Question

1. suppose that a researcher developed a drug that she claims increases extroversion. a sample of ( n = 4 ) participants has a sample mean of ( m = 115 ) on a personality assessment after taking the drug. the personality test has a population mean of ( mu = 100 ) and ( sigma = 30 ). is the sample mean an especially unlikely result based

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{M}=\frac{\sigma}{\sqrt{n}}$.
Substitute $\sigma = 30$ and $n = 4$ into the formula:
$\sigma_{M}=\frac{30}{\sqrt{4}}=\frac{30}{2}=15$.

Step2: Calculate the z - score

The formula for the z - score of a sample mean is $z=\frac{M-\mu}{\sigma_{M}}$.
Substitute $M = 115$, $\mu=100$ and $\sigma_{M}=15$ into the formula:
$z=\frac{115 - 100}{15}=\frac{15}{15}=1$.

Step3: Determine the likelihood

In a normal distribution, about 68% of the data lies within $z=\pm1$, 95% within $z = \pm2$ and 99.7% within $z=\pm3$.
A z - score of $z = 1$ means that the sample mean is 1 standard error above the population mean. This is a relatively common result as it is within 1 standard error of the mean.

Answer:

The sample mean is not an especially unlikely result.