In an experiment designed to study the effects of illumination level on task performance (“Performance of Complex Tasks Under Different Levels of

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.
$$\begin{array}{|c|c|} \hline Subject & (1) & (2) & (3) & (4) & (5) &(6) & (7) & (8) & (9) \\ \hline Black & 25.85 & 28.84 & 32.05 & 25.74 & 20.89 & 41.05 & 25.01 & 24.96 & 27.47 \\ \hline White & 18.28 & 20.84 & 22.96 & 19.68 & 19.509 & 24.98 & 16.61 & 16.07 & 24.59 \\ \hline \end{array}$$
Does the data indicate that the higher level of illumination yields a decrease of more than 5 sec in true average task completion time? Test the appropriate hypotheses using the P-value approach.

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Given:
$n=\text{Sample size}=9$
Given claim: Mean difference is more than 5
The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.
${H}_{0}:{\mu }_{d}=5$
${H}_{a}:{\mu }_{d}>5$
Let us first determine the difference between each pair of data values:
$\begin{array}{|ccc|}\hline Population1& Population2& Differenced\\ 25.85& 18.23& 25.85—18.23=7.62\\ 28.84& 20.84& 28.84—20.84=8.00\\ 32.05& 22.96& 32.05—22.96=9.09\\ 25.74& 19.68& 25.74—19.68=6.06\\ 20.89& 19.50& 20.89—19.50=1.39\\ 41.05& 24.98& 41.05—24.98=16.07\\ 25.01& 16.61& 25.01—16.61=8.40\\ 24.96& 16.07& 24.96—16.07=8.89\\ 27.47& 24.59& 27.47—24.59=2.88\\ \hline\end{array}$
Determine the sample
Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.
$\overline{d}=\frac{7.62+8.00+9.09+\cdots +8.89+2.88}{9}\approx 7.6$
Determine the sample standard deviation of the differences:
${s}_{d}=\sqrt{\frac{\left(7.62-7.6{\right)}^{2}+\cdots +\left(2.88-7.6{\right)}^{2}}{9-1}}\approx 4.1779$
Determine the value of the test statistic:
$t=\frac{\overline{d}-d}{\frac{{s}_{d}}{\sqrt{n}}}=\frac{7.6-5}{\frac{4.1779}{\sqrt{9}}}\approx 1.867$
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number (or interval) in the column title of the Students T distribution in the appendix containing the t-value in the row $df=n-1=9-1=8$
$0.025
If the P-value is less than the significance level, reject the null hypothesis.

There is sufficient evidence to support the claim that the higher level of illumination yields a decrease of more than 5 sec in true average task completion time.