why how do you get the s a right to get the s a uh that's a division five by four right you add all the columns going downwards all right going downward and then the totals uh for block a totals for block b and totals four right so the twos for block a you square them and then you divide by 4 you add them to the totals of the square of block b and c like that square and then you divide everything by 4 and then you subtract cf and what you get there is um 197 comma 637 7 and then you do the same for uh the sum of squares for the fertilizer here but now you are doing it across so we're adding the column the totals of the column dividing now by three you add them all and then you get two right this is your s sb you get 218 .1 1 933 there okay which is this figure so then uh to get now the error you just add the two you add these two right you separate them from the total so that you get this figure there which is 7 1 .4 017 now to get now to get the decrease of freedom right for these uh these ones are four we have four right four columns going downwards they are four so you subject one you get three and the rows going downwards there are three rows there you subject one then you get two there that's the second two degrees of freedom and then uh everything everything inside there are 12 data sets 12 status minus one you get the totals of the data such which is 11 there then you add these two you add these two which is five separate from 11 you get your your error there okay right now to get the mean squares what you do for this fertilizer you divide 218 1 .193 divided by 3 you get 7 2 0 .7 3 right you do that for the blocks again you divide by 2 then you do that for the error now to get this six one what do you do you divide the sum the mean squares of fertilizer there you divide by 11 there and then you get six comma six one that's the f that we we computed there so now the conclusion was in the p value the p value that we get there is 0 .0296 there so right from that figure that we get there okay right so what do we say right, the decision that we have to make, looking at the p -vard there, we are going to say reject the hypothesis that the fertilizer is no effect there.

07:25 The fertilizer effect is zero there.

07:28 So it means the means are not equal.

07:31 You're saying the means are not equal there.

07:34 So there is an effect of the fertilizer, isn't it? then question b, the results of testing the contrast is shown.

07:43 Right as eight the sources of variation the source of variation that is f1 and f2 versus f2 and f4 the sum of squares right if you do the tables there's sum of squares you get 2 .06 there and then for the error it's 71 .40 and then f1 versus f2 there you get four and then the degrees of freedom there is one that this is two right uh two columns two columns then you separate one then it comes two then these are two rows right which is one then the error there six there decrease of freedom so what you do here you divide everything by one you divide this summer squares by one you get two and this one and then this for error i divide by one now to get a 17 there what you do you divide two six two zero six comma six seven divided by the error the mean square for the error there and then you get 17 .37...

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