Question

Assignment 3 Winter 2020 You will have a chance to analyze taxic response data using observational data from Planeria. You will use Pearson's chi square test which allows you to determine whether your planario are showing a taxic response to environmental variables. Pearson's chi-squared test is a statistical test applied to sets of categorical data to evaluate how likely it is that any observed difference between the sets arose by chance. It is the most widely used of many chi-squared tests For each environmental category (5) you will create a table displaying observed behavior and expected behavior. This data is from 3 groups. You will be testing the H, null hypothesis that states that Planaria don't demonstrate taxic responses and that their distribution is random. Treatment Positive taxis Intermediate Negative taxis Light Bright (6) Dark (4) Temperature High (4) Low (8) pH Alkaline (0) Acidic (6) Food Lots (4) None (11) Rheo Flowing (3) Stationary (5) Here is an example for Salinity. Formula used is ye statistic = (0-E) /E Negative non saline Intermediate Positive saline O-E (O-E) (0-E)/E EX statistic Observed 11 12.4 Expected 5 x'statistie (11.5/5(4-574/5(0-51/5.7.2 +0.2 5.0 -12.4 y critical d.f. 2 (0.05) = 5.991 Therefore statistic critical conclusion reject the null hypothesis, Planaria are exhibiting a negative taxic response towards salinity based on observed data. For the environmental responses conduct a chi squire statistical analyses. You will either except or reject the null hypothesis based on the chi square statistic. Each analysis used 15 Planaria in total. (15 marks)

Answer 1

Answer-

According to the given question- taxis response was observed based on several factors such as light, temperature, food, pH and Rheo on the planeria species and chi square test was performed to determine the response of planeria towards the environmental variables.

Here the number of Planeria are 15 so expected frequency will be 1 / 3 of the total.

Treatment	Positive taxis	Intermediate	Negative Taxis
Light	Bright (6)	(5)	dark (4)
temperature	High (4)	(3)	Low (8)
pH	Alkaline (0)	(9)	Acidic (6)
Food	Lost (4)	(0)	None (11)
Rheo	Flowering (3)	(7)	Stationary (5)

First calculating for Light-

	Observed frequency (Oi )	Expected frequency (Ei)	(Oi - Ei )	(Oi - E i)2	(Oi - Ei)2 / Ei
Positive Taxis	6	5	6- 5 = 1	1	1 / 5 = 0.2
Intermediate	5	5	5 - 5 = 0	0	0 / 5 = 0
Negative taxis	4	5	4 - 5 = - 1	1	1 / 5 = 0.2
	Total = 15	Total = 15			X2 = 0.4

here the valve of   X² statistics = 0.4

but X² critical for two degree of freedom at the probability of 0.05 = 5.991

so the   X² statistics <  X² critical i.e. 0.4 < 5.991

so we accept the null hypothesis. this means that the Planaria are exhibiting the positive taxic response towards the light based on observed data.

Calculating for temperature-

	Observed frequency (Oi )	Expected frequency (Ei)	(Oi - Ei )	(Oi - E i)2	(Oi - E i)2 / Ei
Positive Taxis	4	5	4 - 5 = - 1	1	1 / 5 = 0.2
Intermediate	3	5	3 - 5 = - 2	4	4 / 5 = 0.8
Negative taxis	8	5	8 - 5 = 3	9	9 / 5 = 1.8
	Total = 15	Total = 15			X2 = 2.8

here the valve of   X² statistics = 2.8

but X² critical for two degree of freedom at the probability of 0.05 = 5.991

so the   X² statistics <  X² critical i.e. 2.8  < 5.991

so we accept the null hypothesis. this means that the Planaria are exhibiting the positive taxic response towards the temperature based on observed data.

Calculating for pH-

	Observed frequency (Oi )	Expected frequency (Ei)	(Oi - Ei )	(Oi - E i)2	(Oi - Ei)2 / Ei
Positive Taxis	0	5	0 - 5 = - 5	25	25 / 5 = 5
Intermediate	9	5	9 - 5 = 4	16	16 / 5 = 3.2
Negative taxis	6	5	6 - 5 = 1	01	1 / 5 = 0.2
	Total = 15	Total = 15			X2 = 8.4

here the valve of   X² statistics = 8.4

but X² critical for two degree of freedom at the probability of 0.05 = 5.991

so the   X² statistics >  X² critical i.e. 8.4 >  5.991

so we reject  the null hypothesis. this means that the Planaria are exhibiting the negative taxic response towards the pH based on observed data.

Calculating for Food-

	Observed frequency (Oi )	Expected frequency (Ei)	(Oi - Ei )	(Oi - E i)2	(Oi - Ei)2 / Ei
Positive Taxis	4	5	4 - 5 = -1	1	1 / 5 = 0.2
Intermediate	0	5	0 - 5 = -5	25	25 / 5 = 5
Negative taxis	11	5	11 - 5 = 6	36	36 / 5 = 7.2
	Total = 15	Total = 15			X2 = 12.4

here the valve of   X² statistics = 12.4

but X² critical for two degree of freedom at the probability of 0.05 = 5.991

so the   X² statistics >  X² critical i.e. 12.4 >  5.991

so we reject  the null hypothesis. this means that the Planaria are exhibiting the negative taxic response towards the food based on observed data.

Calculating for Rheo-

	Observed frequency (Oi )	Expected frequency (Ei)	(Oi - Ei )	(Oi - E i)2	(Oi - Ei)2 / Ei
Positive Taxis	3	5	3-5 = - 2	4	4 / 5 = 0.8
Intermediate	7	5	7 - 5 = 2	4	4 / 5 = 0.8
Negative taxis	5	5	5 - 5 = 0	0	0 / 5 = 0
	Total = 15	Total = 15			X2 = 1.6

here the valve of   X² statistics = 1.6

but X² critical for two degree of freedom at the probability of 0.05 = 5.991

so the   X² statistics <  X² critical i.e. 1.6 < 5.991

so we accept the null hypothesis. this means that the Planaria are exhibiting the positive taxic response towards the Rheo based on observed data.