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 $\sigma_M=\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 30$ and $n = 4$, then $\sigma_M=\frac{30}{\sqrt{4}}=\frac{30}{2}=15$.

Step2: Calculate the z - score

The formula for the z - score of a sample mean is $z=\frac{M-\mu}{\sigma_M}$.
Given $M = 115$, $\mu=100$ and $\sigma_M = 15$, then $z=\frac{115 - 100}{15}=\frac{15}{15}=1$.

Step3: Determine the probability

Using the standard normal distribution table, the probability of getting a z - score of $z = 1$ or more extreme (in both tails for a two - tailed test, but since the claim is about an increase, we can consider a one - tailed test here). For a one - tailed test, $P(Z\geq1)=1 - P(Z < 1)$.
From the standard normal table, $P(Z < 1)=0.8413$, so $P(Z\geq1)=1 - 0.8413 = 0.1587$.

Answer:

Since the probability of getting a sample mean of \(M = 115\) (or more extreme in the direction of the claim) is \(p=0.1587\), which is greater than the common significance levels (e.g., \(0.05\) or \(0.01\)), the sample mean is not an especially unlikely result.