A university is interested in promoting graduates of its honors program by establishing that the mean GPA of these graduates exceeds 3.50. A sample of 36 honors students is taken and is found to have a mean GPA equal to 3.60. The population standard deviation is assumed to equal 0.40. The value of the test statistic is ________. A) z = -1.50 B) t₃₅= -1.50 C) z = 1.50 D) t₃₅ = 1.50

Question

Quizplus · Accepted Answer

The test statistic is calculated using the formula $z = \frac{\bar{x} - \mu}{(\sigma/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values ($\bar{x} = 3.60$, $\mu = 3.50$, $\sigma = 0.40$, $n = 36$), we get $z = \frac{3.60 - 3.50}{(0.40/\sqrt{36})} = \frac{0.10}{(0.40/6)} = \frac{0.10}{0.0667} = 1.50$.

A University Is Interested in Promoting Graduates of Its Honors