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 (1.645).

Step-by-step solution

Define the null hypothesis and alternative hypothesis

We want to test if the new lighting had an impact on student productivity. Let's define the null hypothesis (H0) as the probability of a student's productivity improving (P) is equal to 0.5. The alternative hypothesis (H1) is that the probability of a student's productivity improving (P) is greater than 0.5. \(H_0: P=0.5\) \(H_1: P>0.5\)

Find the test statistic using the normal approximation to the sign test

Let n be the number of students observed. The mean and variance of a Bernoulli random variable can be calculated as follows: Mean: \(\mu=n * p\), where p is the probability of a student's productivity increase Variance: \(\sigma^2 = n * p * (1-p)\) In this case, there are 35 students observed, and 21 had an increase in productivity. Using the formula under the null hypothesis, we get \(\mu = n * p = 35 * 0.5 = 17.5\) and \(\sigma^2 = 35 * 0.5 * (1-0.5) = 8.75\). Now we will use the normal approximation to find the test statistic, which is calculated using the formula: \(Z = \frac{X - \mu}{\sigma}\), where X is the number of successful outcomes (21) First, we need to calculate the standard deviation, \(\sigma =\sqrt{8.75}=2.958\) Then, we can find the test statistic: \(Z = \frac{21 - 17.5}{2.958} = 1.18\)

Find the critical value and compare with the test statistic

The problem asks us to determine whether the new lighting was effective in increasing student productivity at the 5% level of significance. This means that we need to find the critical value for a one-tailed test at 5% or 0.05: \(Z_{0.05} = 1.645\) Since the test statistic (1.18) is less than the critical value (1.645), we fail to reject the null hypothesis.

Conclusion

Given the observed data, and using the normal approximation to the sign test at a 5% level of significance, we cannot conclude that the new lighting was effective in increasing student productivity. The test statistic (1.18) is less than the critical value (1.645), therefore we fail to reject the null hypothesis.

Key concepts

Normal Approximation

Normal approximation is a statistical technique that simplifies the complexity of certain problems, making it easier to solve. It becomes especially handy when dealing with large sample sizes. In the context of hypothesis testing, normal approximation can convert a binomial distribution to a normal distribution, which is easier to analyze.
For the given exercise, the productivity of students can either increase or not due to new lighting. This represents a binary outcome that can be modeled using a binomial distribution. However, working directly with binomial distributions may be complex when sample sizes are large. Through normal approximation, we can use the standard normal distribution. This involves calculating the mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution, and then applying these to the normal distribution context.

• Mean \(\mu = n * p\) considers the total number of trials (students) and the probability of success (increase in productivity).
• Variance \(\sigma^2 = n * p * (1-p)\) accounts for the variability in outcomes.

The transformation of data in this way helps us apply the more manageable normal distribution methods to solve the hypothesis test.

Sign Test

The sign test is a non-parametric test that is used to determine if there is a significant difference between paired observations. It's particularly useful when you can't assume the underlying distribution of the data.
In this exercise, the sign test helps determine if the new classroom lighting significantly improved student productivity. Each student’s performance is evaluated by comparing 'before' and 'after' scenarios. If the productivity improved, it contributes a positive sign; if not, a negative sign.

• We observe 35 students: 21 indicate increased productivity.
• The sign test evaluates if the proportion of positive signs (increased productivity) is significantly greater than what would occur by random chance.

Ultimately, the sign test does not require the productivity differences to be normally distributed, making it robust and flexible to this setup.

Null Hypothesis

A null hypothesis represents the default position that there is no effect or no difference. It serves as a starting point for statistical testing.
For this particular problem, the null hypothesis \(H_0\) states that the probability of a student showing improved productivity is 0.5, indicating no effect from the new lighting. In other words, the new lighting is presumed not to make a difference in productivity:\[H_0: P=0.5\]
The goal of a hypothesis test is to determine if there is enough statistical evidence to reject the null hypothesis. In this case, the null hypothesis assumes any observed improvement is due to random chance rather than the lighting effect.
By failing to reject \(H_0\), the test suggests insufficient evidence to prove the lighting effectively increases productivity.

Alternative Hypothesis

An alternative hypothesis is what you may believe to be true or hope to prove true. It proposes that there is indeed an effect or a difference.In the context of the exercise, the alternative hypothesis \(H_1\) posits that the new lighting actually improves productivity to a degree greater than what would happen by random chance:\[H_1: P>0.5\]
The objective is to find statistical evidence supporting the alternative hypothesis, thereby refuting the null hypothesis. In hypothesis testing, providing enough evidence to accept the alternative hypothesis suggests the lighting installation positively affects productivity.While the exercise concludes that we fail to reject the null hypothesis, keeping the alternative hypothesis clearly defined ensures clarity regarding the intended investigation of potential effects.