Question

A farmer conducts a test on 3 types of superfeed. The weight gain (in lbs.) are as follows: Type A Type B Type C 12 11 14 17 21 19 17 13 12 10 18 Test the claim that the mean weight gain is the same for all three types of superfeed at the 99% level.

Answer 1

12 11 14 17 12 21 19 10 17 13 18 Sum = 61 33 70 Average = 15.25 11 17.5 Σ; Χ2 = - 963 365 1250 St. Dev. = 3.304 1 2.887 SS = 32.75 2 25 η = 4 3 4

The total sample size is N = 11. Therefore, the total degrees of freedom are: dftotal = 11 -1 = 10 Also, the between-groups degrees of freedom are dfbetween = 3 – 1 = 2, and the within-groups degrees of freedom are: df within = d ftotal dfbetween = 10 – 2 = 8 First, we need to compute the total sum of values and the grand mean. The following is obtained ΣΧ;, 61 +33 + 70 = 164 Also, the sum of squared values is Σx? = 963 + 365 + 1250 = 2578 Based on the above calculations, the total sum of squares is computed as follows 1642 S Stotal = x = 2578 = = 132.909 11 The within sum of squares is computed as shown in the calculation below: S Swithin SSwithingroups = 32.75 + 2 + 25 = 59.75

M Sbetween S Sbetween dfbetween 73.159 2 = 36.58 M Swithin S Swithin df within 59.75 8 = 7.469 Now that sum of squares are computed, we can proceed with computing the mean sum of squares: M Sbetween SSbetween dfbetween 73.159 2 = 36.58 59.75 MS within = S Swithin df within = = 7.469 8 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: 36.58 F = M Sbetween M Swithin = 4.898 7.469

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃

Ha: Not all means are equa

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df1​=2 and df2​=2, therefore, the rejection region for this F-test is R={F:F>Fc​=8.649}

Since it is observed that F=4.898≤Fc​=8.649, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0408, and since p=0.0408≥0.01, it is concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the α=0.01 significance level.

ANOVA Summary Sum of Squares Degrees of Freedom Mean Square Source F-Stat P-Value DF SS MS Between Groups 2 73.1591 36.5795 4.8977 0.0408 Within Groups 8 59.7501 7.4688 Total: 10 132.9092