/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Refer to Exercise 10.91 . Suppos... [FREE SOLUTION] | 91Ó°ÊÓ

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Refer to Exercise 10.91 . Suppose that the word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows: \begin{tabular}{ccc} Person & Stimulus 1 & Stimulus 2 \\ \hline 1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 2 & 1 \\ 5 & 1 & 2 \\ 6 & 2 & 3 \\ 7 & 3 & 3 \\ 8 & 2 & 3 \end{tabular} Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)

Short Answer

Expert verified
Answer: Yes, there is a significant difference in mean reaction times for the two stimuli, as we reject the null hypothesis with a significance level of 0.05.

Step by step solution

01

Calculate the differences between the reaction times

For each person, subtract the reaction time of stimulus 1 from the reaction time of stimulus 2: 1: 4 - 3 = 1 2: 2 - 1 = 1 3: 3 - 1 = 2 4: 1 - 2 = -1 5: 2 - 1 = 1 6: 3 - 2 = 1 7: 3 - 3 = 0 8: 3 - 2 = 1 Resulting in: 1, 1, 2, -1, 1, 1, 0, 1
02

Calculate the mean and standard deviation of the differences

The mean of the differences \((\bar{d})\) is given by: \(\bar{d} = \frac{\sum d_i}{n}\) where \(d_i\) represents the difference for each person, and \(n\) is the number of participants (8 in this case): \(\bar{d} = \frac{1+1+2-1+1+1+0+1}{8} = \frac{6}{8} = 0.75\) The standard deviation of the differences \((s_d)\) is given by: \(s_d = \sqrt{\frac{\sum(d_i - \bar{d})^2}{n-1}}\) \(s_d = \sqrt{\frac{(1-0.75)^2+(1-0.75)^2+(2-0.75)^2+(-1-0.75)^2+(1-0.75)^2+(1-0.75)^2+(0-0.75)^2+(1-0.75)^2}{8-1}}\) \(s_d = \sqrt{\frac{0.0625+0.0625+1.5625+3.0625+0.0625+0.0625+0.5625+0.0625}{7}} = \sqrt{\frac{6.5}{7}} \approx 0.86\)
03

Calculate the t-test statistic and degrees of freedom

The t-test statistic is calculated using the following formula: \(t = \frac{\bar{d}}{\frac{s_d}{\sqrt{n}}}\) \(t = \frac{0.75}{\frac{0.86}{\sqrt{8}}} \approx 2.48\) Degrees of freedom (df) is given by: \(df = n - 1 = 8 - 1 = 7\)
04

Determine the critical t-value and make a conclusion

We need to find the critical t-value for a two-tailed test, with \(\alpha=0.05\) and \(df=7\). Using a t-table or calculator, we find that the critical t-value is around \(\pm 2.365\). Since our calculated t-value (\(2.48\)) is greater than the positive critical t-value (\(2.365\)), we can reject the null hypothesis. Thus, the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli, and the significance level of \(0.05\) is met.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Statistical Significance
When we talk about statistical significance in the realm of experiments and data analysis, we're discussing the likelihood that the results we observed are due to something other than mere chance. In the case of our reaction time experiment, we use statistical tests to determine if the observed differences in reaction times between the two stimuli are significant, or if they could have occurred by random variation alone.

To assess this, we compare a calculated statistic—in this case, the t-statistic—to a critical value that corresponds to a chosen significance level. A commonly used significance level is \( \alpha = 0.05 \), which implies a 5% risk of concluding that a difference exists when there's actually none. If our test statistic exceeds the critical value at this significance level, we declare the findings statistically significant and reject the null hypothesis—a baseline assumption that there is no effect or difference. The smaller the value of \( \alpha \), the stronger the evidence must be to declare significance. Through precise calculation and comparison, statistical significance tells us when our results are likely to be meaningful and not just the product of statistical noise.
Reaction Time Experiment
A reaction time experiment is designed to measure how quickly individuals respond to a stimulus. In our example, we presented subjects with two different stimuli and recorded the time it took for them to react. The purpose of such experiments could range from psychological research to ergonomics assessment.

A key consideration in reaction time experiments is the experimental design. By using each participant as their own control and randomizing the order of the stimuli, we can reduce the impact of external variables and internal biases. It's crucial to ensure that the conditions are as consistent as possible across trials to obtain reliable data. Also, increasing the sample size could improve the validity of the findings, as it allows for a more accurate estimation of the population parameters. Hence, in an in-depth experimental analysis, addressing the subtleties of design and execution is as important as the statistical analysis in drawing meaningful conclusions.
Hypothesis Testing
Hypothesis testing is a formal procedure used by statisticians to accept or reject hypotheses based on sample data. In the context of our exercise, the hypothesis testing framework enabled us to decide whether the reaction times for two different stimuli were significantly different.

The process starts by stating the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). For the reaction time experiment, \( H_0 \) might claim that there's no difference in reaction times between the stimuli, while \( H_a \) suggests that there is a difference. After performing the experiment and collecting the data, we went through calculated statistical measures like mean differences, standard deviation, t-statistics, and degrees of freedom, leading us to a point where we can reject or fail to reject the null hypothesis based on the comparison with the critical t-value.

Bridging the gap between theory and practical application, hypothesis testing allows for making informed decisions grounded in statistical theory. It is a fundamental process across various types of research, ensuring that we make data-driven conclusions rather than arbitrary judgments.

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Most popular questions from this chapter

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