Question

Reaction Times A comparison of reaction times (in seconds) for two different stimuli in a psychological word-association experiment produced the following results when applied to a random sample of 16 people: $$ \begin{array}{l|llllllll} \text { Stimulus 1 } & 1 & 3 & 2 & 1 & 2 & 1 & 3 & 2 \\ \hline \text { Stimulus 2 } & 4 & 2 & 3 & 3 & 1 & 2 & 3 & 3 \end{array} $$ Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)

Short answer

Answer: No, there is not sufficient evidence to indicate a difference in mean reaction times for the two stimuli using a significance level of \(\alpha = 0.05\).

Step-by-step solution

Identify the null and alternative hypotheses

We want to test if there is a difference in mean reaction times for the two stimuli. The null hypothesis (\(H_0\)) states that there is no difference in the means, while the alternative hypothesis (\(H_a\)) states that there is a difference. Formally, these hypotheses can be written as: \(H_0: \mu_{1} = \mu_{2}\) \(H_a: \mu_{1} \neq \mu_{2}\)

Calculate the descriptive statistics

We will calculate the mean and standard deviation for each stimulus group. For Stimulus 1: \(X_1 = \{1, 3, 2, 1, 2, 1, 3, 2\}\) Mean: \(\bar{X}_1 = \frac{1+3+2+1+2+1+3+2}{8} = \frac{15}{8} = 1.875\) Standard deviation: \(s_1 = \sqrt{\frac{\Sigma(X_i- \bar{X}_1)^{2}}{n-1}} \approx 0.991\) For Stimulus 2: \(X_2 = \{4, 2, 3, 3, 1, 2, 3, 3\}\) Mean: \(\bar{X}_2 = \frac{4+2+3+3+1+2+3+3}{8} = \frac{21}{8} = 2.625\) Standard deviation: \(s_{2} = \sqrt{\frac{\Sigma(X_i- \bar{X}_2)^{2}}{n-1}} \approx 0.744\)

Conduct a t-test

Since we do not know the population standard deviation, we will use a two-sample t-test to compare the means of the two stimuli. The t-test statistic is: $$t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$$ Plugging in the values, we get: $$t = \frac{1.875 - 2.625}{\sqrt{\frac{0.991^2}{8}+\frac{0.744^2}{8}}} \approx -1.622$$

Find the critical value(s) and p-value

We will use a significance level of \(\alpha = 0.05\). Since this is a two-tailed test, we will split the alpha value in half, and we need to find the critical values, \(t_{critical}\), corresponding to the probability in each tail. The degrees of freedom for our t-distribution are: $$df = n_1 + n_2 - 2 = 8 + 8 - 2 = 14$$ Using a t-table or a calculator, we find the critical values: $$t_{critical} \approx \pm2.145$$ Now, we calculate the p-value associated with the observed t-statistic, \(t = -1.622\). Using a calculator or statistical software, we find: $$p\text{-value} \approx 0.127$$

Draw conclusions

Since the p-value (\(0.127\)) is greater than the significance level (\(0.05\)), we fail to reject the null hypothesis. Additionally, our t-statistic (\(-1.622\)) falls within the range of the critical values (\(\pm2.145\)), further supporting that we should not reject the null hypothesis. We do not have sufficient evidence to indicate a difference in mean reaction times for the two stimuli using a significance level of \(\alpha = 0.05\).

Key concepts

Two-sample t-test

The two-sample t-test is used to compare the means of two independent groups to determine if there is a statistically significant difference between them. In this case, it's used to compare reaction times for two different stimuli. The test assumes that the data samples are drawn from normally distributed populations with unknown but equal variances.

When performing a two-sample t-test, the test statistic, denoted as \( t \), measures the difference between the sample means relative to the variation in the data. Here’s how it works:

• Calculate the means and standard deviations of the two samples.
• Use these values to find the t-statistic with the formula: \[ t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
• Compare the calculated t-statistic to a critical t-value from the t-distribution table to decide if the difference is significant.

In this exercise, the calculated t-value was approximately \(-1.622\), which was compared against critical t-values for 14 degrees of freedom, illustrating that the sample means were not significantly different at the \(\alpha = 0.05\) level.

Null and alternative hypotheses

In hypothesis testing, formulating the null and alternative hypotheses is crucial as they define what you are testing. The null hypothesis, denoted as \( H_0 \), typically suggests that there is no effect or no difference. Conversely, the alternative hypothesis \( H_a \) suggests that there is an effect or difference.

For this exercise, the hypotheses were:

• \( H_0: \mu_1 = \mu_2 \) - This states that the means of reaction times for the two stimuli are equal.
• \( H_a: \mu_1 eq \mu_2 \) - This states that the means are not equal, implying that a difference exists.

The outcome of the hypothesis test depends on the p-value or critical values obtained from the t-test. In this scenario, since the p-value is greater than the level of significance \( \alpha = 0.05 \), the null hypothesis was not rejected, indicating insufficient evidence to claim a difference.

Descriptive statistics

Descriptive statistics are used to summarize and organize the characteristics of a dataset. In this exercise, they were essential in computing the preliminary values needed for the t-test. Key descriptive statistics include the mean and standard deviation, which provide a snapshot of the data's central tendency and spread, respectively.

For Stimulus 1, the reaction times were \( \{1, 3, 2, 1, 2, 1, 3, 2\} \). The mean, or average, calculated as \( \bar{X}_1 = 1.875 \), depicts the central value. The standard deviation, \( s_1 = 0.991 \), describes the variability around the mean.
Similarly, for Stimulus 2, with reaction times \( \{4, 2, 3, 3, 1, 2, 3, 3\} \), the mean was \( \bar{X}_2 = 2.625 \) and standard deviation \( s_2 = 0.744 \).
Collectively, these statistics provided a foundation for carrying out the two-sample t-test, effectively evaluating whether or not to reject the null hypothesis based on mean comparison.