In: Statistics and Probability
Problem 3:
In a study to determine the effectiveness of a synthetically produced mate attractant chemical (pheromone), the synthetic hormone was compared with a natural male pheromone for its ability to attract female fruit fly. Fifteen synthetic baits and 14 natural baits were placed at random in a group of peach trees and the number of female fruit fly attracted to the baits over a 2 hour period were counted (data shown below).
Is there a difference in the attractiveness of the synthetic and natural pheromones?
Synthetic: 5, 7, 11, 13, 25, 35, 22, 17, 9, 20, 18, 31, 12, 16, 10
Natural: 6, 47, 41, 19, 21, 14, 8, 23, 32, 28, 29, 24, 15, 27
The data is as follows:
Synthetic | Natural |
5 | 6 |
7 | 47 |
11 | 41 |
13 | 19 |
25 | 21 |
35 | 14 |
22 | 8 |
17 | 23 |
9 | 32 |
20 | 28 |
18 | 29 |
31 | 24 |
12 | 15 |
16 | 27 |
10 |
Let's see the descriptive statistics of both the variables:
Synthetic | Natural | |
Mean | 16.73333333 | 23.85714286 |
Variance | 75.20952381 | 132.9010989 |
To check the difference in the attractiveness of the synthetic and natural pheromones, we will run a two sample t-test assuming equal variances (one sample variance is no larger than twice the other)
Let's run the two-sample t-test:
To calculate Sp (Pooled variance):
Pooled variance = 102.98
n1 = 15, n2 = 14
t-statistic = -1.89
The t-critical value at df = 15 + 14 - 2 = 27 and 0.05 significance level is +-2.05
As t-calculated = -1.89 is not less than -2.05, we cannot reject the null hypothesis. The p-value = 0.069
Therefore, we can conclude that there is no difference in the attractiveness in the baits