Answer:
We use z-test for this hypothesis.
[tex]z_{stat} = 2.23[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 4.88
Sample mean, [tex]\bar{x}[/tex] = 5.91
Sample size, n = 48
Alpha, α = 0.05
Population standard deviation, σ = 3.2
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 4.88\\H_A: \mu > 4.88[/tex]
The null hypothesis states that the mean score of successful managers on a psychological test is 4.88 and the alternate hypothesis says that the mean score of successful managers on a psychological test is greater than 4.88.
We use One-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{5.91 - 4.88}{\frac{3.2}{\sqrt{48}} } = 2.23[/tex]