Q 1.1. - The Practice of Statistics for AP

Question

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

$$\begin{gathered} H_0: \mu =0 \\ {H}_a: \mu is not equal to 0 \end{gathered}$$

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.

Step-by-Step Solution

Answer

The test statistic is $1.273$

Step 1: Given information

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

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

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

Step 2: 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

Step 3: 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 $1.273$