Question

Using the sample data from Exercise 1, find the z score corresponding to the prediction error of 0 min. Is that prediction error significantly low or high? Why or why not?

Short answer

The z-score is –0.13, and the error is insignificant.

Step-by-step solution

Given information

With reference to Exercise 1, the errors (minutes) are listed below.

4, –7, 0, 1, –1, 1, –4, –7, 22, 7, –5, 1

The predicted error is 0 minutes.

Express the idea for z-score

The Z-score is required to find the standard value of the specific raw score. It is the resultant of the quotient of two values, the difference of the raw score (x) and the mean measure $\left(\overline{x}\right)$; and the standard deviation value (s).

The formula of z-score is shown below.

$z=\frac{x-\overline{x}}{s}$

Compute the sample mean and sample standard deviation

The formulae for sample mean and sample standard deviation are given below.

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ s=\sqrt{\frac{\sum {\left(x-\overline{x}\right)}^2}{n-1}} \end{gathered}$$

For n values, substitute the values in both formulae.

The mean of the observation is calculated below.

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{4+\left(-7\right)+0+...+1}{12} \\ =\frac{12}{12} \\ =1 \end{gathered}$$

The sample standard deviation is calculated below.

$$\begin{gathered} s=\sqrt{\frac{\sum {\left(x-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(-7-1\right)}^2+{\left(-7-1\right)}^2+{\left(-5-1\right)}^2+...+{\left(22-1\right)}^2}{12-1}} \\ =\sqrt{\frac{64+64+36+...+441}{11}} \\ =7.862 \end{gathered}$$

Thus, the mean and the standard deviations are 1.0 minute and 7.9 minutes, respectively.

Find the components of the z-score 

The difference between the observed and mean predicted errors is

$$\begin{gathered} x-\overline{x}=0-1 \\ =-1 \end{gathered}$$

The quotient of the difference over the standard deviation measure is computed below.

$$\begin{gathered} z=\frac{x-\overline{x}}{s} \\ =\frac{0-1}{7.9} \\ =-0.1266 \\ \approx -0.13 \end{gathered}$$

Thus, the z-score is –0.13.

Significance of the error value

When the z-score computed for the observed values is less than 2 and greater than –2, the value is not unusual and the result is insignificant.

Since the z-score lies within the range of –2 and 2, the error is insignificant—neither low nor high.