91Ó°ÊÓ

Given that of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish .
Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish

we have to test the homogeneity

The null hypothesis is given below:

$H_0$ : The distribution of fish caught in Green Valley Lake is the same as fish caught in Echo Lake.

Against the alternative hypothesis as given below:

$H_a:$The distribution of fish caught in Green Valley Lake is different from fish caught in Echo Lake.

The degrees of freedom can be calculated by the formula given below:

$df=(\text{number of columns}-1)$

Therefore,

$df=(\text{number of columns}-1)$

$=4-1$

$=3$

From above calculation, it is clear that the distribution for the test is ${\mathrm{χ}}_3^2$.

The observed value table is already given in the textbook. Now we have to find the expected frequencies by using the formula shown below:

$E=\frac{\left(\text{row total}\right)\left(\text{column total}\right)}{\text{overall total}}$

All calculations can be done in excel worksheet. Hence, the expected (E) values table is shown below:

The test statistic of the independence test is given below:

$\text{Test statistic}=\underset{(i×j)}{∑}\frac{(O-E{)}^2}{E}$

To calculate$\frac{(O-E{)}^2}{E}$ apply the formula$=\left((B4-B11{)}^2\right)/B11$

Hence the test statistic is 11.75

The $p$-value can be calculated in excel by using CHIDIST ( ) formula as shown below:

Hence, the $p$-value is

0.008

i. Alpha:$0.05$

ii: Decision: Reject the null hypothesis $H_0$iii. Reason for decision: Because$p$-value$<\mathrm{Î\pm }$

iv. Conclusion: There is sufficient evidence to ensure that the distribution of fish caught in Green Valley Lake is different from fish caught in Echo Lake.