In: Statistics and Probability
On your fish farm, you have been growing tilapia in four ponds. Each pond gets a different feed (a, b, c, and d). After a few months you harvest the fish in each pond and get a total weight. The data are given below. Use a Chi-Square test to see if there is a difference between feeds, use α = 0.05, write the H0, give the P-value, and write a conclusion.
feed pounds of fish produced
a 39
b 44
c 50
d 41
Category | Observed Frequency (O) | Proportion, p | Expected Frequency (E) | (O-E)²/E |
a | 39 | 0.25 | 174 * 0.25 = 43.5 | (39 - 43.5)²/43.5 = 0.4655 |
b | 44 | 0.25 | 174 * 0.25 = 43.5 | (44 - 43.5)²/43.5 = 0.0057 |
c | 50 | 0.25 | 174 * 0.25 = 43.5 | (50 - 43.5)²/43.5 = 0.9713 |
d | 41 | 0.25 | 174 * 0.25 = 43.5 | (41 - 43.5)²/43.5 = 0.1437 |
Total | 174 | 1.00 | 174 | 1.5862 |
Null and Alternative hypothesis:
Ho: Proportions are same.
H1: Proportions are different.
Test statistic:
χ² = ∑ ((O-E)²/E) = 1.5862
df = n-1 = 3
p-value = CHISQ.DIST.RT(1.5862, 3) =
0.6625
Decision:
p-value > α, Do not reject the null hypothesis
There is not enough evidence to conclude that there is a difference
between feeds at 0.05 significance level.