Question

Before contracting to have stereo music piped into each of his suites of offices, an executive had his office manager randomly select seven offices in which to have the system installed. The average time (in minutes) spent outside these offices per excursion among the employees involved was recorded before and after the music system was installed with the following results. $$ \begin{array}{lccccccc} \text { Office Number } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { No Music } & 8 & 9 & 5 & 6 & 5 & 10 & 7 \\ \text { Music } & 5 & 6 & 7 & 5 & 6 & 7 & 8 \end{array} $$ Would you suggest that the executive proceed with the installation? Conduct an appropriate test of hypothesis. Find the approximate \(p\) -value and interpret your results.

Short answer

Explain your answer. Answer: No, the executive should not proceed with installing the music system to reduce the average time spent outside offices. The paired t-test results showed a p-value of 0.1376, which is greater than the common significance level of 0.05. This means there is insufficient evidence to suggest that the music system reduced the average time spent outside the offices.

Step-by-step solution

Calculate Differences

First, calculate the differences in time spent outside the offices for each pair (before and after installing the music system). $$ \begin{array}{lccccccc} \text { Office Number } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { Difference (No Music - Music) } & 3 & 3 & -2 & 1 & -1 & 3 & -1 \end{array} $$

Calculate Sample Mean and Variance of Differences

Next, calculate the sample mean and variance of these differences. Sample Mean, \(\bar{d} = \frac{\sum_{i=1}^7 d_i}{7} = \frac{3+3-2+1-1+3-1}{7} = 1\) Sample Variance, \(s_d^2 = \frac{\sum_{i=1}^7 (d_i - \bar{d})^2}{7-1} = \frac{(2+2+9+0+4+2+4)}{6} = 3.5\)

Determine the Null and Alternative Hypotheses

In this case, the null hypothesis, \(H_0\), is that there is no difference in the average time spent outside the offices before and after the music system installation, i.e., \( \mu_d = 0\). The alternative hypothesis, \(H_1\), is that there is a difference (the average time spent outside offices has reduced due to the music system), i.e., \( \mu_d < 0\).

Calculate the Test Statistic

Compute the t-statistic using the following formula: $$t = \frac{\bar{d} - \mu_d}{s_d/\sqrt{n}} = \frac{1 - 0}{\sqrt{3.5/7}} = 1.1781$$

Find the p-value

As this is a one-tailed test (we only care if the average time spent outside offices has reduced), we need to determine the one-tailed p-value. With 6 degrees of freedom, we can use a t-distribution table or online calculator to find the p-value. The p-value for \(t = 1.1781\) and \(df = 6\) is approximately \(0.1376\).

Interpret Results

In this case, we can compare the p-value to the significance level (commonly set at \(\alpha = 0.05\)). If the p-value is less than \(\alpha\), we would reject the null hypothesis and accept the alternative hypothesis. Since \(0.1376 > 0.05\), we cannot reject the null hypothesis. Therefore, there is insufficient evidence to suggest that the music system reduced the average time spent outside the offices. Thus, based on this data, it is not recommended to proceed with the installation of the music system.

Key concepts

t-test

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups.
In hypothesis testing, the t-test helps assess whether changes observed in a sample hold true in the larger population.

sample mean

The sample mean is a crucial concept used to estimate the population mean based on a sample from that population.
To calculate it, you add up all the sample values and then divide by the number of observations.
For instance, the sample mean of differences in our example is calculated as follows:

• Sum of differences: 3 + 3 - 2 + 1 - 1 + 3 - 1 = 6
• Sample size: 7
• Sample mean, \( \bar{d} = \frac{6}{7} = 1 \)

This average is used to perform further calculations in the hypothesis testing process.

p-value

The p-value in hypothesis testing helps determine the statistical significance of the observed effect.
If the p-value is lower than the predetermined significance level (like 0.05), it suggests that the result is statistically significant.
In plain terms, the smaller the p-value, the stronger the evidence against the null hypothesis.
For the current scenario, with a p-value of approximately 0.1376, this suggests that the observed data is not strongly against the null hypothesis.
Thus, it implies that the music system installation did not signifcantly reduce employees' time outside of the offices.

null hypothesis

The null hypothesis, denoted as \( H_0 \), is a statement positing that there is no effect or difference in that particular experiment or test.
In the context of the example, the null hypothesis states that the music system has no impact on the time spent outside offices, i.e., \( \mu_d = 0 \).
The goal of the hypothesis test is to find evidence to reject or fail to reject this assumption based on the data.

alternative hypothesis

The alternative hypothesis, \( H_1 \), represents what the researcher aims to prove. It is the opposite of the null hypothesis.
In our example, the alternative hypothesis suggests that the music system does reduce the time spent outside the offices, i.e., \( \mu_d < 0 \).
The testing process evaluates this hypothesis against the data, aiming to accept it if there is significant evidence that contradicts the null hypothesis.
Although, in this particular test, the data failed to provide enough evidence to accept \( H_1 \), indicating the music system might not have the desired effect.