Question

Analysis of variance (ANOVA)

An investigator compares the difference in the amount of 20% sucrose solution removed by pollinators between unadulterated solution and solutions with 50, 100, 150, and 200 parts per million of caffeine. The results are expressed as differences in grams for the following caffeine concentrations:

50 ppm caffeine: -0.4 0.34. 0.19, 0.05, -0.14

100 ppm caffeine: 0.01, -0.39, -0.08, -0.09, -0.31

150 ppm caffeine: 0.65, 0.53, 0.39, -0.15, 0.46

200 ppm caffeine: 0.24, 0.44, 0.13, 1.03, 0.05

Does the mean amount of nectar taken depend on the concentration of caffeine in the nectar?

Carry out an analysis of variance (ANOVA) to find out.

a. State the null and alternative hypotheses appropriate for this question.

b. Calculate the following summary statistics for each group: n_i, Y-bar_i, and s_i.

c. Set up an ANOVA table to keep track of your results.

d. What is the mean square error MS_error?

e. How many degrees of freedom are associated with error? How many with groups?

f. Calculate the estimate of the grant mean.

g. Calculate the group sum of squares.

h. Calculate the group degrees of freedom and the group mean square.

i. What is F for this example?

j. Use a statistical table or Excel to find the P-value for this test.

Answer 1

a. To test: H₀: Vs H_a: At least one pair of means differ. where, are the mean differences in concentration of sucrose of unadulterated solution and solutions with 50, 100, 150, and 200 parts per million of caffeine.

b. Summary Statistics:

Grand mean  

For 50 ppm group,

No. of observations, = 5 Group mean   =

Similarly, for all the groups:

c. ANOVA table:

We need to construct a table of the format:

Here, k = no. of treatments = 4 n = no. of observations = 20 SST = Sum of squares due to treatment (Between groups) = ....(where are the group means and is the grand mean)

   = 1.134

SSE = Error sum of squares = =    = 1.448

SST = Total sum of squares = SST + SSE = 1.134 + 1.448 = 2.582 Substituting in the ANOVA table,

d. From the ablove table, MSE = SSE / (n-k) = 1.134 / (20 - 4) =

e. Degrees of freedom are associated with error = n - k = 20 - 4 = 16 Degrees of freedom are associated with groups = k - 1 = 4 - 1 = 3

f. Estimate of the grand mean:

Grand mean  

g. Group SS = SST = (Calculated in detail in part c.)

h. Group degrees of freedom = k - 1 = 4 - 1 = 3 Group mean square = MST = SST / k-1 = 1.134 / 3 =

i. F =

j. Using excel, p value for the F test can be obtained using the formula:

We get, p value =

Since this p value < 0.05, we do not have sufficient evidence to support the null hypothesis. Hence reject H₀ at 5% level of significance. We may conclude that all group means are not equal.