Question

Lighting in the Classroom The productivity of 35 students was measured both before and after the installation of new lighting in their classroom. The productivity of 21 of the 35 students was improved, whereas the others showed no perceptible gain from 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 system was effective in increasing student productivity at the 5% level of significance.

Step-by-step solution

Calculate proportions of improved and unchanged students

Count the number of improved students (21) and unchanged students (14). Calculate the proportion of improved students \(p = \frac{21}{35} \approx 0.6\) and the proportion of unchanged students \(q = 1 - p = 0.4\).

Determine the null and alternative hypotheses

The null hypothesis (\(H_0\)) states that there is no difference in student productivity due to the new lighting system, meaning the probability of improvement is \(0.5\). The alternative hypothesis (\(H_1\)) states that the new lighting system has a positive effect on student productivity, meaning the probability of improvement is greater than \(0.5\): \(H_0: p = 0.5\) \(H_1: p > 0.5\)

Calculate the expected number of improved students under the null hypothesis

Under the null hypothesis, we would expect half of the students to show improvement and half not to show improvement. Thus, the expected number of improved students is: \(E = 0.5 \times 35 = 17.5\)

Calculate the standard deviation for the binomial distribution

The standard deviation \(\sigma\) for a binomial distribution is given by the formula \(\sigma = \sqrt{n \times p \times q}\), where \(n\) is the total number of students, \(p\) and \(q\) are the proportions calculated in step 1. Under the null hypothesis, plug in the values: \(\sigma = \sqrt{35 \times 0.5 \times 0.5} = \sqrt{8.75} \approx 2.958\)

Calculate the Z-value

Calculate the Z-value, which represents how many standard deviations the observed value is from the expected value: \(Z = \frac{21 - 17.5}{2.958} = 1.182\)

Determine the rejection region and conclusion

For a 5% level of significance and a one-tailed test, we look up the critical value in the Z-table. The critical value is \(Z_{critical} = 1.645\). Since the calculated Z-value (1.182) is less than the critical value (1.645), we fail to reject the null hypothesis. Therefore, we cannot conclude that the new lighting was effective in increasing student productivity at the 5% level of significance.

Key concepts

Binomial Distribution

Understanding the binomial distribution is crucial when analyzing data where the outcome can fall into one of two categories, like a success or a failure. In the classroom lighting example, a student's productivity either improved or did not after a change was made—a textbook scenario for binomial distribution.

In general, if we have a certain number of trials (in this case, 35 students), each trial with only two possible outcomes (improved or not), a fixed probability of success (here we initially assume it to be 0.5 under the null hypothesis), and each trial is independent, then the random variable representing the number of successes (students whose productivity improved) follows a binomial distribution.

The probability of exactly 'k' successes in 'n' trials is given by the binomial formula: \( P(X = k) = {n \choose k}p^k(1-p)^{n-k} \), where \( {n \choose k} \) is the binomial coefficient.

Null Hypothesis

The null hypothesis, denoted by \(H_0\), is a statistical statement that indicates no effect or no difference. It is essentially a presumption of innocence in the world of statistics. In the context of the exercise, the null hypothesis assumes that the new lighting had no impact on student productivity.

By stating \(H_0: p = 0.5\), we assert that there is a 50% chance that any individual student's productivity would improve, which corresponds to the probability we'd expect by random chance alone. In order to claim that the new lighting is beneficial, evidence strong enough to reject this hypothesis must be presented.

Standard Deviation

The standard deviation is a measure of the amount of variation or dispersion in a set of values. In a binomial distribution, it quantifies the expected fluctuation in the number of successes. In the classroom lighting study, standard deviation tells us how much variation we might expect in the number of students who improved productivity if the null hypothesis were true.

The formula for the standard deviation of a binomial distribution, \( \sigma = \sqrt{n \times p \times q} \), harnesses the number of trials (n), and the probability of success (p) and failure (q). Calculating this value helps us determine the likelihood of observing a specific number of improved cases just by chance.

Z-value

The Z-value, also known as the Z-score, quantifies how many standard deviations an observed data point is from the mean under the null hypothesis. It's a tool used in hypothesis testing to compare the observed result to what we would typically expect if the null hypothesis were true.

In our exercise, the calculation \(Z = \frac{21 - 17.5}{2.958}\) gauges the distance between the actual number of students who improved (21) and the number expected by chance (17.5), in units of standard deviation. If this Z-value is large enough, it suggests that the observed outcome is not just a product of random variation.

Statistical Significance

Statistical significance acts as the threshold for determining when the observed results are unlikely to have occurred under the null hypothesis, indicating a potential real effect. In the scenario with the students, a significance level of \(5\text{%}\) was chosen, corresponding to a Z-value below which 95% of all possible sample outcomes would fall if the null hypothesis were true.

In hypothesis testing, if the calculated Z-score is beyond the critical Z-value from the Z-table corresponding to the selected significance level (here, \(Z_{critical} = 1.645\)), we would reject the null hypothesis, indicating a statistically significant result. However, since the calculated Z-value (1.182) did not exceed the critical value, the result is not statistically significant, leading to the conclusion that there is insufficient evidence to claim the lighting improved productivity.