Question

After randomly assigning subjects to treatments in a randomized comparative experiment, we can compare the treatment groups to see how well the randomization worked. We hope to find no significant differences among the groups. A study of how to provide premature infants with a substance essential to their development assigned infants at random to receive one of four types of supplement, called PBM, NLCP, PL-LCP, and TG- LCP. We are interested in whether the randomization results in equal proportions of boys and girls for each treatment. (Use a 0.05) (a) The subjects where 77 premature infants. Outline the design of the experiment if 20 are assigned to the PBM group and 19 to each of the other treatments. o We use test for goodness of fit to test the null hypothesis Ho: Pi = P2 = P3 = P4 = 1/4 against the alternative hypothesis Ha: not all pa = 1/4. We use test for goodness of fit to test the null hypothesis Ho: Pi = P2 = P3 = P4 = 1/4 against the alternative hypothesis Ha: P2 = P3 = P4 and Pı > P2. We use significance test for comparing two proportions. O Ho: There is no relationship between treatment assignment and gender of the infant (or genders were equally represented among the treatment groups) and Ha: Genders were not equally assigned to the treatments. (b) The random assignment resulted in 9 females in the TG-LCP group and 11 females in each of the other groups. Make a two-way table of group by sex. NLCP PL-LCP TG-LCP The test whether there are significant differences among the groups has which of the following? or = 0.0519, 3 degrees of freedom, p > 0.25 OX? = 6.46, 3 degrees of freedom, 0.05 < p < 0.1 Ox? = 0.56813, 3 degrees of freedom, p > 0.25 Ox? = 0.56813, 6 degrees of freedom, p > 0.25 There are significant differences among the groups. The main differences are the higher proportion of Male infants and lower proportion of Female infants who were assigned for TG-LCP. True or False?

Answer 1

a)

answer: option D)

b)

	male	female
PBM	9	11
NLCP	8	11
PL-LCP	8	11
TG-LCP	10	9

Chi-Square Test of independence
	Observed Frequencies
	0
0	PBM	NLCP	PL-LCP	TG-LCP			Total
women	11	11	11	9			42
men	9	8	8	10			35
Total	20	19	19	19			77
	Expected frequency of a cell = sum of row*sum of column / total sum
	Expected Frequencies
	PBM	NLCP	PL-LCP	TG-LCP			Total
women	20*42/77=10.909	19*42/77=10.364	19*42/77=10.364	19*42/77=10.364			42
men	20*35/77=9.091	19*35/77=8.636	19*35/77=8.636	19*35/77=8.636			35
Total	20	19	19	19			77
	(fo-fe)^2/fe
women	0.0008	0.0391	0.0391	0.1794
men	0.0009	0.0469	0.0469	0.2153

Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =   0.56813

Number of Rows =   2
Number of Columns =   4
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 4- 1 ) =   3
p-Value >0.25   

so, answer: option C)

False