Question

Two random samples with sizes 100 and n are chosen from the populations with the means 85.6 and 82.1. They have standard deviations 12.4 and 8.9, respectively. Which of these values of n would result in the smallest SE?
a. 100
b. 120
c. 90
d. 50
e. 70

Answer 1

Answer:

$SE= \sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$

And for this case if we have the same sample size we got the minimum value when we have the higher value fo n for each one and for this case would be the answer:

b. 120

Step-by-step explanation:

For this case we have the following info given:

$n_1 = n_2 = 100$

$\mu_1 = 85.6$

$\mu_2 = 82.1$

$\sigma_1 =12.4$

$\sigma_2 = 8.9$

We assume that the variable of interest is the linear combination of the two means and for this case the standard error would be given by:

$SE= \sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$

And for this case if we have the same sample size we got the minimum value when we have the higher value fo n for each one and for this case would be the answer:

b. 120