Answer:
1)
2)
And we got the same decision reject the null hypothesis at 5% of significance.
Step-by-step explanation:
Previous concepts
The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:
Where O rpresent the observed values and E the expected values.
State the null and alternative hypothesis
Null hypothesis: The distribution is 30% catfish, 15% bass, 40% bluegill, and 15% pike
Alternative hypothesis: The distribution is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike
The observed values are given by the table given:
Catfish =112, BAss = 95, Bluegill=210, Pike=83
Calculate the expected values
In order to calculate the expected values we can use the following formula for each cell of the table
![E_{Catfish}=500*0.3=150](https://tex.z-dn.net/?f=E_%7BCatfish%7D%3D500%2A0.3%3D150)
![E_{Bass}=500*0.15=75](https://tex.z-dn.net/?f=E_%7BBass%7D%3D500%2A0.15%3D75)
![E_{Bluegill}=500*0.4=200](https://tex.z-dn.net/?f=E_%7BBluegill%7D%3D500%2A0.4%3D200)
![E_{Pike}=500*0.15=75](https://tex.z-dn.net/?f=E_%7BPike%7D%3D500%2A0.15%3D75)
Part 1: Calculate the statistic
Calculate the critical value
First we need to calculate the degrees of freedom given by:
Since the confidence provided is 95% the significance would be
and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be
We can calculate also the p value:
And we got the same decision reject the null hypothesis at 5% of significance.