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Chapter 13: Problem 14

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.

Short Answer

Expert verified
In the given problem, we have performed a paired t-test to determine if there is a significant difference in the mean swimming speeds between water and syrup. We started by identifying the null and alternative hypotheses, and by using the provided data, we calculated the mean and standard deviation of the differences in swimming speeds. We then computed the t-value and degrees of freedom, found the corresponding p-value, and compared it to the given significance level of 0.01. If the p-value 鈮� 0.01, we reject the null hypothesis (H鈧�), suggesting a significant difference in mean swimming speeds between water and syrup. If the p-value > 0.01, we fail to reject the null hypothesis (H鈧�), indicating no significant difference in the mean swimming speeds between water and syrup. Further calculations are required to determine the actual p-value and make our conclusion.

Step by step solution

01

Identify hypotheses

Let 碌鈧� be the mean swimming speed in water and 碌鈧� be the mean swimming speed in syrup. The null hypothesis (H鈧�) will be that there is no significant difference in the means of the swimming speeds in both environments, i.e., 碌鈧� = 碌鈧�. The alternative hypothesis (H鈧�) will be that the means are different, i.e., 碌鈧� 鈮� 碌鈧�.
02

Calculate the difference in swimming speeds

For each swimmer, calculate the difference in swimming speeds between water and syrup. Denote this difference as d.
03

Calculate the mean and standard deviation of the differences

Find the mean of the differences by summing up all the differences and dividing by the number of swimmers (n=20). Denote this mean as \(\bar{d}\). Calculate the standard deviation of the differences. Denote this standard deviation as s_d.
04

Determine the t-value and degrees of freedom

Use the mean of the differences and the standard deviation to calculate the t-value. \[ t = \frac{\bar{d}}{s_d / \sqrt{n}} \] The degrees of freedom for this test will be n-1 = 19.
05

Calculate the p-value

Using a t-distribution table or statistical software, find the p-value for the calculated t-value and the degrees of freedom (19).
06

Compare the p-value to the significance level

Compare the p-value from step 5 to the given significance level of 0.01. If the p-value 鈮� 0.01, reject the null hypothesis (H鈧�) and conclude that there is a significant difference in the mean swimming speeds between water and syrup. If the p-value > 0.01, fail to reject the null hypothesis (H鈧�) and conclude that there is no significant difference in the mean swimming speeds between water and syrup.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, commonly represented as \( H_0 \), serves as a default statement that there is no effect or no difference. It's the assertion that any observations are the result of pure chance. For instance, in the context of our swimming speeds example, the null hypothesis posits that the mean swimming speed in water (\( \mu_1 \)) is equal to the mean swimming speed in syrup (\( \mu_2 \)), or \( \mu_1 = \mu_2 \).

To assess this null hypothesis, statistical tests are performed, and its validity is judged based on the data. If evidence suggests that the null hypothesis is unlikely, then it is rejected in favor of the alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis, represented as \( H_1 \) or \( H_a \), is the statement we accept if the null hypothesis is rejected. It proposes that there is an effect or a difference, and in the study at hand, it suggests that swimming speeds are indeed affected by the viscosity of the liquid; meaning the swimming speed in water is different from that in syrup (\( \mu_1 eq \mu_2 \)).

Essentially, the alternative hypothesis is what the researcher is trying to prove. Testing aims to find enough evidence to support the alternative hypothesis, indicating that the null hypothesis may not adequately explain the observation.
T-Value
The t-value, or t-score, is a ratio calculated from a given sample during a t-test. It reflects how many standard deviations the observed values are from the mean. The formula used to compute the t-value, as seen in the swimming speeds study, is:
\[ t = \frac{\bar{d}}{s_d / \sqrt{n}} \]
where \(\bar{d}\) is the average of the differences between swimming speeds in the two conditions, \(s_d\) is the standard deviation of these differences, and \(n\) is the sample size. A large absolute value of the t-score suggests that a significant difference exists between the two groups being compared.
P-Value
The p-value quantifies the probability of obtaining test results at least as extreme as the ones observed during the test, assuming that the null hypothesis is true. It鈥檚 a crucial part of hypothesis testing as it helps us determine the statistical significance of our results. A small p-value (usually 鈮� 0.05) indicates strong evidence against the null hypothesis, leading researchers to reject it in favor of the alternative hypothesis.

In our exercise, the p-value is derived from the calculated t-value and the degrees of freedom. If this p-value is less than or equal to the significance level (0.01 in this case), it suggests that the difference in swimming speeds is significant, not just due to random chance.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold set by the researcher to determine the criterion for rejecting the null hypothesis. Common choices for alpha are 0.01, 0.05, or 0.10, corresponding to 1%, 5%, and 10% significance levels respectively. The lower the alpha, the more stringent and conservative the test.

In the swimming speed test, an alpha of 0.01 means that the researcher allows for a 1% chance of incorrectly rejecting the null hypothesis (a Type I error). Decisions about the null hypothesis are made by comparing the p-value to this chosen significance level. If the p-value is less than or equal to the significance level, the results are statistically significant, and the null hypothesis is rejected.

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Most popular questions from this chapter

Can moving their hands help children learn math? This question was investigated in the paper "Gesturing Gives Children New Ideas About Math" (Psychological Science [2009]: \(267-272\) ). Eighty-five children in the third and fourth grades who did not answer any questions correctly on a test with six problems of the form \(3+2+8=-8\) were participants in an experiment. The children were randomly assigned to either a no-gesture group or a gesture group. All the children were given a lesson on how to solve problems of this form using the stratcgy of trying to make both sides of the equation equal. Children in the gesture group were also taught to point to the first two numbers on the left side of the equation with the index and middle linger of one hand and then to point at the blank on the right side ol the equation. This gesture was supposed to emphasize that grouping is involved in solving the problem. The children then practiced udditional problems of this type. All children were then given a test with six problems to solve, and the number of correct answers was recorded for each child. Summary statistics are given below. Is there evidence that learning the gesturing approach to solving problems of this type results in a significantly higher mean number of correct responses? Test the relevant hypotheses using \(\alpha=0.05\).

Many people now turn to the Internet to get information on health-related topics. The paper "An Examination of Health, Medical and Nutritional Information on the Internet: A Comparative study of Wikipedia, WebMD and the Mayo Clinic Websites" (The International Journal of Communication and Health [2015]: 30-38) used Flesch reading ease scores (a measure of reading difficulty based on factors such as sentence length and number of syllables in the words used) to score pages on Wikipedia and on WebMD. Higher Flesch scores correspond to more difficult reading levels. The paper reported that for a representative sample of health-related pages on Wikipedia, the mean Flesch score was 26.7 and the standard deviation of the Flesch scores was \(14.1 .\) For a representative sample of pages from WebMD, the mean score was 43.9 and the standard deviation was 19.4 . Suppose that these means and standard deviations were based on samples of 40 pages from each site. Is there convincing evidence that the mean reading level for health-related pages differs for Wikipedia and WebMD? Test the relevant hypotheses using a significance level of \(\alpha=0.05\)

Use the data given in the previous exercise to complete the following: a. Estimate the mean difference in motion between dominant and nondominant arms for position players using a \(95 \%\) confidence interval. b. The authors asserted that pitchers have a greater difference in mean motion of their shoulders than do position players. Do you agree? Explain.

The article "An Alternative Vote: Applying Science to the Teaching of Science" (The Economist, May 12,2011 ) describes an experiment conducted at the University of British Columbia. A total of 850 engineering students enrolled in a physics course participated in the experiment. These students were randomly assigned to one of two experimental groups. The two groups attended the same lectures for the first 11 weeks of the semester. In the twelfth week, one of the groups was switched to a style of teaching where students were expected to do reading assignments prior to class and then class time was used to focus on problem solving, discussion and group work. The second group continued with the traditional lecture approach. At the end of the twelfth week, the students were given a test over the course material from that week. The mean test score for students in the new teaching method group was 74 and the mean test score for students in the traditional lecture group was 41 . Suppose that the two groups each consisted of 425 students and that the standard deviations of test scores for the new teaching method group and the traditional lecture method group were 20 and 24 , respectively. Can you conclude that the mean test score is significantly higher for the new teaching method group than for the traditional lecture method group? Test the appropriate hypotheses using a significance level of \(\alpha=0.01\).

To determine if chocolate milk is as effective as other carbohydrate replacement drinks, nine male cyclists performed an intense workout followed by a drink and a rest period. At the end of the rest period, each cyclist performed an endurance trial in which he exercised until exhausted, and the time to exhaustion was measured. Each cyclist completed the entire regimen on two different days. On one day, the drink provided was chocolate milk, and on the other day the drink provided was a carbohydrate replacement drink. Data consistent with summary quantities in the paper "The Efficacy of Chocolate Milk as a Recovery Aid" (Medicine and Science in Sports and Exercise [2004]: S126) are given in the table at the bottom of the page. Is there evidence that the mean time to exhaustion is greater after chocolate milk than after a carbohydrate replacement drink? Use a significance level of \(\alpha=0.05\).
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