The mean score of adult men on a psychological test that measures "masculine stereotypes" is 4.88. A researcher studying hotel managers suspects that successful managers score higher than adult men in general. A random sample of 48 managers of large hotels has mean x-bar = 5.91. Assume the population standard deviation is sigma = 3.2.Using the null and alternative hypotheses that you set up in problem 5, the value of the test statistic for this hypothesis test is ______

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 = 4.88Sample mean, [tex]\bar{x}[/tex] = 5.91Sample size, n = 48Alpha, α = 0.05Population standard deviation, σ = 3.2First, 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]