Find the value of the standard score, z, and determine whether the alternative hypothesis is supported or not supported at a 0.05 significance level.
-$$\mathrm { H } _ { \mathrm { a } } : \mu \neq 23 , \mathrm { n } = 60 , \overline { \mathrm { x } } = 25 , \sigma = 1.5$$
A) $z = 10.33$; not supported
B) $z = 10.33$; supported
C) $\mathrm { z } = - 1.96$; not supported
D) $z = 1.5$; not supported

Question

Quizplus · Accepted Answer

The standard score (z) is calculated using the formula $z = \frac{\overline{x} - \mu}{\sigma / \sqrt{n}}$, where $\overline{x}$ 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, $z = \frac{25 - 23}{1.5 / \sqrt{60}} \approx 10.33$. Since the z-score is significantly larger than the critical value for a 0.05 significance level (which is approximately $\pm1.96$ for a two-tailed test), the alternative hypothesis ($\mu \neq 23$) is supported.

Find the Value of the Standard Score, Z, and Determine Ha:μ≠23,n=60,x‾=25,σ=1.5\mathrm { H } _ { \mathrm { a } } : \mu \neq 23 , \mathrm { n } = 60 , \overline { \mathrm { x } } = 25 , \sigma = 1.5Ha​:μ=23,n=60,x=25,σ=1.5

Find the Value of the Standard Score, Z, and Determine $\mathrm { H } _ { \mathrm { a } } : \mu \neq 23 , \mathrm { n } = 60 , \overline { \mathrm { x } } = 25 , \sigma = 1.5$