Question

5. Kate Sanders, a researcher in the department of biology at IPFW University, studied the effect of agriculture contaminants on the stream fish population in Northeastern Indiana (April 2012). Specially designed traps collected samples of fish at each of four stream locations. A research question was, Did the differences in agricultural contaminants found at the four locations alter the proportion of the fish population by gender? Observed frequencies were as follows. Stream Locations Gender Male Female 49 36 39 49 41 46

Use the data from Prob. 5 with a 0.05 level of significance and answer the following: a) Are the proportions of male fish equal across the stream locations? Explain. b) If there is a difference in proportions of male fish across stream locations, do a Marascuilo pairwise comparison for each pairing and report results, i.e., is there difference in male proportions between streams A and B, A and C, A and D, B and C, B and D, and C and D?

Answer 1

Answer:

Use the data from Prob. 5 with a 0.05 level of significance and answer the following:

1. Are the proportions of male fish equal across the stream locations? Explain.

Ho: proportions of male fish equal across the stream locations

H1: proportions of male fish not equal across the stream locations

Chi-Square Test
Observed Frequencies
	Column variable						Calculations
Row variable	C1	C2	C3	C4	Total		fo-fe
R1	49	44	49	39	181		2.1897	-2.8103	4.7902	-4.1695
R2	41	46	36	44	167		-2.1897	2.8103	-4.7902	4.1695
Total	90	90	85	83	348
Expected Frequencies
	Column variable
Row variable	C1	C2	C3	C4	Total		(fo-fe)^2/fe
R1	46.8103	46.8103	44.2098	43.1695	181		0.1024	0.1687	0.5190	0.4027
R2	43.1897	43.1897	40.7902	39.8305	167		0.1110	0.1829	0.5625	0.4365
Total	90	90	85	83	348
Data
Level of Significance	0.05
Number of Rows	2
Number of Columns	4
Degrees of Freedom	3
Results
Critical Value	7.815
Chi-Square Test Statistic	2.4858
p-Value	0.4779
Do not reject the null hypothesis

Calculated chi square =2.4858

DF=3

Critical Chi square value at 0.05 level =7.815

Calculated chi square =2.4858 < 7.815 the table value.

The null hypothesis is  not rejected.

We conclude the proportions of male fish equal across the stream locations.

b) If there is a difference in proportions of male fish across stream locations, do a Marascuilo pairwise comparison for each pairing and report results, i.e., is there difference in male proportions between streams A and B, A and C, A and D, B and C, B and D, and C and D?

Since there is no difference in proportions of male fish across stream locations, Marascuilo pairwise comparison is not appropriate here.