Answer:
Null hypothesis = H₀ = There food preferences among vole species are independent of one another.
Alternate hypothesis = H₁ = There is a relationship between voles and food preference.
Expected meadow vole/apple slices = 29.983051
Expected common vole/apple slices = 28.016949
Expected meadow vole/peanut butter-oatmeal = 31.016949
Expected common vole/peanut butter-oatmeal = 28.983051
Chi-square value = χ² = 2.154239
Degree of freedom = 1
Critical value = 3.841
χ² < Critical value
We failed to reject H₀
We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.
Step-by-step explanation:
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.
Null hypothesis = H₀ = There food preferences among vole species are independent of one another.
Alternate hypothesis = H₁ = There is a relationship between voles and food preference.
Data collected by Dr. Pagels:
meadow voles common voles Row Total
apple slices 26 32 58
peanut butter-oatmeal 35 25 60
Column Total 61 57 118
Where 118 is the grand total.
The expected number is given by
Expected = (row total)×(column total)/grand total
Expected meadow vole/apple slices = 58×61/118
Expected meadow vole/apple slices = 29.983051
Expected common vole/apple slices = 58×57/118
Expected common vole/apple slices = 28.016949
Expected meadow vole/peanut butter-oatmeal = 60×61/118
Expected meadow vole/peanut butter-oatmeal = 31.016949
Expected common vole/peanut butter-oatmeal = 60×57/118
Expected common vole/peanut butter-oatmeal = 28.983051
The chi-square statistic value is given by
χ² = Σ(Observed - Expected)²/Expected
χ² = (26 - 29.983051)²/29.983051 + (32 - 28.016949)²/28.016949 + (35 - 31.016949)²/31.016949 + (25 - 28.983051)²/28.983051
χ² = 2.154239
The degrees of freedom is given by
DoF = (row - 1)×(col - 1)
For the given case, we have 2 rows and 2 columns
DoF = (2 - 1)×(2 - 1)
DoF = 1
The given level of significance = 0.05
The critical value from the chi-square table at α = 0.05 and DoF = 1 is found to be
Critical value = 3.841
Conclusion:
Reject H₀ If χ² > Critical value
We reject the Null hypothesis If the calculated chi-square value is more than the critical value.
For the given case,
χ² < Critical value
We failed to reject H₀
We do not have significant evidence at the given significance level to show that there is a relationship between voles and food preference.
