Chi-Square Test Hypothesis

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Results
The duration of the invader cricket in the resident’s home allowed data to be collect for all groups. When the ten minuet time frame had elapsed, and after all the experiments were conducted, there was a sample size of 478 crickets that were used. The total number of trials that were completed was 239. The number of expected wins between the resident cricket and intruder cricket was expected to be equal (50/50). The expected frequency was 50 percent and the chi-square test resulted in a p-value of .0015 which rejected the null hypothesis and accepted the alternative hypothesis. The degree of freedom was one.
Discussion
The calculated result from the chi-square test gave a p-value of .0015. The null hypothesis was rejected and the alternative hypothesis was accepted because the p-value was less than .05. The …show more content…

The resident cricket approached the intruder cricket and would kick or bite the intruder. No crickets that were examined showed all five levels of aggression. Majority of time, the resident and intruder crickets avoided each other. The most aggressive attack that was directed towards the intruder cricket was one of the resident crickets tore off the leg of the intruder. The crickets were not evenly matched due to the limited choices of crickets that were able to be chosen from.
There are alternative explanations for the data collected and some limitations to the experiment. Sizes varied among the house crickets which may have caused some of the pairs of house crickets to be unequal in size. The cricket that was larger had a better chance to attack and win more often compared to the smaller cricket. Another limitation for this experiment is that many crickets died before the experiment could be conducted. Due to the small container size, the house crickets were also stressed out because they were not contained in their natural

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