Question

1. a random sample is selected from a normal population with a mean of \\( \mu=20 \\) and a standard deviation of \\( \sigma=10 \\). after a treatment is administered to the individuals in the sample, the sample mean is found to be \\( m = 25 \\).

a. if the sample consists of \\( n = 25 \\) scores, is the sample mean sufficient to conclude that the treatment has a significant effect? use a two - tailed test with \\( \alpha=.05 \\).

Explanation:

Step1: Calculate the standard error

The formula for the standard error $\sigma_{M}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 10$ and $n = 25$, then $\sigma_{M}=\frac{10}{\sqrt{25}}=\frac{10}{5}=2$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{M-\mu}{\sigma_{M}}$. Substitute $M = 25$, $\mu=20$ and $\sigma_{M}=2$ into the formula. So $z=\frac{25 - 20}{2}=\frac{5}{2}=2.5$.

Step3: Determine the critical z - values

For a two - tailed test with $\alpha=0.05$, the critical z - values are $z=\pm1.96$.

Since $|z| = 2.5>1.96$.

Answer:

Yes, the sample mean is sufficient to conclude that the treatment has a significant effect.