Researchers want to survey the mass of American alligators in a certain region. They want to maximize the efficiency of this effort, and get as many measurements as possible. Capturing alligators, putting them on a scale, and releasing them involves risk and time. They decide to survey alligators from the air. In this manner they can measure the length of the animals. They want to use these measurements to estimate the mass of the alligators. They have access to data about length and mass of captive alligators.

Which statistical test should they use to evaluate the usefulness of this method?

a. Correlation analysis
b. Linear regression with t-test for significance of slope
c. Chi Squared Test
d. ANOVA

Option B is correct.

An appropriate statistical test that can be used to evaluate the usefulness of this method is; Linear regression with t-test for significance of slope.

Step-by-step explanation:

For this research, they want to measure the lengths of alligators and use the already obtained data relating the lengths of captured and released alligators to their masses, to obtain the masses of alligators whose lengths have been measured.

So, in this research survey, the masses of the alligators is the dependent variable which depends on the lengths.

Basically, the researchers measure the length of an alligator and use the data to work out its mass.

So, an appropriate method to relate the two variables is to use regression analysis on the already obtained data, to obtain the linear relationship between mass of the alligators and their lengths. So that, just by putting on the measured lengths of the alligators surveyed into this expression, their corresponding masses would be immediately obtained.

But the correctness of this expression obtained will be checked by using t-test for the significance of the slope obtained in the linear regression.

So, the appropriate statistical test that can be used to evaluate the usefulness of this method is: Linear regression with t-test for significance of slope.

Hope this Helps!!!

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17