91Ó°ÊÓ

To calculate the test statistic, we need to first find the proportion of success (improved productivity) and the proportion of failures (no perceptible gain). We have 21 improved students out of 35, so the proportion of success is given as: \( \frac{21}{35} = 0.6 \) Now, we can calculate the test statistic using the normal approximation formula: \(Z = \frac{(x - n\pi_0) - \frac{1}{2}}{\sqrt{n\pi_0(1 - \pi_0)}}\) where x is the number of successes (improved students), n is the total number of trials (number of students), and \(\pi_0\) is the proportion of success under the null hypothesis. In this case, we assume equal probability of success and failure under the null hypothesis, so \(\pi_0 = 0.5\). Plugging the values into the formula, we get: \(Z = \frac{(21 - 35*0.5) - \frac{1}{2}}{\sqrt{35*0.5(1 - 0.5)}}\) \(Z = \frac{(21 - 17.5) - 0.5}{\sqrt{35*0.5(0.5)}}\) \(Z = \frac{(3.5)}{\sqrt{8.75}}\) \(Z \approx 1.18\)

We now need to compare the test statistic to the critical value at the 5% level of significance. Since this is a two-tailed test, we will use the z-score for \(2.5 \%\) on each side: \(Z_{0.025} = 1.96\) Since our calculated test statistic, \(Z \approx 1.18 < 1.96\), we fail to reject the null hypothesis.

As we failed to reject the null hypothesis, we cannot conclude that the new lighting was effective in increasing student productivity at the 5% level of significance. However, this does not mean that the lighting had no effect, only that the evidence is not strong enough to show a significant difference in productivity.