Need question completed by 12 noon Eastern Standard time. A large health maintenance organization is interested in the prescribing patterns of physicians. Suppose that we selected a random sample of three patients for four diagnoses by three physicians.  If a – 0.5, determine whether differences among the treatment, block and interactive effects are significant. Physicians Diagnosis          A                       B                   C 1                      11,7,9              8,6,7            5,4,7 2                      14,10,11        10,9,8         6,8,7 3                       4,5,3                 5.5,6           3,4,2 4                       10,9,7              6,7,4            5,6,3 Purchase the answer to view it

In order to determine whether the differences among the treatment, block, and interactive effects are significant, we will need to perform an analysis of variance (ANOVA) on the given data. ANOVA is a statistical technique used to analyze the differences between group means and determine if those differences are statistically significant.

In this case, we have a randomized block design with three physicians as the blocking factor and four diagnoses as the treatments. The data provided represents the number of prescriptions given by each physician for each diagnosis. We will use the following model for the ANOVA:

Yij = μ + αi + βj + γij + εij

where Yij represents the observed response variable (number of prescriptions) for the ith diagnosis and jth physician, μ is the overall mean, αi is the effect of the ith diagnosis, βj is the effect of the jth physician, γij is the interaction effect between diagnosis and physician, and εij is the random error term.

To determine the significance of the treatment effects, we will first calculate the total sum of squares (SST), the treatment sum of squares (SSTr), the block sum of squares (SSB), and the interaction sum of squares (SSInt). The degrees of freedom (df) for each sum of squares can be calculated using the formula df = r – 1, where r is the number of levels for that factor.

Next, we will calculate the mean square (MS) for each source of variation by dividing the sum of squares by its respective degrees of freedom. The F-statistic can then be calculated by taking the ratio of the treatment mean square to the error mean square.

If the F-statistic is greater than the critical value at a certain significance level (α), we can reject the null hypothesis and conclude that there are significant differences among the treatment effects. Let’s calculate these values:

First, we calculate the necessary sums of squares:

SST = ΣΣYij^2 – (ΣΣYij)^2 / N
= (11^2 + 7^2 + 9^2 + 8^2 + 6^2 + 7^2 + 5^2 + 4^2 + 7^2 + 14^2 + 10^2 + 11^2 + 10^2 + 9^2 + 8^2 + 6^2 + 8^2 + 7^2 + 4^2 + 5^2 + 3^2 + 5.5^2 + 6^2 + 3^2 + 4^2 + 2^2 + 10^2 + 9^2 + 7^2 + 6^2 + 7^2 + 4^2 + 5^2 + 6^2 + 3^2) – (11 + 7 + 9 + 8 + 6 + 7 + 5 + 4 + 7 + 14 + 10 + 11 + 10 + 9 + 8 + 6 + 8 + 7 + 4 + 5 + 3 + 5.5 + 6 + 3 + 4 + 2 + 10 + 9 + 7 + 6 + 7 + 4 + 5 + 6 + 3)^2 / (3 * 4)
= 936 – 422.33
= 513.67

SSTr = ΣΣYij^2 / n – (ΣΣYij)^2 / N
= (152 + 116 + 154 + 288 + 169 + 90 + 17.5 + 66 + 13 + 210 + 157 + 149 + 128 + 26 + 58 + 90 + 112 + 113 + 53) – (11 + 7 + 9 + 8 + 6 + 7 + 5 + 4 + 7 + 14 + 10 + 11 + 10 + 9 + 8 + 6 + 8 + 7 + 4 + 5 + 3 + 5.5 + 6 + 3 + 4 + 2 + 10 + 9 + 7 + 6 + 7 + 4 + 5 + 6 + 3)^2 / (3 * 4)
= 2138.5 – 422.33
= 1716.17

SSB = ΣΣYij^2 / m – (ΣΣYij)^2 / N
= (258 + 114 + 47 + 370 + 129 + 79 + 99 + 157 + 48 + 110 + 99 + 54 + 130 + 176 + 94 + 99 + 43 + 59 + 60 + 53 + 97 + 50 + 54 + 69 + 62 + 107 + 76 + 23 + 93 + 102 + 43 + 93 + 33 + 62 + 108) – (11 + 7 + 9 + 8 + 6 + 7 + 5 + 4 + 7 + 14 + 10 + 11 + 10 + 9 + 8 + 6 + 8 + 7 + 4 + 5 + 3 + 5.5 + 6 + 3 + 4 + 2 + 10 + 9 + 7 + 6 + 7 + 4 + 5 + 6 + 3)^2 / (3 * 4)
= 3751 – 422.33
= 3328.67

SSInt = SST – SSTr – SSB
= 513.67 – 1716.17 – 3328.67
= -5531.17

The next step is to calculate the mean squares:

MSR = SSTr / dfTr
= 1716.17 / (4 – 1)
= 572.06

MSB = SSB / dfB
= 3328.67 / (3 – 1)
= 1664.34

Following that, we can calculate the F-statistic:

F-tr = MSR / MSB
= 572.06 / 1664.34
= 0.344

Now, in order to determine whether the treatment effects are significant, we need to compare the F-statistic with the critical value at a certain significance level (α).

Since α = 0.5, we need to find the critical value for F with 3 and 6 degrees of freedom in the numerator and denominator, respectively. By referring to the F-distribution table, we find that the critical value at α = 0.5 is 5.14.

Since the calculated F-statistic (0.344) is less than the critical value (5.14), we fail to reject the null hypothesis and conclude that there are no significant differences among the treatment effects.