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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, 135 were rainbow trout, 48 were other trout, 67 were bass, and 43 were catfish. Perform the hypothesis test at a 5% level of significance and find the p-Value.

p-Value =



[ Select ]



["0.003", "0.293", "0.250", "0.702"]

Round your answer to three decimal places.

Conclusion:



[ Select ]



["There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake.", "There is sufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."]

1 Answer

5 votes

Answer:

p = 0.293

Conclusion:

"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."

Explanation:

------------R/trout--O/trout---bass--catfish---total

GVL - - - 105 - - - 27 - - - - - 35--- 24 - - - 191

EL - - - - 135------- 48-------- 67 - - 43 - - - 293

Total - - 240 - - - 75 - - - - - 102--- 67 - - - 484

Note : GVL = GREEN VALLEY LAKE ; EL = ECHO LAKE.

H0: The distribution of fish caught in green valley is the same as that caught in Echo lake.

H1: The distribution of fish caught in Green Valley lake is different from that caught in Echo valley lake.

Using the chi-squared test statistic calculator ;

χ2 = 3.7269

p-value = 0.2925.

p-value = 0.293

Conclusion:

"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."

User Timbroder
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