/*! 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 15 To illustrate the effects of dri... [FREE SOLUTION] | 91Ó°ÊÓ

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To illustrate the effects of driving under the influence (DUI) of alcohol, a police officer brought a DUI simulator to a local high school. Student reaction time in an emergency was measured with unimpaired vision and also while wearing a pair of special goggles to simulate the effects of alcohol on vision. For a random sample of nine teenagers, the time (in seconds) required to bring the vehicle to a stop from a speed of 60 miles per hour was recorded. $$ \begin{array}{lccccccccc} \text { Subject } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} \\ \hline \text { Normal, } X_{i} & 4.47 & 4.24 & 4.58 & 4.65 & 4.31 & 4.80 & 4.55 & 5.00 & 4.79 \\ \hline \text { Impaired, } Y_{i} & 5.77 & 5.67 & 5.51 & 5.32 & 5.83 & 5.49 & 5.23 & 5.61 & 5.63 \\ \hline \end{array} $$ (a) Whether the student had unimpaired vision or wore goggles first was randomly selected. Why is this a good idea in designing the experiment? (b) Use a \(95 \%\) confidence interval to test if there is a difference in braking time with impaired vision and normal vision where the differences are computed as "impaired minus normal." Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Short Answer

Expert verified
Random selection reduces biases. The 95% confidence interval (0.694, 1.206) indicates impaired vision significantly increases braking time.

Step by step solution

01

Understand Random Selection

Randomly selecting whether students had unimpaired vision or wore goggles first is important because it helps eliminate biases. This randomization ensures that the order in which the conditions are presented does not systematically favor one condition over the other, thus reducing potential confounding variables.
02

Calculate the Differences

To assess the impacts of impaired vision, calculate the differences between the impaired and normal braking times for each subject. Let the differences be denoted as: \[ D_i = Y_i - X_i \] The differences are: \[ 5.77-4.47, 5.67-4.24, 5.51-4.58, 5.32-4.65, 5.83-4.31, 5.49-4.80, 5.23-4.55, 5.61-5.00, 5.63-4.79 \] This simplifies to: \[ 1.30, 1.43, 0.93, 0.67, 1.52, 0.69, 0.68, 0.61, 0.84 \]
03

Find the Mean and Standard Deviation of Differences

Now, compute the mean \( \bar{D} \) and the standard deviation \( s_D \) of the differences: \[ \bar{D} = \frac{\sum_{i=1}^{n} D_i}{n} = \frac{1.30 + 1.43 + 0.93 + 0.67 + 1.52 + 0.69 + 0.68 + 0.61 + 0.84}{9} \] \[ \bar{D} = 0.95 \] The standard deviation is given by: \[ s_D = \sqrt{\frac{\sum_{i=1}^{n} (D_i - \bar{D})^2}{n-1}} = 0.332 \]
04

Compute the Confidence Interval

Use the formula for the 95% confidence interval for the mean difference: \[ \bar{D} \pm t_{\alpha/2, n-1} \left( \frac{s_D}{\sqrt{n}}\right) \] \( t_{\alpha/2, n-1} \) is the t-value corresponding to a 95% confidence level with 8 degrees of freedom (which is approximately 2.306). Calculate the standard error: \[ SE = \frac{s_D}{\sqrt{n}} = \frac{0.332}{\sqrt{9}} = 0.111 \] Hence, the confidence interval is: \[ 0.95 \pm 2.306 \cdot 0.111 \] \[ 0.95 \pm 0.256 \] \[ (0.694, 1.206) \]
05

Interpret the Results

The 95% confidence interval for the mean difference in braking time (impaired minus normal) is (0.694, 1.206). Since this interval does not contain 0, it indicates that there is a statistically significant difference in braking times, with impaired vision increasing the braking time.

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Key Concepts

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

Randomization in Experiments
In experiments, randomization is crucial for ensuring unbiased results. When it comes to studying the effects of different conditions on a variable, randomizing the order of these conditions can help eliminate biases.
For instance, in this exercise, randomly selecting whether students had unimpaired vision or wore goggles first helps ensure that the order does not systematically favor one condition over the other. This is important because factors like learning effects or fatigue could otherwise skew the results.
By randomizing the order, we reduce the influence of confounding variables, making sure the results reflect the true impact of the conditions being tested.
Mean and Standard Deviation
Mean and standard deviation are fundamental concepts in statistics that help us understand and summarize data. The mean (or average) gives us a central value, while the standard deviation tells us how spread out the values are around the mean.
In this exercise, we calculate the differences between braking times under impaired and normal vision for each subject. To compute the mean difference \(\bar{D}\), we sum up all the differences and divide by the number of observations:\
\[ \bar{D} = \frac{\sum_{i=1}^{n}{D_i}}{n} = \frac{0.95}{9} = 0.95 \]
Similarly, the standard deviation of the differences \(s_D\) is calculated to see the variability in the differences:\
\[ s_D = \sqrt{\frac{\sum_{i=1}^{n} (D_i - \bar{D})^2}{n-1}} = 0.332 \]
Understanding both the mean and standard deviation helps us not only to gauge the central tendency of the differences but also how much these differences vary among subjects.
Statistical Significance
Statistical significance helps us determine if the observed results are likely due to chance or if they reflect a true difference between groups. In this context, we use a 95% confidence interval to assess the mean difference in braking times between impaired and normal vision.
The confidence interval is computed using the formula:
\[ \bar{D} \pm t_{\alpha/2, n-1} \left( \frac{s_D}{\sqrt{n}} \right) \]
For a 95% confidence level with 8 degrees of freedom, the critical t-value is approximately 2.306. Therefore, the confidence interval is:\
\[ 0.95 \pm 2.306 \cdot 0.111 \]
Resulting in:\
\[ (0.694, 1.206) \]
Since this interval does not include 0, it indicates a statistically significant difference in braking times, implying that impaired vision likely increases the time to stop. This means the difference is unlikely to be due to random chance, but instead, it genuinely reflects the impact of impaired vision.

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Most popular questions from this chapter

The drug Prevnar is a vaccine meant to prevent certain types of bacterial meningitis, typically administered to infants around 2 months. In randomized, double-blind clinical trials of Prevnar, infants were randomly divided into two groups. Subjects in group 1 received Prevnar, while subjects in group 2 received a control vaccine. After the first dose, 107 of 710 subjects in the experimental group (group 1) experienced fever as a side effect. After the first dose, 67 of 611 of the subjects in the control group (group 2) experienced fever as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced fever as a side effect than subjects in group 2 at the a = 0.05 level of significance?

For each study, explain which statistical procedure (estimating a single proportion; estimating a single mean; hypothesis test for a single proportion; hypothesis test for a single mean; hypothesis test or estimation of two proportions, hypothesis test or estimation of two means, dependent or independent) would most likely be used for the research objective given. Assume all model requirements for conducting the appropriate procedure have been satisfied. Is the mean IQ of the students in Professor Dang's statistics class higher than that of the general population, \(100 ?\)

MythBusters In a MythBusters episode, the question was asked,"Which is better? A four-way stop or a roundabout?" "Better" was determined based on determining the number of vehicles that travel through the four-way stop over a 5 -minute interval of time. Suppose the folks at MythBusters conducted this experiment 15 times for each intersection design. (a) What is the variable of interest in this study? Is it qualitative or quantitative? (b) Explain why the data might be analyzed by comparing two independent means. Include in this explanation what the null and alternative hypotheses would be and what each mean represents (c) A potential improvement on the experimental design might be to identify 15 groups of different drivers and ask each group to drive through each intersection design for the 5 - minute time interval. Explain why this is a better design and explain the role randomization would play. What would be the null and alternative hypotheses here and how would the mean be computed?

\( \begin{array}{ccccccccc} \text { Observation } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\ \hline X_{i} & 19.4 & 18.3 & 22.1 & 20.7 & 19.2 & 11.8 & 20.1 & 18.6 \\ \hline Y_{i} & 19.8 & 16.8 & 21.1 & 22.0 & 21.5 & 18.7 & 15.0 & 23.9 \end{array} \) (a) Determine \(d_{i}=X_{i}-Y_{i}\) for each pair of data. (b) Compute \(\bar{d}\) and \(s_{d}\) (c) Test if \(\mu_{d} \neq 0\) at the \(\alpha=0.01\) level of significance. (d) Compute a \(99 \%\) confidence interval about the population mean difference \(\mu_{d}\).

Perform the appropriate hypothesis test. If \(n_{1}=61, s_{1}=18.3, n_{2}=57,\) and \(s_{2}=13.5,\) test whether the population standard deviations differ at the \(\alpha=0.05\) level of significance.

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