Question

5. Presenting means in graphs Aa Aa Suppose you randomly decide whether to administer Treatment A or Treatment B to each of 120 volunteers, such that 40 female volunteers received Treatment A, 20 female volunteers received Treatment B, 20 male volunteers received Treatment A, and 40 male volunteers received Treatment B. Treatments A and B are two different doses of the same drug intended to improve memory. After volunteers receive their treatment, they take a memory test and are given a memory score on a scale from 1 to 12, where higher scores indicate better memory. The table that follows includes the mean memory scores for each of the four groups. Treatment A Treatment B Females M-9 n-40 n-20 Males M-9 n20 n 40 M= Compute the overall mean memory scores for the females, the males, those who received Treatment A, and those who received Treatment B. Enter each of the means to the right of the appropriate row or below the appropriate column in the previous table. Enter all values with one decimal point. (Hint: You will need to compute weighted means because the sample sizes are not the same for each group.)

Answer 1

Weighted mean is calculated by the formula:

$Weighted \hspace{.1cm}mean = \frac{n_1\bar{x_1}+n_2\bar{x_2}}{n_1+n_2}$

Here, $n_1 \& n_2$ are the samples sizes and $\bar{x_1} \& \bar{x_2}$ are the means of respective groups.

a) Overall mean memory score for females:

Here, $n_1 = 40, n_2 = 20, \bar{x_1}=9 , \bar{x_2}=5$ . Thus,

$Overall \hspace{.1cm}mean \hspace{.1cm} memory \hspace{.1cm} score \hspace{.1cm} for \hspace{.1cm} females = \frac{40*9+20*5}{40+20} = 7.7$

b) Overall mean memory score for males:

$n_1 = 20, n_2 = 40, \bar{x_1}=9 , \bar{x_2}=5$. Thus,

$Overall \hspace{.1cm}mean \hspace{.1cm} memory \hspace{.1cm} score \hspace{.1cm} for \hspace{.1cm} males = \frac{20*9+40*5}{40+20} = 6.3$

c) Overall mean memory score for volunteers receiving Treatment A:

$n_1 = 40, n_2 = 20, \bar{x_1}=9 , \bar{x_2}=9$.

$Overall \hspace{.1cm} mean \hspace{.1cm} memory \hspace{.1cm} score \hspace{.1cm} for \hspace{.1cm} treatment \hspace{.1cm} A = \frac{40*9+20*9}{40+20} = 9.0$

d) Overall mean memory score for volunteers receiving Treatment B:

$n_1 = 20, n_2 = 40, \bar{x_1}=5 , \bar{x_2}=5$.

$Overall \hspace{.1cm} mean \hspace{.1cm} memory \hspace{.1cm} score \hspace{.1cm} for \hspace{.1cm} treatment \hspace{.1cm} B= \frac{20*5+40*5}{40+20} = 5.0$

After puting these results in the table given in question the table becomes:

	Treatment A	Treatment B
Females	M = 9 n = 40	M = 5 n = 20	M = 7.7
Males	M = 9 n = 20	M = 5 n = 40	M = 6.3
	M = 9.0	M = 5.0

The line graph for the mean score of females and males is given below:

Line graph showing mean scores 10 Females Males Treatment A Treatment B Treatment

The lines for males and females completely overlap so only one line is visible.

Now, considering the overall mean for females and males calculated earlier suggests that females have better memories than males, as the overall mean for females is 7.7 and that for males is 6.3.

However, the line graph shows that there is no difference between memories of females and males.

This is because, in the line graph the sample size for each treatment group is not considered while comparing memories of males and females. These sample sizes act as weights while computing weighted means in order to make the two groups comparable. So, if the line graph was made for overall scores, it would have given the same conclusion as the calculation.