Given:

\(\displaystyle{n}=\text{Sample size}={9}\)

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:

\(\begin{array}{|c|c|} \hline Population 1 & Population 2 & Difference d \\ \hline 25.85 & 18.23 & 25.85 — 18.23 = 7.62 \\ \hline 28.84 & 20.84 & 28.84 — 20.84 = 8.00 \\ \hline 32.05 & 22.96 & 32.05 — 22.96 = 9.09\\ \hline 25.74 & 19.68 & 25.74 — 19.68 = 6.06\\ \hline 20.89 & 19.50 & 20.89 — 19.50 = 1.39\\ \hline 41.05 & 24.98 & 41.05 — 24.98 = 16.07\\ \hline 25.01 & 16.61 & 25.01 — 16.61 = 8.40\\ \hline 24.96 & 16.07 & 24.96 — 16.07 = 8.89\\ \hline 27.47 & 24.59 & 27.47 — 24.59 = 2.88\\ \hline \end{array}\)

Determine the sample

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar{d}=\frac{7.62+8.00+9.09+\cdots+8.89+2.88}{9}\approx7.6\)

Determine the sample standard deviation of the differences:

\(s_{d}=\sqrt{\frac{(7.62-7.6)^{2}+\cdots+(2.88-7.6)^{2}}{9-1}}\approx4.1779\)

Determine the value of the test statistic:

\(t=\frac{\bar{d}-d}{\frac{s_{d}}{\sqrt{n}}}=\frac{7.6-5}{\frac{4.1779}{\sqrt{9}}}\approx1.867\)

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 \(df=n-1=9-1=8\)

\(0.025<P<0.05\)

If the P-value is less than the significance level, reject the null hypothesis.

\(P<0.05\Rightarrow Reject\ H_{0}\)

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.