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 385 randomly selected fish caught in Green Valley Lake, 206 were rainbow trout, 67 were other trout, 45 were bass, and 67 were catfish. Of the 519 randomly selected fish caught in Echo Lake, 248 were rainbow trout, 108 were other trout, 81 were bass, and 82 were catfish. Conduct the appropriate hypothesis test using an α = 0.10 level of significance. What is the correct statistical test to use? Paired t-test Homogeneity Independence Goodness-of-Fit What are the null and alternative hypotheses? H 0 : The distribution of fish is not the same for Green Valley Lake and Echo Lake. The distribution of fish is the same for Green Valley Lake and Echo Lake. Fish breed and lake are independent. Fish breed and lake are dependent. H 1 : Fish breed and lake are dependent. The distribution of fish is the same for Green Valley Lake and Echo Lake. The distribution of fish is not the same for Green Valley Lake and Echo Lake. Fish breed and lake are independent. The test-statistic for this data = (Please show your answer to three decimal places.) The p-value for this sample = (Please show your answer to four decimal places.) The p-value is α Based on this, we should reject the null fail to reject the null accept the null Thus, the final conclusion is... There is insufficient evidence to conclude that the distribution of fish is not the same for Green Valley Lake and Echo Lake. There is insufficient evidence to conclude that fish breed and lake are independent. There is sufficient evidence to conclude that the distribution of fish is not the same for Green Valley Lake and Echo Lake. There is sufficient evidence to conclude that the distribution of fish is the same for Green Valley Lake and Echo Lake. There is sufficient evidence to conclude that fish breed and lake are dependent.

Answer 1

We need to determine whether frequency counts are distributed identically across different populations. So, the correct statistical test to use is Chi Square test of Homogeneity.

The appropriate hypothesis are

H0: The distribution of fish is the same for Green Valley Lake and Echo Lake.

H1: The distribution of fish is not the same for Green Valley Lake and Echo Lake.

The contingency table is,

	rainbow trout	other trout	bass	catfish	Total
Green Valley Lake	206	67	45	67	385
Echo Lake	248	108	81	82	519
Total	454	175	126	149	904

The test statistic is a chi-square random variable (Χ²) defined by the following equation.

Χ² = Σ [ (O_r,c - E_r,c)² / E_r,c ]

where O_r,c is the observed frequency count E_r,c is the expected frequency count for row r and column c.

E_r,c = (n_r * n_c) / n
E_1,1 = (454 * 385) / 904 = 193.3518

E_1,2 = (175 * 385) / 904 = 74.52987

E_1,3 = (126 * 385) / 904 = 53.6615

E_1,4 = (149 * 385) / 904 = 63.45686

E_2,1 = (454 * 519) / 904 = 260.6482

E_2,2 = (175 * 519) / 904 = 100.4701

E_2,3 = (126 * 519) / 904 = 72.3385

E_2,4 = (149 * 519) / 904 = 85.54314

Χ² = (206 - 193.3518)² / 193.3518 + (67 - 74.52987)² / 74.52987 + (45 - 53.6615)² / 53.6615 + (67 - 63.45686)² / 63.45686 + (248 - 260.6482)² / 260.6482 + (108 - 100.4701)² / 100.4701 + (81 - 72.3385)² / 72.3385 + (82 - 85.54314)² / 85.54314  

= 5.546

Degree of freedom = (r - 1) * (c - 1) = (2 - 1) * (4 - 1) = 3

P-value = P(Χ² > 5.546, df = 3) = 0.1359

The p-value is greater than α

Based on this, we should fail to reject the null.

Thus, the final conclusion is - There is insufficient evidence to conclude that the distribution of fish is not the same for Green Valley Lake and Echo Lake.