Question

A researcher wanted to see if giving a selective serotonin reuptake inhibitor (anti-depressant) would decrease the number of self-injurious behaviors (SIB’s) found in adolescents. They gave the anti-depressant to a group of 8 participants and counted the numbers of SIBs a person performed for a month after taking the drug for 6 months. Here are the number of SIB’S performed by the 8 people, 27, 25, 32, 40, 43, 37, 35, 38.
a. Find the mean variance, and standard deviation of this group.
b. Suppose the u=45, what is the z score?
c. What is the SEM for this particular group?

Answer 1

Answer:

a. The mean of the sample is M=35.

The variance of the sample is s^2=39.125.

The standard deviation of the sample is s=6.255.

b. z=-1.6

c. SEM = 2.212

Step-by-step explanation:

The mean of the sample is M=35.

The variance of the sample is s^2=39.125.

The standard deviation of the sample is s=6.255.

Sample mean

$M=\dfrac{1}{8}\sum_{i=1}^{8}(27+25+32+40+43+37+35+38)\\\\\\ M=\dfrac{277}{8}=35$

Sample variance and standard deviation

$s^2=\dfrac{1}{(n-1)}\sum_{i=1}^{8}(x_i-M)^2\\\\\\s^2=\dfrac{1}{7}\cdot [(27-(35))^2+(25-(35))^2+(32-(35))^2+(40-(35))^2+(43-(35))^2+(37-(35))^2+(35-(35))^2+(38-(35))^2]$

$s^2=\dfrac{1}{7}\cdot [(58.141)+(92.641)+(6.891)+(28.891)+(70.141)+(5.64)+(0.14)+(11.39)]\\\\\ s^2=\dfrac{273.875}{7}=39.125\\\\\\s=\sqrt{39.125}=6.255$

b. If the population mean is 45, the z-score for M=35 would be:

$z=\dfrac{M-\mu}{\sigma}=\dfrac{35-45}{6.255}=\dfrac{-10}{6.255}=-1.6$

c. The standard error of the mean (SEM) of this group is calculated as:

$SEM=\dfrac{s}{\sqrt{n}}=\dfrac{6.255}{\sqrt{8}}=\dfrac{6.255}{2.828}=2.212$