A horticulturist is designing a study to investigate the effectiveness of five m

Question

A horticulturist is designing a study to investigate the effectiveness of five methods for the irrigation of blueberry shrubs. The methods are surface, trickle, center pivot, lateral move, and subirrigation. There are 10 blueberry farms available for the study representing a wide variety of types of soil, terrains, and wind gradients. The horticulturist wants to use each of the five methods of irrigation on all 10 farms to moderate the effect of the many extraneous sources of variation that may impact the blueberry yields. Each farm is divided into five and the response variable will be the weight of the harvested fruit from each of the plots of blueberry shrubs. Method of Irrigation Farm Surface Trickle Center Point Lateral Subirrigation Farm Mean 391 350 370 248 401.9 636 382 591 348 603 366 649 258 512 32 588 423 689 406 492 452 343 493.9 430.9 408.7 449 406 550 526 500.0 507.3 502 419 458 389.6 550 Mcthod Mean 616.3 353.2 464.3 497.6 464.2 Tvo way ANOVA: Yield veraus Method, Far Source DF Mathod 4 421560 105390 62.10 0.000 Parm Error 36 62097 1725 Total 49 549975 9 66318 7369 4.270.001 S" 41.53 R. Sq. 88.729 R-Sqtaaj) -84.630

Explanation / Answer

a) From the residual plot we can say that the assumption of constant variance is satisfied.

For normality we have to test the hypothesis that,

H0 : The data follows normal distribution.

H1 : The data does not follows normal distribution.

Assume alpha = level of significance = 0.05

We have given that P-value for the normal probability plot is > 0.100

Accept H0 at 5% level of significance.

Conclusion : The data follows normal distribution.

b) Now we have to find standard error for each five methods of irrigation.

standard error = sd / sqrt(n)

where sd is standard deviation.

The five standard deviations and standard errors are :

d) Here we have to test the hypothesis that,

H0 : All the means are equal.

H1 : Atleast one of the mean is differ than 0.

Test statistic follows F-distribution.

This is the one way anova.

We can do one way anova in excel.

steps :

ENTER data into excel sheet --> Data --> Data Analysis --> Anova : Single factor --> ok --> Labels in first row --> Alpha : 0.05--> Output range : select one empty cell --> ok

Test statistic = 36.97

P-value = 0.000

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : Atleast one of the mean is differ than 0.

sd se 52.82266 16.70399 60.32836 19.0775 48.23369 15.25283 52.00684 16.44601 52.73635 16.6767

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