Dr. Pagels is a mammalogist who studies meadow and common voles. He frequently t

Question

Dr. Pagels is a mammalogist who studies meadow and common voles. He frequently traps the moles and has noticed what appears to be a preference for a peanut butter-oatmeal mixture by the meadow voles vs apple slices are usually used in traps, where the common voles seem to prefer the apple slices. So he conducted a study where he used a peanut butter-oatmeal mixture in half the traps and the normal apple slices in his remaining traps to see if there was a food preference between the two different voles Indicate which of the following is the null hypothesis, and which is the alternate hypothesis There food preferences among vole species are independent of one another. Null Hypothesis There is a relaionship between voles and food preference. Alternative Hypothesis To test for independence, we need to calculate the Chi-square statistic. These are the data that Dr. Pagels collected meadow voles common voles apple slices peanut butter-oatmeal 21 16 15 25 You will want to use Excel (or Google sheets) to do the calculations for you, so that your answers are exact (they need to be exact to get credit. Here is a Google sheet to get . When transferring your answers, make sure you carry them out to AT LEAST SIX SIGNIFICANT FIGURES unless otherwise stated. expected meadow volelapple slices expected common volelapple slices expected meadow vole/peanut butter-oatmeal expected common vole/peanut butter-oatmeal - chi-square value degrees of freedom (whole number only) -using Statistical Table A (pg 704 of your textbook), what is the chi-square critical value with significance level of =0.057(report exactly the value in the table) will you reject or fail to reject the null hypothesis? (answer either reject or fail to reject)

Explanation / Answer

Ans:

Chi square vlue=2.863073

degree of freedom=(2-1)*(2-1)=1

critical chi square=CHIINV(0.05,1)=3.841459

As,calculated chi square value=2.863073<3.841459,we fail to reject null hypothesis.

Observed(O) meadow voles common voles Total apple slices 15 21 36 peanut butter 25 16 41 Total 40 37 77 Expected=row sum*colum sum/77 meadow voles common voles Total apple slices 18.701299 17.298701 36 peanut butter 21.298701 19.701299 41 Total 40 37 77 Chi square=(O-E)^2/E meadow voles common voles Total apple slices 0.732549 0.791945 1.524493 peanut butter 0.643213 0.695366 1.338579 Total 1.375762195 1.487310481 2.863073

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