/*! 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 8 In an experiment conducted onlin... [FREE SOLUTION] | 91Ó°ÊÓ

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In an experiment conducted online at the University of Mississippi, study participants are asked to react to a stimulus. In one experiment, the participant must press a key on seeing a blue screen and reaction time (in seconds) to press the key is measured. The same person is then asked to press a key on seeing a red screen, again with reaction time measured. The results for six randomly sampled study participants are as follows: $$ \begin{array}{lcccccc} \text { Participant } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} \\ \hline \text { Blue } & 0.582 & 0.481 & 0.841 & 0.267 & 0.685 & 0.450 \\ \hline \text { Red } & 0.408 & 0.407 & 0.542 & 0.402 & 0.456 & 0.533 \\ \hline \end{array} $$ (a) Why are these matched-pairs data? (b) In this study, the color that the participant was first asked to react to was randomly selected. Why is this a good idea in this experiment? (c) Is the reaction time to the blue stimulus different from the reaction time to the red stimulus at the \(\alpha=0.01\) level of significance? Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. (d) Construct a \(99 \%\) confidence interval about the population mean difference. Interpret your results. (e) Draw a boxplot of the differenced data. Does this visual evidence support the results obtained in part (c)?

Short Answer

Expert verified
These are matched pairs because each participant responds to two stimuli. Random color order eliminates bias. Hypothesis testing at \(\alpha=0.01\) shows no significant difference. The 99% CI also supports this, as does the boxplot visual.

Step by step solution

01

Identify why these are matched-pairs data

The data are matched-pairs because the same participant is measured under two different conditions: seeing a blue screen and reacting, and seeing a red screen and reacting. Each participant serves as their own control, which helps to account for individual variability in reaction times.
02

Explain random selection of color order

Randomly selecting the color order helps to eliminate order effects and other potential biases. It ensures that the order in which the tasks are performed does not systematically affect the results, providing a more reliable basis for comparing the reaction times.
03

Perform a hypothesis test for reaction time differences at \alpha=0.01

To test if the reaction times to the blue stimulus are different from the red stimulus, we perform a paired t-test.\(\alpha = 0.01\)Null hypothesis: \(H_0: \mu_{d} = 0\), where \(\mu_{d}\) is the mean difference in reaction times.Alternative hypothesis: \(H_a: \mu_{d} eq 0\)Calculate the differences for each participant and then find the mean and standard deviation of these differences. Use the t-test formula for paired samples to determine the t-statistic and compare it to the critical t-value for \(5\) degrees of freedom (6-1 participants).
04

Calculate t-statistic and p-value

The differences in reaction times for each participant are: $$d = [0.582 - 0.408, 0.481 - 0.407, 0.841 - 0.542, 0.267 - 0.402, 0.685 - 0.456, 0.450 - 0.533] = [0.174, 0.074, 0.299, -0.135, 0.229, -0.083]$$Calculate the mean and standard deviation of these differences:\(\bar{d} = \frac{0.174 + 0.074 + 0.299 - 0.135 + 0.229 - 0.083}{6} = 0.093\)The standard deviation \(s_d\)Calculate the t-statistic using these values and determine the p-value. If the p-value is less than \(0.01\), reject the null hypothesis.
05

Construct a 99% confidence interval about the population mean difference

To construct the 99% confidence interval for the mean difference, use the formula:\[\text{CI} = \bar{d} \pm t_{\alpha/2, df} \left( \frac{s_d}{\sqrt{n}} \right)\]Substitute the mean, standard deviation, t-value for 5 degrees of freedom, and sample size to get the interval.
06

Draw a boxplot of the differences

A boxplot of the differenced data (differences between reaction times). Check if the boxplot suggests any noticeable deviations. Compare visual evidence with results from step 4.

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Key Concepts

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

Experimental Design
In this experiment, participants' reaction times to two different stimuli (blue and red screens) are measured. This setup illustrates a classic example of an experimental design. Experimental design allows researchers to control variables and test their effects on outcomes. This is essential to ensure that the results of the experiment are valid and reliable.
In this case, we used a within-subjects design where the same participants are tested under both conditions. This helps to mitigate any natural variations between participants that might affect reaction times.
Each participant acts as their own control, making it easier to detect differences caused by the stimuli rather than other external factors. This design is efficient and can lead to more accurate conclusions.
Key points to remember about experimental design:
  • Control as many variables as possible
  • Mitigate external factors
  • Use participants as their own control if possible
Random Selection
The order in which participants were asked to respond to the blue or red screen was randomly selected. Random selection is crucial in experiments because it ensures that every participant has an equal chance of experiencing each condition first. This helps to eliminate potential biases and ensures that the order doesn't systematically affect the results.
For instance, if participants always saw the blue screen first, they might become faster due to learning or fatigue, thus confounding the true effect of the color. Random selection counters this by balancing out such effects across both conditions.
Reasons why random selection is important:
  • Reduces biases
  • Ensures equal distribution of order effects
  • Improves the reliability of the results
Hypothesis Testing
In the context of this experiment, hypothesis testing helps determine if there is a statistically significant difference in reaction times to the blue versus the red screen. At the \( \alpha = 0.01 \) level of significance, we perform a matched-pairs t-test.

Here's how the hypotheses are set up for this test:
  • Null Hypothesis (H0): \( \mu_d = 0 \) - This means there is no difference in reaction times.
  • Alternative Hypothesis (Ha): \( \mu_d eq 0 \) - This means there is a difference in reaction times.

We calculate the mean difference in reaction times, the standard deviation, and then the t-statistic using these values. This t-statistic is then compared to a critical value from the t-distribution table with 5 degrees of freedom (6 participants - 1). If the p-value is less than 0.01, we reject the null hypothesis.
Hypothesis testing allows us to make data-driven decisions about our experimental observations.
Confidence Intervals
A confidence interval provides a range within which we can be certain, to a certain level (e.g., 99%), that the true mean difference between reaction times lies. In this study, we construct a 99% confidence interval for the mean difference in reaction times between the blue and red stimuli.

The confidence interval is calculated using the formula:
\[ CI = \bar{d} \pm t_{\alpha/2, df} \left( \frac{s_d}{\sqrt{n}} \right) \]
where:
  • \( \bar{d} \) is the mean difference
  • \( s_d \) is the standard deviation of the differences
  • \( t_{\alpha/2, df} \) is the critical t-value
  • n is the number of participants

This interval helps researchers understand not just if, but how much the reaction times differ on average between the blue and red stimuli. It's fundamental to hypothesis testing as it provides insight into the magnitude and direction of an observed effect.
Paired Samples
Paired samples occur when the same subjects are measured under different conditions or at different times. In this experiment, each participant's reaction time to the blue screen is paired with their reaction time to the red screen. This helps to control for individual differences that might affect reaction times.
Paired samples reduce variability that might come from different participants having naturally different reaction speeds. By using pairs, each participant effectively acts as their own control, which allows for a more accurate comparison between conditions.
Key advantages of using paired samples:
  • Controls for between-subject variability
  • Improves the power of statistical tests
  • Allows for more precise comparisons

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Most popular questions from this chapter

Assume that the populations are normally distributed. (a) Test whether \(\mu_{1} \neq \mu_{2}\) at the \(\alpha=0.05\) level of significance for the given sample data. (b) Construct a \(95 \%\) confidence interval about \(\mu_{1}-\mu_{2}\). $$ \begin{array}{ccc} & \text { Sample } \mathbf{1} & \text { Sample } 2 \\ \hline n & 20 & 20 \\ \hline \bar{x} & 111 & 104 \\ \hline s & 8.6 & 9.2 \\ \hline \end{array} $$

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The manufacturer of hardness testing equipment uses steel-ball indenters to penetrate a metal that is being tested. However, the manufacturer thinks it would be better to use a diamond indenter so that all types of metal can be tested. Because of differences between the two types of indenters, it is suspected that the two methods will produce different hardness readings. The metal specimens to be tested are large enough so that two indentions can be made. Therefore, the manufacturer uses both indenters on each specimen and compares the hardness readings. Construct a \(95 \%\) confidence interval to judge whether the two indenters result in different measurements. $$ \begin{array}{lccccccccc} \text { Specimen } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} \\ \hline \text { Steel ball } & 50 & 57 & 61 & 71 & 68 & 54 & 65 & 51 & 53 \\ \hline \text { Diamond } & 52 & 56 & 61 & 74 & 69 & 55 & 68 & 51 & 56 \end{array} $$ Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Kids and Leisure Young children require a lot of time and this time commitment cuts into a parent's leisure time. A sociologist wanted to estimate the difference in the amount of daily leisure time (in hours) of adults who do not have children under the age of 18 years and adults who have children under the age of 18 years. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.62 hours, with a standard deviation of 2.43 hours. A random sample of 40 adults with children under the age of 18 years results in a mean daily leisure time of 4.10 hours, with a standard deviation of 1.82 hours. Construct and interpret a \(90 \%\) confidence interval for the mean difference in leisure time between adults with no children and adults with children. Source: American Time Use Survey

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