Question

Fifteen trainees in a technical program are randomly assigned to three different types of instructional approaches, all of which are concerned with developing a specified level of skill in computer-assisted design. The achievement test scores at the conclusion of the instructional unit are reported in Table along with the mean performance score associated with each instructional approach. Use the analysis of variance procedure to test the null hypothesis that the three-sample means were obtained from the same population, using the 5 percent level of ificance for the test.

Instrumental

Method

A1

A2

A3

Test Scores

81

88

73

Total Scores

400

425

375

Mean Test Scores

80 85

75

86

90

82

79

76

68

70

82

71

84

89

81

20

Answer 1

We will carry out an ANOVA test, assuming the samples were normally distributed and the variances in the three populations were homogeneous. He selected Anova: Single Factor, then entered the data ranges, selected Grouped By: Rows and set Alpha to 0.05. He selected a suitable output option and then OK.

This returned a table of data as shown below.

The F value calculated was 3.34. This is less than the stated critical value (Fcrit ) of 3.88, and the probability of obtaining this result by chance (P-value) was calculated as 0.0699 (6.99% to three significant figures). We conclude that there was not a significant difference in means and the three-sample means were obtained from the same population, since P >0.05.

Alternative method by manual calculations:

n = 5 replications

a = 3 treatment

alpha, a = 0.05

overall mean $\mu =\frac{80+85+75}{3} = 80$

SS treatment $= 5 \times [(80-80)^2 + (85-80)^2 + (75-80)^2]$

SS treatment = 250

SS total = $[(86-80)^2 + (79-80)^2 + (81-80)^2 + (70-80)^2 + (84-80)^2 + (90-80)^2 + (76-80)^2 + (88-80)^2 + (82-80)^2 + (89-80)^2 + (82-80)^2 + (68-80)^2 + (73-80)^2 + (71-80)^2 + (81-80)^2]$

SS total = 698

SSE = SS total - SS treatment

SSE = 448

MS treatment $= \frac{SS \;treatment }{a-1} = 125$

$MSE = \frac{SSE }{a(n-1)} = 37.333 \\ F_0 = \frac{MS \;treatment }{ MSE} \\ F_0 = 3.348$

critical value = F(a,a-1,a(n-1)) = F(0.05,2,12) = 3.89

since F₀<3.89 hence null hypothesis can not be rejected

Hence we conclude that the three sample means were obtained from the same population.