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Research has shown that, for baseball players, good hip range of motion results in improved performance and decreased body stress. The article "Functional Hip Characteristics of Baseball Pitchers and Position Players" (The American Journal of Sports Medicine, \(2010: 383-388\) ) reported on a study of independent samples of 40 professional pitchers and 40 professional position players. For the pitchers, the sample mean hip range of motion was 75.6 degrees and the sample standard deviation was 5.9 degrees, whereas the sample mean and sample standard deviation for position players were 79.6 degrees and 7.6 degrees, respectively. Assuming that the two samples are representative of professional baseball pitchers and position players, test hypotheses appropriate for determining if mean range of motion for pitchers is less than the mean for position players.

Short Answer

Expert verified
After calculating and comparing the test statistic and critical value, we can make a decision whether to reject or not reject the null hypothesis. This will indicate whether there is a statistically significant difference in mean hip range of motion between pitchers and position players.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)) are defined as follows: \(H_0\): \(\mu_1 - \mu_2 = 0\) (There is no difference in mean range of motion between pitchers and position players). \(H_a\): \(\mu_1 - \mu_2 < 0\) (The mean range of motion for pitchers is less than that for position players). Where, \(\mu_1\) and \(\mu_2\) represent the population means for pitchers and position players, respectively.
02

Compute the Test Statistic

The test statistic for two independent samples is calculated as: \( z = \frac {(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{{s_1}^2}{n_1} + \frac{{s_2}^2}{n_2}}}\) Substituting the given values: \( z = \frac {(75.6 - 79.6) - 0}{\sqrt{\frac{{5.9}^2}{40} + \frac{{7.6}^2}{40}}}\) After calculation, we get the z-score.
03

Determine the Critical Value

Assuming a 0.05 level of significance (typical in social sciences), and because we stated a one-tailed (less than) hypothesis, we refer to the standard normal (Z) distribution table and find that the critical value is -1.645.
04

Make a Decision

We compare the calculated test statistic with the critical value. If the test statistic is less than the critical value, we reject the null hypothesis; otherwise, we do not reject it. This will show whether the mean hip range of motion for pitchers is statistically significantly less than that for position players.

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Key Concepts

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

Independent Samples
In statistical analysis, when we talk about independent samples, we're referring to two groups of subjects that are not connected to one another. Each group constitutes its own separate group of individuals or observations. In the context of our baseball study, this means that the data for pitchers and position players are collected separately and do not influence each other. For example, the hip range of motion for one pitcher does not affect the hip range of motion for any position player, and vice versa. Why Use Independent Samples?
  • They allow us to compare two distinct groups in our study.
  • They make it possible to make inferences about the differences between the two groups.
  • They ensure that the results are not mixed up by some common factor influencing both groups.
When conducting hypothesis testing with independent samples, it's crucial to confirm that the samples are indeed independent. Any dependency, such as shared influences or overlapping members, can bias the results and lead to incorrect conclusions. In our exercise, it is supposed that the samples of pitchers and players are correctly representing the populations of interest.
Test Statistic
The test statistic is a vital part of hypothesis testing. It's a standardized value that results from the sample data, and it is used to decide whether to support or reject the null hypothesis. In simpler terms, it's a tool that helps you understand how much your sample data deviates from the null hypothesis.Calculation of the Test Statistic
For our baseball study, we use the formula for the z-test statistic for two independent samples:\[ z = \frac {(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{{s_1}^2}{n_1} + \frac{{s_2}^2}{n_2}}} \]Where:
  • \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means.
  • \(\mu_1\) and \(\mu_2\) are the population means hypothesized to be equal if the null hypothesis is true.
  • \(s_1\) and \(s_2\) are the sample standard deviations.
  • \(n_1\) and \(n_2\) are the sample sizes.
The result, called the z-score, will tell us how far away — in terms of standard deviations — our sample mean difference is from the null hypothesis of no difference.
Critical Value
The critical value is a threshold that helps determine the decision rule for hypothesis tests. It is derived from the significance level we choose for our test. Understanding the Critical Value
In hypothesis testing:
  • The critical value acts as a cutoff point.
  • It is determined by the preselected significance level (often 0.05 in social sciences), which defines the probability of rejecting a true null hypothesis.
For a one-tailed test, like in the baseball study, we identify a critical value from the z-distribution table. With a 0.05 level of significance, the critical value corresponds to -1.645 for a left-tailed test, indicating that there is a 5% chance of observing a sample mean as extreme as ours, or more extreme, under the null hypothesis. This value ultimately helps decide whether our test statistic shows a statistically significant result.
Z-Score
In hypothesis testing, the z-score is an essential concept when comparing sample data against a population value (or in this case, against another sample). This score measures how many standard deviations an element is from the mean. Roles of the Z-Score
  • It standardizes the difference so that we can determine the likelihood of a difference as extreme as our observed data.
  • Helps in assessing how unusual our data is under the assumption of the null hypothesis being true.
In our calculations, the z-score is compared to the critical value. If the z-score is beyond this critical point, it implies that our observed difference is unusual enough under the null hypothesis to suggest that the null hypothesis may not hold. Thus, providing evidence that supports the alternative hypothesis. For our example, if the calculated z-score is less than -1.645, it indicates that the pitchers' mean hip range of motion is indeed statistically significantly less than that of the position players.

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Most popular questions from this chapter

The paper referenced in the Preview Example of this chapter ("Mood Food: Chocolate and Depressive Symptoms in a Cross-Sectional Analysis," Archives of Internal Medicine [2010]: \(699-703\) ) describes a study that investigated the relationship between depression and chocolate consumption. Participants in the study were 931 adults who were not currently taking medication for depression. These participants were screened for depression using a widely used screening test. The participants were then divided into two samples based on their test score. One sample consisted of people who screened positive for depression, and the other sample consisted of people who did not screen positive for depression. Each of the study participants also completed a food frequency survey. The researchers believed that the two samples were representative of the two populations of interest-adults who would screen positive for depression and adults who would not screen positive. The paper reported that the mean number of servings per month of chocolate for the sample of people that screened positive for depression was 8.39 , and the sample standard deviation was \(14.83 .\) For the sample of people who did not screen positive for depression, the mean was \(5.39,\) and the standard deviation was \(8.76 .\) The paper did not say how many individuals were in each sample, but for the purposes of this exercise, you can assume that the 931 study participants included 311 who screened positive for depression and 620 who did not screen positive. Carry out a hypothesis test to confirm the researchers' conclusion that the mean number of servings of chocolate per month for people who would screen positive for depression is higher than the mean number of chocolate servings per month for people who would not screen positive.

The article "Plugged In, but Tuned Out" (USA Today, January 20,2010 ) summarizes data from two surveys of kids ages 8 to 18 . One survey was conducted in 1999 and the other was conducted in \(2009 .\) Data on the number of hours per day spent using electronic media, consistent with summary quantities given in the article, are given in the following table (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years the surveys were conducted. Construct and interpret a \(98 \%\) confidence interval estimate of the difference between the mean number of hours per day spent using electronic media in 2009 and \(1999 .\) $$ \begin{array}{llllllllllllllll} 2009 & 5 & 9 & 5 & 8 & 7 & 6 & 7 & 9 & 7 & 9 & 6 & 9 & 10 & 9 & 8 \\ 1999 & 4 & 5 & 7 & 7 & 5 & 7 & 5 & 6 & 5 & 6 & 7 & 8 & 5 & 6 & 6 \end{array} $$

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be for a difference in population means. If not, explain why not. Scenario 1: A researcher at the Medical College of Virginia conducted a study of 60 randomly selected male soccer players and concluded that players who frequently "head" the ball in soccer have a lower mean IQ (USA Today, August 14,1995 ). The soccer players were divided into two samples, based on whether they averaged 10 or more headers per game, and IQ was measured for each player. You would like to determine if the data support the researcher's conclusion. Scenario 2: A credit bureau analysis of undergraduate students" credit records found that the mean number of credit cards in an undergraduate's wallet was 4.09 ("Undergraduate Students and Credit Cards in \(2004,{ }^{n}\) Nellie Mae, May 2005 ). It was also reported that in a random sample of 132 undergraduates, the mean number of credit cards that the students said they carried was 2.6. You would like to determine if there is convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09 .\) Scenario 3: Some commercial airplanes recirculate approximately \(50 \%\) of the cabin air in order to increase fuel efficiency. The authors of the paper "Aircraft Cabin Air Recirculation and Symptoms of the Common Cold" (Journal of the American Medical Association \([2002]: 483-486)\) studied 1,100 airline passengers who flew from San Francisco to Denver. Some passengers traveled on airplanes that recirculated air, and others traveled on planes that did not. Of the 517 passengers who flew on planes that did not recirculate air,

Do girls think they don't need to take as many science classes as boys? The article "Intentions of Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences" (Science Education [1999]: \(55-76\) ) describes a survey of randomly selected children in grades \(4,5,\) and 6 . The 224 girls participating in the survey each indicated the number of science courses they intended to take in the future, and they also indicated the number of science courses they thought boys their age should take in the future. For each girl, the authors calculated the difference between the number of science classes she intends to take and the number she thinks boys should take. a. Explain why these data are paired. b. The mean of the differences was -0.83 (indicating girls intended, on average, to take fewer science classes than they thought boys should take), and the standard deviation was 1.51 . Construct and interpret a \(95 \%\) confidence interval for the mean difference.

Two proposed computer mouse designs were compared by recording wrist extension in degrees for 24 people who each used both mouse designs ("Comparative Study of Two Computer Mouse Designs," Cornell Human Factors Laboratory Technical Report RP7992). The difference in wrist extension was calculated by subtracting extension for mouse type \(\mathrm{B}\) from the wrist extension for mouse type \(\mathrm{A}\) for each person. The mean difference was reported to be 8.82 degrees. Assume that this sample of 24 people is representative of the population of computer users. a. Suppose that the standard deviation of the differences was 10 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. b. Suppose that the standard deviation of the differences was 26 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. c. Briefly explain why different conclusions were reached in the hypothesis tests of Parts (a) and (b).

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