Question

The productivity of 35 students was observed and measured both before and after the installation of new lighting in their classroom. The productivity of 21 of the 35 students was observed to have improved, whereas the productivity of the others appeared to show no perceptible gain as a result of the new lighting. Use the normal approximation to the sign test to determine whether or not the new lighting was effective in increasing student productivity at the \(5 \%\) level of significance.

Short answer

Answer: No, we cannot conclude that the new lighting was effective in increasing student productivity at the 5% level of significance, as the test statistic (1.18) is less than the critical value (1.96) and we fail to reject the null hypothesis.

Step-by-step solution

Calculate test statistic

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: \(Z = \frac{(21 - 35*0.5) - \frac{1}{2}}{\sqrt{35*0.5(1 - 0.5)}}\) \(Z = \frac{(21 - 17.5) - 0.5}{\sqrt{35*0.5(0.5)}}\) \(Z = \frac{(3.5)}{\sqrt{8.75}}\) \(Z \approx 1.18\)

Compare to critical value

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.

Interpret the result

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.