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Chapter 10: Problem 95

Refer to Exercise 10.94. 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: $$ \begin{array}{ccc} \text { Person } & \text { Stimulus } 1 & \text { Stimulus 2 } \\ \hline 1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 2 & 1 \\ 5 & 1 & 2 \\ 6 & 2 & 3 \\ 7 & 3 & 3 \\ 8 & 2 & 3 \end{array} $$ Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)

Short Answer

Expert verified
The decision based on the t-statistic and 伪=0.05 is to reject the null hypothesis that there is no difference in mean reaction times for the two stimuli. Our calculated t-statistic is -3.184, which falls in the rejection region (t < -2.365) for a two-tailed test. Therefore, there is significant evidence at 伪 = 0.05 to indicate a difference in mean reaction times for the two stimuli.

Step by step solution

01

Calculate the differences in reaction times for each person

We need to calculate the difference in reaction times for each person between Stimulus 1 (S1) and Stimulus 2 (S2). Create a new column called "Difference" and compute the differences by subtracting the reaction times of Stimulus 2 from Stimulus 1: $$ \begin{array}{cccc} \text{Person} & \text{Stimulus 1} & \text{Stimulus 2} & \text{Difference} \\ \hline 1 & 3 & 4 & -1 \\ 2 & 1 & 2 & -1 \\ 3 & 1 & 3 & -2 \\ 4 & 2 & 1 & 1 \\ 5 & 1 & 2 & -1 \\ 6 & 2 & 3 & -1 \\ 7 & 3 & 3 & 0 \\ 8 & 2 & 3 & -1 \\ \end{array} $$
02

Calculate the mean and standard deviation of the differences

For our sample, we need to calculate the mean (\(\bar{d}\)) and standard deviation (sd) of the differences: Mean (\(\bar{d}\)) = \((-1-1-2+1-1-1+0-1)/8 = -9/8 = -1.125\) Standard deviation (sd) = \(s_d= \sqrt{\frac{(-1-(-1.125))^2+(-1-(-1.125))^2+(-2-(-1.125))^2+(1-(-1.125))^2+(-1-(-1.125))^2+(-1-(-1.125))^2+(0-(-1.125))^2+(-1-(-1.125))^2}{8-1}} = 1.125\)
03

Calculate the t-statistic

Use the formula for the t-statistic: \(t = \frac{\bar{d} - \mu_d}{s_d/\sqrt{n}}\), where \(\bar{d}\) is the mean of the differences, \(\mu_d\) is the population mean difference (0 since we are testing for a difference), \(s_d\) is the standard deviation of differences, and n is the number of samples (8). \(t = \frac{-1.125 - 0}{1.125/\sqrt{8}} = -3.184\)
04

Find the critical t-value and determine the rejection region

The critical t-value can be found using a t-distribution table or a calculator. Our degrees of freedom (df) are calculated as \(n-1 = 8-1 = 7\). With a significance level of \(\alpha = 0.05\), the critical t-value for a two-tailed test is \(t_{critical} = \pm 2.365\). The rejection region consists of two areas: t < -2.365 and t > 2.365.
05

Make a decision based on the t-statistic and rejection region

Our calculated t-statistic is -3.184. Since this falls in the rejection region (t < -2.365), we reject the null hypothesis that there is no difference in mean reaction times for the two stimuli. There is significant evidence at 伪 = 0.05 to indicate a difference in mean reaction times for the two stimuli.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a cornerstone of statistics that helps us make decisions based on data. It allows us to test if a certain claim about a population is true. In this exercise, we test if there is a significant difference in reaction times between two stimuli.

Here's how hypothesis testing works:
  • **Null Hypothesis (H鈧�):** This assumes no effect or no difference. In our case, it suggests the mean reaction time for both stimuli is the same.
  • **Alternative Hypothesis (H鈧�):** This proposes an effect or a difference. Here, it indicates a difference in the mean reaction times for the stimuli.
  • **Significance Level (伪):** Usually set to 0.05 or 5%, it determines the threshold for rejecting the null hypothesis. A smaller 伪 indicates more stringent testing.
  • **Decision Rule:** Based on calculated statistics, decide whether to reject or not reject the null hypothesis.
In our exercise, with 伪 = 0.05, we evaluated statistical evidence to decide if we should accept or reject the hypothesis that there is no difference in reaction times.
Paired Sample t-Test
A paired sample t-test is crucial when comparing two related samples. It helps understand if the mean difference between them is significant. This is particularly useful when each participant is tested under both conditions, like in our reaction time analysis.

**Why Use a Paired Sample t-Test?**
  • **Control for Variability:** Since each person serves as their own control, the variability due to different individuals is minimized.
  • **Increased Sensitivity:** It鈥檚 more sensitive because it's easier to detect differences when variance is controlled.
When performing this test, you calculate:
  • **Differences:** The difference in performance from one condition to another for each person.
  • **Mean of Differences (\(\bar{d}\)):** The average of these differences.
  • **Standard Deviation of Differences:** Measures variability among the differences.
In the exercise, after calculating the mean and standard deviation of the differences, we used them to compute the t-statistic. This helped us decide whether the observed differences in reaction times were statistically significant.
Reaction Time Analysis
Reaction time analysis involves studying how quickly individuals respond to stimuli, which can reveal cognitive processes. In experiments like the one described, measuring reaction times helps understand mental processing speed and attention.

In this exercise, reaction times were measured for two different stimuli with each participant exposed to both. Analyzing reaction times can help determine:
  • **Effectiveness of Stimuli:** Identifying which stimulus generates quicker responses.
  • **Consistency Among Individuals:** Observing variations in response times across different participants.
Reaction time differences were then used to discern patterns or changes in performance from one condition to another.

Through statistical tools like a paired sample t-test, we assess whether the difference in reaction times is just random or if it's meaningful. In this case, the analysis indicated a significant difference, showing that the chosen stimuli affect reaction speed differently.

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Most popular questions from this chapter

A producer of machine parts claimed that the diameters of the connector rods produced by his plant had a variance of at most .03 inch \(^{2}\). A random sample of 15 connector rods from his plant produced a sample mean and variance of .55 inch and .053 inch \(^{2}\), respectively. a. Is there sufficient evidence to reject his claim at the \(\alpha=.05\) level of significance? b. Find a \(95 \%\) confidence interval for the variance of the rod diameters.

The earth's temperature can be measured using either ground-based sensors or infrared-sensing devices mounted in aircraft or space satellites. Ground-based sensoring is very accurate but tedious, while infrared-sensoring appears to introduce a bias into the temperature readings- that is, the average temperature reading may not be equal to the average obtained by ground-based sensoring. To determine the bias, readings were obtained at five different locations using both ground- and air-based temperature sensors. The readings (in degrees Celsius) are listed here: $$ \begin{array}{ccc} \text { Location } & \text { Ground } & \text { Air } \\ \hline 1 & 46.9 & 47.3 \\ 2 & 45.4 & 48.1 \\ 3 & 36.3 & 37.9 \\ 4 & 31.0 & 32.7 \\ 5 & 24.7 & 26.2 \end{array} $$ a. Do the data present sufficient evidence to indicate a bias in the air-based temperature readings? Explain. b. Estimate the difference in mean temperatures between ground- and air-based sensors using a \(95 \%\) confidence interval. c. How many paired observations are required to estimate the difference between mean temperatures for ground- versus air-based sensors correct to within \(.2^{\circ} \mathrm{C}\), with probability approximately equal to \(.95 ?\)

To compare the demand for two different entrees, the manager of a cafeteria recorded the number of purchases of each entree on seven consecutive days. The data are shown in the table. Do the data provide sufficient evidence to indicate a greater mean demand for one of the entrees? Use the Excel printout. $$ \begin{array}{lcc} \text { Day } & \mathrm{A} & \mathrm{B} \\ \hline \text { Monday } & 420 & 391 \\ \text { Tuesday } & 374 & 343 \\ \text { Wednesday } & 434 & 469 \\ \text { Thursday } & 395 & 412 \\ \text { Friday } & 637 & 538 \\ \text { Saturday } & 594 & 521 \\ \text { Sunday } & 679 & 625 \end{array} $$

An experiment is conducted to compare two new automobile designs. Twenty people are randomly selected, and each person is asked to rate each design on a scale of 1 (poor) to 10 (excellent). The resulting ratings will be used to test the null hypothesis that the mean level of approval is the same for both designs against the alternative hypothesis that one of the automobile designs is preferred. Do these data satisfy the assumptions required for the Student's \(t\) -test of Section 10.4 ? Explain.

Lobsters In a study of the infestation of the Thenus orientalis lobster by two types of barnacles, Octolasmis tridens and \(O\). lowei, the carapace lengths (in millimeters) of 10 randomly selected lobsters caught in the seas near Singapore are measured: \(^{2}\) $$ \begin{array}{lllllllll} 78 & 66 & 65 & 63 & 60 & 60 & 58 & 56 & 52 & 50 \end{array} $$ Find a \(95 \%\) confidence interval for the mean carapace length of the \(T\). orientalis lobsters.
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