Question

To test the comparative brightness of two red dyes, nine samples of cloth were taken from a production line and each sample was divided into two pieces. One of the two pieces in each sample was randomly chosen and red dye 1 applied; red dye 2 was applied to the remaining piece. The following data represent a "brightness score" for each piece. Is there sufficient evidence to indicate a difference in mean brightness scores for the two dyes? Use \(\alpha=.05 .\) $$ \begin{array}{lrlrrrrrrr} \text { Sample } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text { Dye 1 } & 10 & 12 & 9 & 8 & 15 & 12 & 9 & 10 & 15 \\ \text { Dye 2 } & 8 & 11 & 10 & 6 & 12 & 13 & 9 & 8 & 13 \end{array} $$

Short answer

Answer: No, there is not sufficient evidence at the 0.05 significance level to indicate a difference in mean brightness scores for the two dyes.

Step-by-step solution

Calculate the differences

Take the difference in brightness scores between dye 1 and dye 2 for each paired sample, by subtracting the dye 2 score from the dye 1 score. Differences: \(2, 1, -1, 2, 3, -1, 0, 2, 2\)

Calculate the mean and standard deviation of the differences

Calculate the mean and standard deviation of the differences. The mean (\(\bar{d}\)) is the sum of the differences divided by the number of samples (9). The standard deviation (s) can be calculated using the formula $$s = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}}$$ where \(d_i\) are the individual differences and \(n\) is the number of samples. Mean: \(\bar{d} = \frac{2 + 1 - 1 + 2 + 3 - 1 + 0 + 2 + 2}{9} = \frac{10}{9} \approx 1.11\) Standard deviation: \(s \approx 1.45\)

Perform a paired t-test

We will perform a paired t-test to determine if there is a significant difference in the mean brightness scores for the two dyes at a significance level of \(\alpha = 0.05\). The null hypothesis (\(H_0\)) states that there is no difference in the mean brightness scores between the two dyes (\(\mu_1 - \mu_2 = 0\)), and the alternative hypothesis (\(H_a\)) states that there is a difference (\(\mu_1 - \mu_2 \neq 0\)). Calculate the t-value using the formula $$t = \frac{\bar{d} - 0}{s / \sqrt{n}}$$ where \(\bar{d}\) is the mean difference, \(s\) is the standard deviation and \(n\) is the number of samples. t-value \(\approx 1.93\)

Calculate the critical t-value and make a decision

Calculate the critical t-value for a two-tailed test with 8 degrees of freedom (since \(df = n-1\) where n is the number of samples) and a significance level of \(\alpha = 0.05\). This value can be found in the t-distribution table or using statistical software. The critical t-value is approximately \(2.306\). Since the calculated t-value (1.93) is less than the critical t-value (2.306), we fail to reject the null hypothesis. Conclusion: There is not sufficient evidence at the \(\alpha = 0.05\) significance level to indicate a difference in mean brightness scores for the two dyes.

Key concepts

null hypothesis

When conducting a statistical test, the null hypothesis plays a central role. It is a statement that assumes there is no effect or difference present. In the context of this exercise, the null hypothesis (\(H_0\)) states that there is no difference in the mean brightness scores between the two dyes, denoted as (\(\mu_1 - \mu_2 = 0\)).
The null hypothesis is the default or initial claim that we test against. If sufficient evidence is found in the data to contradict this hypothesis, it might be rejected in favor of an alternative hypothesis. The alternative hypothesis (\(H_a\)), which suggests otherwise, in this case is (\(\mu_1 - \mu_2 eq 0\)), meaning there is a difference in the mean brightness scores.
In hypothesis testing, there is always some level of uncertainty. Thus, a statistical test is used to determine if the observed data provides convincing evidence to reject the null hypothesis. In this example, the test will help decide if the difference in brightness scores is due to random chance or indicates a real difference between the dyes. Being aware of these hypotheses allows us to understand the framework within which a statistical test operates.

t-value

The t-value is an essential component when performing a t-test. It is a statistic that measures the size of the difference relative to the variation in your sample data. To compute the t-value in the paired t-test, we use the formula:\[t = \frac{\bar{d} - \mu_0}{s / \sqrt{n}}\]where \(\bar{d}\)) is the mean of the differences between paired samples, \(s\)) is the standard deviation of these differences, and \(n\)) is the number of samples.
For this exercise, the t-value is calculated to be approximately 1.93. A higher t-value indicates that the observed difference is more pronounced relative to the expected variation in the sample data. But interpreting this t-value requires comparing it to a critical value, which will determine if the null hypothesis can be rejected.
The t-value helps to quantify and assess the evidence against the null hypothesis. In the context of the dye brightness scores, it tells us how many standard deviations the observed mean difference is away from the assumed population mean difference (which is zero under the null hypothesis). By calculating the t-value, we can determine whether the observed difference is statistically significant or merely due to random chance.

critical value

A critical value is a threshold that the t-value must exceed for a statistically significant result. This value comes from the t-distribution table, which corresponds to the test's degrees of freedom and the chosen significance level. In this test, with 8 degrees of freedom (since \(df = n-1\) where \(n\) is 9), and \(\alpha = 0.05\), the critical t-value for a two-tailed test is approximately 2.306.
If the calculated t-value exceeds the critical value, it indicates evidence against the null hypothesis, suggesting the alternative hypothesis may be true. Conversely, if the t-value is less than the critical value, we fail to reject the null hypothesis.
The notion of critical value helps differentiate between a result that could reasonably occur by chance and one that suggests an underlying true effect. Here, since the calculated t-value of 1.93 does not exceed the critical value of 2.306, we conclude that the evidence is insufficient to suggest a difference in brightness scores between the two dyes.

significance level

The significance level, often denoted as \(\alpha\), is a critical benchmark in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. Commonly set at 0.05, it limits the likelihood of a "false positive."
In this dye brightness exercise, the significance level is set to \(\alpha = 0.05\). This means that we are willing to accept a 5% chance of mistakenly rejecting the null hypothesis when it might be true. It defines the threshold against which we compare the calculated t-value and critical value.
A smaller significance level means stricter criteria for rejecting the null hypothesis, reducing the risk of type I errors (false positives), but possibly increasing type II errors (false negatives). The significance level acts as a risk factor you define in your test, balancing how much evidence you need to confidently conclude that the observed differences in brightness scores are indeed real. By maintaining a clear understanding of how it influences your decision-making, you maintain scientific rigor in your interpretation of results.