9.3-15. Ledolter and Hogg (see References) report that
an operator of a feedlot wants to compare the effective- ness of three different cattle feed supplements. He selects a random sample of 15 one-year-old heifers from his lot of over 1000 and divides them into three groups at random. Each group gets a different feed supplement. Upon not- ing that one heifer in group A was lost due to an accident, the operator records the gains in weight (in pounds) over a six-month period as follows:Group A:
500
650
530
680
Group B:
700
620
780
830
860
Group C:
500
520
400
580
410(a) Test whether there are differences in the mean weight gains due to the three different feed supplements.

To test whether there are differences in the mean weight gains due to the three different feed supplements, we can use a one-way ANOVA test. The null hypothesis is that there is no difference in the mean weight gains between the three groups, while the alternative hypothesis is that at least one group has a different mean weight gain than the others.

Using the formula for one-way ANOVA, we can calculate the F-statistic:

F = (SSbetween / dfbetween) / (SSwithin / dfwithin)

where SSbetween is the sum of squares between groups, dfbetween is the degrees of freedom between groups, SSwithin is the sum of squares within groups, and dfwithin is the degrees of freedom within groups.

We can calculate the necessary values as follows:

SSbetween = [(500+650+530+680)/4 - (700+620+780+830+860)/5]^2 +
          [(500+520+400+580+410)/5 - (700+620+780+830+860)/5]^2 +
          [(500+650+530+680)/4 - (500+520+400+580+410)/5]^2
        = 21682.4

dfbetween = 3 - 1 = 2

SSwithin = (500-575)^2 + (650-575)^2 + (530-575)^2 + (680-575)^2 +
          (700-738)^2 + (620-738)^2 + (780-738)^2 + (830-738)^2 +
          (860-738)^2 + (500-480)^2 + (520-480)^2 + (400-480)^2 +
          (580-480)^2 + (410-480)^2
        = 123610

dfwithin = 15 - 3 = 12

Plugging in the values, we get:

F = (21682.4 / 2) / (123610 / 12) = 2.227

Using a significance level of α = 0.05, we can look up the critical F-value for 2 degrees of freedom for the numerator and 12 degrees of freedom for the denominator in an F-distribution table. The critical value is 3.89.

Since the calculated F-statistic of 2.227 is less than the critical value of 3.89, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there are differences in the mean weight gains due to the three different feed supplements.

To know more about ANOVA:

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