Q. 34 - The Practice of Statistics for AP

Question

After randomly assigning subjects to treatments in a randomized comparative experiment, we can compare the treatment groups to see how well the random assignment worked. We hope to find no significant differences among the groups. A study on how to provide premature infants with a substance essential to their development assigned infants at random to receive one of four types of supplements, called PBM, NLCP, PL-LCP, and TG-LCP. The subjects were $77$ premature infants. In the experiment, $20$ were assigned to the PBM group and $19$ to each of the other treatments.

(a) 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 the group by gender. Calculate the proportion of females in each treatment group. Does it appear that the random assignment roughly balanced the groups by gender? Explain.

(b) Are the differences between the groups statistically significant? Give appropriate evidence to support your answer. 

Step-by-Step Solution

Answer

(a) No, it does not appear that the random assignment roughly balanced the groups by gender. 

(b) No,  the differences between the groups are not statistically significant. 

Part (a) Step 1: Given information

Number of premature infants is $77$

PGM group is $20$

Other treatments is $19$

Part (a) Step 2: Explanation

The number of males is the number in the group decreased by the number of females

The proportions in each group is the number of infants divided by the row total given in the last column:

The proportions appears to be balanced, because there are no large differences between the proportions.

Part (b) Step 1: Given information

The number of premature infants is $77$

PGM group is $20$

Other treatments is $19$

Part (b) Step 2: Explanation

Observed counts

The expected counts are the row total multiplied by the column total, divided by the sample size $n=77$

The chi-square statistic is the sum of squared deviations (between observed and expected counts) divided by the expected count:

$$\begin{gathered} {\chi }^2=\frac{(9-9.0909{)}^2}{9.0909}+\frac{(8-8.6364{)}^2}{8.6364}+\frac{(8-8.6364{)}^2}{8.6364}+\frac{(10-8.6364{)}^2}{8.6364} \\ +\frac{(11-10.9091{)}^2}{10.9091}+\frac{(11-10.3636{)}^2}{10.3636}+\frac{(11-10.3636{)}^2}{10.3636}+\frac{(9-10.3636{)}^2}{10.3636} \\ =0.5683 \end{gathered}$$

The interval for the P-value can then be found in table C in the column title which has the $X^2$-value in the corresponding interval in the row with

$$\begin{gathered} df=(c-1)(r-1) \\ =(4-1)(2-1) \\ =3 \end{gathered}$$

$p>0.25$

If the P-value is smaller than the significance level, then the null hypothesis is rejected.

$P>0.05\Rightarrow Fail to reject H_0$

There is not sufficient evidence that there is an association between the variables.