Project 6: test Comparison of cell length in two photosynthetic organisms using

Question

Project 6: test Comparison of cell length in two photosynthetic organisms using a t Do a -test to compare the lengths of the Spingnu and Eladea cells. Table 3: Using the class data for Projects 4 and 5, ill in the following chart Spirogyra (Project 4) Elodea (Project 5) cells Cell length across 10x field Cell lengthNumber of across 10x fieldCell length do 0.0q o.06 0.056 2 4 3S 0 0.0300 33 0, 0488 0.01a84 5 3 0.063 ose 14. PLOT: Average cell length vs. plant type using a column graph. Also add standard deviation error bars to the graph. See Lab 3, Introduction to Excel, for instruction on adding error bars. 15. Do a -test to compare the mean lengths of the two types of cells. H: nduction to Microscopy- Page 111

Explanation / Answer

Spirogyra

Mean= .090+.0750+ .051+ .070+.0782/5= .07284

Variance= (.090-.07284)2+ (.0750-.07284)2+ (.051-.07284)2+(.070-.07284)2+ (.0782-.07284)2/5-1=

= .0887+ .0000046656+ .00047+ .00000806+ .0000287/4= 0.0223

standard deviation= square root of variance= 0.149

Elodea

Mean= 0.0644

Variance= (0.09-0.0644)2+ (0.064-.0644)2+ (0.056-.0644)2+ (0.05-.644)2+ (0.062-.0644)2/5-1=

= .000655+ .00000016+ .000070+ .000207+ .00000576/4= .000937/4= .0002344

standard deviation= 0.0153

step 3 t test statistic= .00844/.067= .1259

step 4

Degree of freedom= 5-1=4

t test value at alpha 0.05 for df 4= 2.1319

null hypothesis: no significant diffrence between two measurements

alternate hypothesis : there is significant difference between two measurements

calculated t value is 0.1259 which is much less than table value so we accept the null hypothesis.

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