Question

#1 In an experiment designed to study the effects of illumination level on task performance (“Performance of Complex
Tasks Under Different Levels of Illumination,” J. Illuminating Eng., 1976: 235-242), subjects were required to insert a
fine-tipped probe into the eyeholes of ten needles in rapid succession both for a low light level with a black background
and a higher level with a white background. Each data value is the time (sec) required to complete the task.
Background Subject

1 2 3 4 5 6 7 8 9
Black 25.85 28.84 32.05 25.74 20.89 41.05 25.01 24.96 27.47
White 18.23 20.84 22.96 19.68 19.50 24.98 16.61 16.07 24.59
Does the data indicate that the higher level of illumination yields a decrease of more than 5 sec in true average task
completion time?

Answer 1

Given data suggest that the paired $t$ test should be used.

The paired $t$ test

When $D=X-Y, \mu_D=\mu_1-\mu_2$ and null hypothesis

$$H_0:\mu_D=\triangle_0$$

the test statistic value for testing the hypothesis is

$$t={\bar{d}-\triangle_0 \over s_D/\sqrt{n}},$$

where $\bar{d}$ and $s_D$ are the sample mean and the sample standard deviation of differences $d_i,$ respectively.

In order to use this test assume that the differences $D_i$ are from a normal distribution.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} \begin{array}{cc} Blackground \\ Subject \end{array} & \begin{array}{cc} Black, \\ x_i \end{array} & \begin{array}{cc} White, \\ y_i \end{array} &\begin{array}{cc} Difference, \\ d_i=x_i-y_i \end{array} \\ \hline 1 & 25.85 & 18.23 &7.62 \\ 2 & 28.84 & 20.84 & 8.00 \\ 3 & 32.05 & 22.96 & 9.09 \\ 4 & 25.74 & 19.68 & 6.06 \\ 5 & 20.89 & 19.50 &1.39 \\ 6 & 41.05 & 24.98 & 16.07 \\ 7 & 25.01 & 16.61& 8.40 \\ 8 & 24.96 & 16.07& 8.89 \\ 9 & 27.47 & 24.59 & 2.88 \\ \hline \end{array}$$

The sample mean $\bar{d}$ for the difference

$$\bar{d}={7.62+8.00+9.09+6.06+1.39+16.07+8.40+8.89+2.88\over 9}=7.60$$

The sample variance is

$$s_D^2={1 \over n-1}\displaystyle\sum_{i=1}^n{(d_i-\bar{d})^2}$$

$$s_D^2=$$

$$={1 \over9-1}((7.62-7.60)^2+(8.00-7.60)^2+(9.09-7.60)^2+$$

$$+(6.06-7.60)^2+(1.39-7.60)^2+(16.07-7.60)^2+$$

$$+(8.40-7.60)^2+(8.89-7.60)^2+(2.88-7.60)^2)=$$

$$=17.45495$$

and the sample standard deviation

$$s_D=\sqrt{s_D^2}=\sqrt{17.45495}\approx4.1779$$

The hypotheses are:

$$H_0: \mu_D=5$$

$$H_1: \mu_D>5$$

The test statistic value is

$$t={7.60-5 \over 4.1779/\sqrt{9}}\approx1.8670$$

The degrees of the freedom are: $df=n-1=9-1=8.$

The P value

$$P=P(T>1.8670)=0.049431$$

$$P=0.049431<0.05<0.1$$

Reject the null hypothesis at a significance level 0.1 and 0.05.

$$P=0.049431>0.01$$

Do not reject the null hypothesis at a significance level 0.01.