Having some trouble with this assignment. Any help would be greatly appreciated!

Question

Having some trouble with this assignment. Any help would be greatly appreciated! Thank you Bio 260 (Sp17) Assignment#9 Form C, Due APR 20 Name: Provide answers only in space provided, do not attach additional sheets. Write clearly to ensure full credit for your answers. Write the frst letter of your surname in box above. (1)A researcher has collected mass data for seven fish collected from each of six small ponds in central California. The researcher is interested in whether any of the ponds seem to have larger or smaller fish than the others. The data is Draenor Ewymn Kalimdor Mulgore Tirisfal Zandalar Fish 27 31 31 31 37 35 32 31 35 33 37 31 31 35 (a. 2 pts) What is F for your preliminary variance equality F test? (round to nearest 0.001) (b, 2 pts) What is the critical F value for this F test? Critical F (Use the F table posted on the website and a a 0.05) (c, 1 pt ea Complete the ANOVA table below. Provide a range for the p value using the tables provided on the course website. (round all non-integer values to nearest 0.001) Source Idf RSS MS F p Among Within (d, 3 pts) What is your conclusion? Use the grammar described in lecture and state with what degree of confidence you make your conclusion by providing the most specific range of p values from the table provided in lecture. You must use the phrase 'significantly different" or "not significantly different in your answer. Also, if any means appear to be different, make your best guess as to the most likely pair that differ, Note: no credit will be given for ANY text outside the box or hard to read answers. (2, 1 pt ea)Fillin the 5 missing values to complete the ANOVA table below. Provide an exact p value using R or Excel. Round any non-integer values to the nearest 0.001 IdM SS MS F p Source 62 Among thin Total 53

Explanation / Answer

a) Fmax= 4.4167

b) critical value = 13.7

c) Response: obs
source Df Sum Sq Mean Sq F value Pr(>F)   
Among 5 196 39.200 3.5817 0.009896
Within 36 394 10.944

Total 41 590

At 95% Confodence pvalue is < 0.05 which infer that we reject the null hypothesis i.e there is significant difference between the ponds in terms of fishes.

source Df Sum Sq Mean Sq F value Pr(>F)   
Among 5 310 62 3.1 0.01667
Within 48 960 20

Total 53 1270

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