Question

For the job satisfaction study described in Section 9.1, the hypotheses are

$$\begin{gathered} H_0:\mu =0 \\ {H}_a:\mu isnotequalto0 \end{gathered}$$

where $\mu$is the mean difference in job satisfaction scores (self-paced machine-paced) in the population of assembly-line workers at the company. Data from a random sample of $18$workers gave $\stackrel{⏜}{x}=17$and $s_x=60$

Calculate the test statistic. Show your work.

Short answer

The test statistic is $1.273$

Step-by-step solution

Given information

Sample mean $\left(x\right)=17$

Sample size$\left(n\right)=18$

Sample standard deviation $\left(s\right)=60$

Concept

The test statistic is computed as: $t=\frac{{\displaystyle \frac{\stackrel{⏜}{x}-\mu }{s}}}{\sqrt{n}}$

$\overline{x}$=Sample mean

$\mu =$Population mean

$n=$Sample size

$s=$Sample standard deviation

Explanation

The null and alternative hypotheses are:

$$\begin{gathered} H_0:\mu =0 \\ {H}_a:\mu \ne 0 \end{gathered}$$

The test statistic can be computed as:

$$\begin{gathered} t=\frac{{\displaystyle \frac{\stackrel{⏜}{x}-\mu }{s}}}{\sqrt{n}} \\ t=\frac{{\displaystyle \frac{18-0}{60}}}{\sqrt{18}} \\ =1.273 \end{gathered}$$

Thus, the test statistic is$1.273$