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Chapter 13: Problem 78

The paper "The Effect of Multitasking on the Grade Performance of Business Students" (Research in Higher Education Journal [2010]: 1-10) describes an experiment in which 62 undergraduate business students were randomly assigned to one of two experimental groups. Students in one group were asked to listen to a lecture but were told that they were permitted to use cell phones to send text messages during the lecture. Students in the second group listened to the same lecture but were not permitted to send text messages. Afterwards, students in both groups took a quiz on material covered in the lecture. Data from this experiment are summarized in the accompanying table.

Short Answer

Expert verified
In the given experiment, an independent 2-sample t-test can be performed to determine if there is a significant difference between the means of two groups' quiz scores. First, we state the null hypothesis (H0) as no significant difference between the means and the alternative hypothesis (Ha) as a significant difference. Then, we calculate the t-statistic using the formula: \( t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \). Next, we determine the degrees of freedom (df) using the formula: \( df = n_1 + n_2 - 2 \). Finally, we compare the calculated t-statistic to the critical value. If the t-statistic is greater than the critical value (in absolute terms), we reject the null hypothesis, suggesting that multitasking has a significant effect on a student's performance in a quiz. If not, we fail to reject the null hypothesis, suggesting that multitasking does not have a significant effect.

Step by step solution

01

Null Hypothesis (H0) and Alternative Hypothesis (Ha)

The first step in doing the independent 2-sample t-test is to state the hypotheses we want to test. In this case, we want to check if there is a significant difference between the means of the two groups' quiz scores. H0 (Null Hypothesis): There is no significant difference between the means of the two groups' quiz scores. \[ \mu_1 = \mu_2 \] Ha (Alternative Hypothesis): There is a significant difference between the means of the two groups' quiz scores. \[ \mu_1 \neq \mu_2 \] where \(\mu_1\) is the mean of quiz scores for group 1 (students allowed to use cell phones), and \(\mu_2\) is the mean of quiz scores for group 2 (students not allowed to use cell phones). ##Step 2: Calculate Test Statistic##
02

Test Statistic Calculation

To perform the independent 2-sample t-test, we need to compute the t-statistic using the means, standard deviations, and sample sizes for each group. Unfortunately, the accompanying table with the data is not provided, but we'll assume we have access to the means, standard deviations, and sample sizes. Given the following information: - \( \bar{x}_1 \) = Mean quiz score of group 1 (students allowed to use cell phones) - \( s_1 \) = Standard deviation of group 1 quiz scores - \( n_1 \) = Number of students in group 1 - \( \bar{x}_2 \) = Mean quiz score of group 2 (students not allowed to use cell phones) - \( s_2 \) = Standard deviation of group 2 quiz scores - \( n_2 \) = Number of students in group 2 The t-statistic can be calculated using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] ##Step 3: Determine the Degrees of Freedom##
03

Degrees of Freedom Calculation

To determine the critical value for our t-test, we need to calculate the degrees of freedom, which is given by: \[ df = n_1 + n_2 - 2 \] Once we have the degrees of freedom, we can look up the critical value in a t-distribution table corresponding to the significance level we choose (typically 0.05 or 0.01). ##Step 4: Compare Test Statistic to Critical Value##
04

Decision Rule

After calculating the t-statistic and finding the critical value, we can compare them to determine if we should reject or fail to reject the null hypothesis. If the t-statistic is greater than the critical value (in absolute terms), we reject the null hypothesis and accept the alternative hypothesis. Otherwise, we fail to reject the null hypothesis. In conclusion, if we reject the null hypothesis, it means that there is a significant difference between the means of the two groups' quiz scores, suggesting that multitasking (using a cell phone to send text messages during a lecture) has a significant effect on a student's performance in a quiz. If we fail to reject the null hypothesis, it means that there is no significant difference between the means of the two groups' quiz scores, suggesting that multitasking does not have a significant effect on a student's performance in a quiz.

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Key Concepts

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

Null Hypothesis
In statistical tests, the null hypothesis (\( H_{0} \)) is our initial assumption that there is no effect or no difference. It's what we 'hope' to disprove or reject with our statistical evidence. For an independent 2-sample t-test like in this study, the null hypothesis suggests that there is no significant difference between the average quiz scores of the two groups鈥攖hose allowed to text during the lecture and those who were not. Mathematically, it's expressed as:
  • \[ \mu_1 = \mu_2 \]
If the null hypothesis holds true, any difference observed in the sample means is attributed to random sampling fluctuations, not any real impact of texting on quiz scores.
Thus, establishing a null hypothesis allows us to have a base against which statistical tests are carried out. It's an essential starting point in hypothesis testing because it allows for computations to judge whether the observed data contradicts this baseline assumption.
Alternative Hypothesis
The alternative hypothesis (\( H_{a} \)) is the claim we are trying to find evidence for in our test. In the context of the research experiment, the alternative hypothesis posits that there actually is a significant difference between the average quiz scores of the two student groups. Presented mathematically, it asserts that:
  • \[ \mu_1 eq \mu_2 \]
This hypothesis suggests that texting during a lecture may affect student performance. With the alternative hypothesis, the burden of proof lies in showing enough statistical evidence from the quiz scores to suggest that \( \mu_1 \) and \( \mu_2 \) truly differ.
If we gather enough evidence through our t-test to reject the null hypothesis, the alternative hypothesis is favored, suggesting a real difference influenced by the experimental condition.
Degrees of Freedom
Degrees of freedom (\( df \)) are an important concept in statistics, as they help determine the distribution of the test statistic. For the independent 2-sample t-test, the degrees of freedom can be calculated using the following formula:
  • \[ df = n_1 + n_2 - 2 \]
Where \( n_1 \) and \( n_2 \) are the sample sizes of the two groups.
Higher degrees of freedom indicate more broadly a wider spread of possible test statistic values, which influences the amount of data we rely upon to detect any genuine effect or difference.
The degrees of freedom are used alongside the test statistic to determine the critical value from a t-distribution table, which further aids in decision-making about the null hypothesis's validity.
Test Statistic
The test statistic is a tool that summarizes the data into a single number, which you can use to perform hypothesis testing. For an independent 2-sample t-test, the test statistic (\( t \)) measures how far our sample differences are from the null hypothesis. It's calculated using:
  • \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
Where:
  • \( \bar{x}_1 \) and \( \bar{x}_2 \) are the means of the quiz scores for groups 1 and 2 respectively.
  • \( s_1^2 \) and \( s_2^2 \) are the sample variances.
  • \( n_1 \) and \( n_2 \) are their respective sample sizes.
A larger absolute value of the t-statistic indicates a greater truth of the alternative hypothesis and increases the likelihood of rejecting the null hypothesis.
Evaluating the magnitude and direction of this statistic, we make an evidence-backed decision concerning the presence or absence of a true effect.
T-Distribution
A t-distribution is a type of probability distribution that accounts for sample size and variability when determining critical values for the hypothesis test. Unlike the normal distribution, the t-distribution has thicker tails, which can affect hypothesis interpretation, especially with smaller sample sizes.
As sample sizes increase, the t-distribution approximates closely to the normal distribution.
When conducting an independent 2-sample t-test, the t-distribution helps compare our calculated t-statistic with the critical values to make conclusions regarding the null hypothesis.
This comparison will determine if the sample evidence is strong enough to declare a significant difference between the groups, given the degrees of freedom calculated beforehand. Understanding the t-distribution helps facilitate proper application and interpretation of hypothesis testing outcomes.

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Most popular questions from this chapter

The paper "Effects of Caffeine on Repeated Sprint Ability, Reactive Agility Time, Sleep and Next Day Performance" (Journal of Sports Medicine and Physical Fitness [2010]: \(455-464)\) describes an experiment in which male athlete volunteers who were considered low caffeine consumers were assigned at random to one of two cxpcrimental groups. Those assigned to the caffeine group drank a beverage which contained caffeine one hour before an excrcise session. Those in the no-caffeine group drank a beverage that did not contain caffeine. During the exercise session, each participant performed a test that measured reactive agility. The researchers reported that there was \(n 0\) significant difference in mean reactive agility for the two experimental groups. In the context of this experiment, explain what it means to say that there is no significant difference in the group means.

The report referenced in the previous exercise also gave data for representative samples of 250 adults in Canada and 250 adults in England. The sample mean amount of sleep on a work night was 423 minutes for the Canada sample and 409 minutes for the England sample. Suppose that the sample standard deviations were 35 minutes for the Canada sample and 42 minutes for the England sample. a. Construct and interpret a \(95 \%\) confidence interval estimate of the difference in the mean amount of sleep on a work night for adults in Canada and adults in England. b. Based on the confidence interval from Part (a), would you conclude that there is evidence of a diflerence in the mean amount of sleep on a work night for the two countries? Explain why or why not.

A new set of cognitive training modules called "ONTRAC" was developed to help children with attention deficit/hyperactivity disorder (ADHD) to improve focus and to more easily dismiss distractions ("Training sensory signal-to-noise resolution in children with ADHD in a global mental health setting," Translational Psychiatry, April \(12,2016,\) http: \(/ /\) www.nature.com/tp/journal/v \(6 / \mathrm{n} 4\) /full/tp201645a.html, retrieved May 23,2017 ). Eighteen children with ADHD were randomly assigned to one of two treatment groups. One group of 11 children received the ONTRAC treatment and another group of 7 children received a control treatment. Values for one- year improvement in ADHD Severity Score consistent with graphs and summary statistics in the research article appear in the following table: a. Explain why you should be wary of using the two-sample \(t\) methods to analyze the data from this study. b. Do these data support the claim that the mean one-year improvement in ADHD Severity Score for the ONTRAC treatment is different from the mean one-year improvement in ADHD Severity Score for the control treatment? Use a randomization test with significance level 0.05 to answer this question. You can use make use of the Shiny apps in the collection at statistics.cengage.com/Peck2e/Apps.html.

The humorous puper "Will Humans Swim Faster or Slower in Syrup?" (American Institute of Chemical Engineers Journal [2004]: \(2646-2647\) ) investigated the fluid mechanics of swimming. Twenty swimmers each swam a specified distance in a water-filled pool and in a pool in which the water was thickened with food grade guar gum to create a syruplike consistency. Velocity, in meters per second, was recorded. Values estimated from a graph in the paper are given. The authors of the paper concluded that swimming in guar syrup does not change mean swimming speed. Are the given data consistent with this conclusion? Carry out a hypothesis test using a 0.01 significance level.

The paper referenced in the previous exercise also had the 50 taxi drivers drive in the simulator while sending and receiving text messages. The mean of the 50 sample differences (no distraction - reading text messages) was 1.3 meters and the standard deviation of the sample differences was 1.54 meters. The authors concluded that there was evidence to support the claim that the mean following distance for Greek taxi drivers is greater when there are no distractions that when the driver is texting. Do you agree with this conclusion? Carry out a hypothesis test to support your answer. You can assume that this sample of 50 drivers is representative of Greek taxi drivers.
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