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Three males and three females are given 5 minutes to memorize a list of 25 words, and then asked to recall as many of them as possible. The three males recalled \(10,12,\) and 14 of the words, for an average of 12 words; the three females recalled \(11,14,\) and 17 of the words, for an average of 14 words. a. In comparing males' and females' ability for memorization, what would be reasonable null and alternative hypotheses to test in this situation? b. The difference in means for the two groups in this sample was two words. Using that as the "test statistic," explain how you could carry out a permutation test in this situation. c. Give one example of a permutation from the method you explained in part (b), and what the "test statistic" would be for that permutation.

Short Answer

Expert verified
Test the null hypothesis that there is no difference in recalled means by permuting and comparing test statistics.

Step by step solution

01

Formulate Hypotheses

Identify the null hypothesis (H_0) as there being no difference in the average number of words recalled between males and females, formally: \( H_0 : \mu_m = \mu_f \). The alternative hypothesis (\(H_1\)) is that there is a difference: \( H_1 : \mu_m eq \mu_f \).
02

Define Permutation Test Plan

To carry out a permutation test, combine all the recalled words data: 10, 12, 14, 11, 14, 17. Randomly divide these six observations into two groups of three and find the average recall for each group.
03

Calculate Test Statistic

For each permutation, compute the test statistic, which is the absolute difference in the means of the two groups. Compare this result to the observed test statistic of 2 (from males and females' average difference).
04

Example of a Permutation

Take one example permutation such as (10, 11, 12) for one group and (14, 14, 17) for the other. Calculate the test statistic: \(\text{mean of group 1} = \frac{10+11+12}{3} = 11\) and \(\text{mean of group 2} = \frac{14+14+17}{3} = 15\). The test statistic is \(|11 - 15| = 4\).
05

Interpret Results

Perform a large number of permutations. In each case, check how often the test statistic is as extreme or more extreme than the observed \(2\). If this occurs rarely, the null hypothesis may be rejected, suggesting a meaningful difference in memorization ability between sexes.

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Key Concepts

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

Statistical Hypotheses
When comparing male and female memorization abilities, statistical hypotheses help us set a framework for understanding our analysis. The *null hypothesis*, often denoted as \(H_0\), is a claim we assume to be true unless evidence suggests otherwise. In this case, the null hypothesis is that there is no difference in the average number of words recalled by males and females, mathematically expressed as \(H_0 : \mu_m = \mu_f\). This implies that any observed difference is due to random chance, not a systematic difference in performance. On the other hand, we have the *alternative hypothesis*, \(H_1\), which is the statement we want to test against the null. Here, it posits that there is a difference in recall ability between the two genders, represented as \(H_1 : \mu_m eq \mu_f\). This suggests that any difference in mean recall isn't just a fluke, but indicates a true distinction in memory abilities between males and females.
Mean Differences
The difference in means is a simple yet powerful statistic in our analysis of memorization ability. To calculate this, we subtract the mean number of words recalled by one group from the other. In our scenario, the males recalled an average of 12 words, while the females averaged 14 words. Therefore, the difference in means is \(14 - 12 = 2\). This difference is our observed test statistic and serves as a crucial point in the permutation test. A permutation test leverages this difference by examining if such a statistic could have arisen simply by chance. By shuffling or "permuting" the data and subsequently calculating new mean differences from the rearranged data, we assess the likelihood that our observed difference could happen randomly. If we frequently see mean differences of 2 or more in these hypothetical scenarios, the original observed difference might not substantiate claims of differing memorization capacities. But if such results are rare, it lends weight to the evidence that there truly is a difference.
Memorization Ability Analysis
The exercise aims to determine whether there is a real difference between males and females in their ability to memorize words. The focus is on conducting a permutation test, which provides a non-parametric method to test the hypothesis. This means it's less reliant on assumptions about the data, such as normal distribution, making it versatile and robust. So, how do we go about this test? Follow these basic steps to perform the analysis:
  • First, pool all test scores together: 10, 12, 14, 11, 14, 17.
  • Next, randomly split this data into two groups of three, treating each split as a unique permutation.
  • Calculate the mean for each group and find the difference in these means; this is your test statistic for that permutation.
  • Repeat this process many times, often thousands, to create a distribution of test statistics under the assumption that the null hypothesis is true.
An example permutation might assign (10, 11, 12) to one group and (14, 14, 17) to the other, resulting in a test statistic of 4, since the means are 11 and 15. Compare all obtained test statistics to the observed difference of 2. The larger the number of times a computed difference meets or exceeds 2, the more support is drawn for random chance causing any differences; otherwise, if rare, it suggests a true discrepancy in memorization capabilities.

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Most popular questions from this chapter

In performing a randomization test, explain why the samples are simulated by using the assumption that the null hypothesis is true.

An intersection has a four-way stop sign but no traffic light. Currently, about 1200 cars use the intersection a day, and the rate of accidents at the intersection is about one every two weeks. The potential benefit of adding a traffic light was studied using a computer simulation by modeling traffic flow at the intersection if a light were to be installed. The simulation included 100,000 repetitions of 1200 cars using the intersection to mimic 100,000 days of use. Of the 100,000 simulations, an accident occurred in 5230 of them, and no accident occurred in the rest. (There were no simulated days with two or more accidents.) a. Based on the results of the simulation, what is the estimated number of accidents in a 2 - week period if the traffic light were to be installed? b. Does the simulation show that adding the traffic light would be a good idea? Explain.

Suppose that a randomization distribution resulting from the simulation of a chi-square test had \(7 \%\) of the values at or above the chi-square statistic observed for the real sample. a. What is the estimated \(p\) -value for the test? b. What decision would you make for the test, using a level of \(0.05 ?\) c. What decision would you make for the test, using a level of \(0.10 ?\)

Remember that the \(p\) -value corresponding to a chi-square statistic of 3.84 for a table with 2 rows and 2 columns is \(0.05 .\) If you were to simulate 10,000 samples for a \(2 \times 2\) table under the assumption that the null hypothesis is true, about how many of them would you expect to result in a chi-square statistics of less than \(3.84 ?\)

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