Answer:
a) For this case we can use the definition of weighted average given by:
![M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B%20%5Cbar%20X_1%20n_1%20%2B%20%5Cbar%20X_2%20n_2%7D%7Bn_1%20%2Bn_2%7D)
And if we replace the values given we have:
![M = \frac{8*4 + 16*4}{4+4}= 12](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B8%2A4%20%2B%2016%2A4%7D%7B4%2B4%7D%3D%2012)
b) ![M = \frac{8*3 + 16*5}{3+5}= 13](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B8%2A3%20%2B%2016%2A5%7D%7B3%2B5%7D%3D%2013)
c) ![M = \frac{8*5 + 16*3}{5+3}= 11](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B8%2A5%20%2B%2016%2A3%7D%7B5%2B3%7D%3D%2011)
Step-by-step explanation:
Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.
a) What is the mean for the combined set if both of the original samples have n=4 scores
"
For this case we can use the definition of weighted average given by:
![M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B%20%5Cbar%20X_1%20n_1%20%2B%20%5Cbar%20X_2%20n_2%7D%7Bn_1%20%2Bn_2%7D)
And if we replace the values given we have:
![M = \frac{8*4 + 16*4}{4+4}= 12](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B8%2A4%20%2B%2016%2A4%7D%7B4%2B4%7D%3D%2012)
b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5
Using the definition we have:
![M = \frac{8*3 + 16*5}{3+5}= 13](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B8%2A3%20%2B%2016%2A5%7D%7B3%2B5%7D%3D%2013)
c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3
Using the definition we have:
![M = \frac{8*5 + 16*3}{5+3}= 11](https://tex.z-dn.net/?f=%20M%20%3D%20%5Cfrac%7B8%2A5%20%2B%2016%2A3%7D%7B5%2B3%7D%3D%2011)