Question

5. A researcher measures the mean time (in seconds) it takes two groups of children to complete an activity task. She finds that Group A (M = 22 seconds) completed the task more quickly than Group B (M = 36 seconds). She then computes a weighted mean for both groups combined and calculates = 26. Based on the information provided, which group had a larger sample size?

Answer 1

When calculating the weighted mean for different means of groups of data, the total data for each group is multiplied to the mean of the group and the sum of results for all the groups is divided by the overall total number of the data.

Let the sample size of the group with a mean of 22 be x and the sample size of the group with a mean of 36 be y.

The weighted mean is given by

$\frac{22x+36y}{x+y} =26 \\ \\ \Rightarrow22x+36y=26(x+y)=26x+26y \\ \\ \Rightarrow26x-22x=36y-26y \\ \\ \Rightarrow4x=10y \\ \\ \Rightarrow \frac{x}{y} = \frac{10}{4} = \frac{5}{2}$

From the above, it can be seen that the ratio of the sample size of the group with a mean of 22 to the sample size of the group with a mean of 36 is 5 : 2.

Therefore, the group with a mean of 22 had the larger sample size.