Question

A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. 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. 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. Perform a test for homogeneity at a 5% level of significance.

Short answer

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

Step-by-step solution

Given Information

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

Null Hypothesis

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.

Alternative Hypothesis 

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  

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

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

Therefore,

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

$=4-1$

$=3$

Distribution Identification

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

Calculation of Expected Frequencies

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:

 Test Statistic 

The test statistic of the independence test is given below:

$\text{Test statistic}=\sum _{(i\times j)}\frac{(O-E{)}^2}{E}$

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

and drag the same formula up to cell E19. After that, take the total of columns total and rows total. The table of test statistics is shown below:

Hence the test statistic is 11.75

Excel calculation of p value

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

p-value

Hence, the $p$-value is

0.008

Step 10Chi-square Sketch

Decision and conclusion

i. Alpha:$0.05$

ii: Decision: Reject the null hypothesis $H_0$iii. Reason for decision: Because$p$-value$<\alpha$

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.