Q 1.2. - The Practice of Statistics for AP

Question

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

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

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 $\overline{x}=17$ and $s_x=60$

Use Table B to find the P-value. What conclusion would you draw?

Step-by-Step Solution

Answer

The P -value is $0.221$

Step 1: Given information

Sample mean $\left(\overline{x}\right) = 17$

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

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

Test statistic $\left(t\right) = 1.273$

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: Calculation

The degree of freedom is computed as:

$$\begin{gathered} Df = n - 1 \\ = 18 - 1 \\ = 17 \end{gathered}$$

The $P$-value can be calculated as:  

$$\begin{gathered} P - value = 2 \times P (t > |1.2) \\ =0.221 \end{gathered}$$

$P$ is bigger than the significance level in this case $(0.221). (0.05).$ The null hypothesis is rejected in this case.