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Chapter 11: Problem 20

A researcher at the Medical College of Virginia conducted a study of 60 randomly selected male soccer players and concluded that frequently "heading" the ball in soccer lowers players' IQs (USA Today, August 14 1995). The soccer players were divided into two groups, based on whether they averaged 10 or more headers per game. Mean IQs were reported in the article, but the sample sizes and standard deviations were not given. Suppose that these values were as given in the accompanying table. $$ \begin{array}{l|ccc} & & \text { Sample } & \text { Sample } \\ & n & \text { Mean } & \text { sd } \\ \hline \text { Fewer Than 1O Headers } & 35 & 112 & 10 \\ 10 \text { or More Headers } & 25 & 103 & 8 \\ \hline \end{array} $$ Do these data support the researcher's conclusion? Test the relevant hypotheses using \(\alpha=.05 .\) Can you conclude that heading the ball causes lower \(\mathrm{IQ}\) ?

Short Answer

Expert verified
The short answer to the question will be determined on the basis of the p-value compared with the significance level. If the p-value is less than the significance level, the data does support the researcher's conclusion that frequently 'heading' the ball in soccer lowers players' IQs. However, it is important to remember that this test only provides evidence, it does not definitively prove that heading the ball causes a lower IQ.

Step by step solution

01

Set up hypotheses

The first step is to set up the null hypothesis and the alternative hypothesis. The null hypothesis (\(H_0\)) will state no difference in means, and the alternative hypothesis (\(H_A\)) will state that there is a difference. In this case, the null hypothesis is: \(\mu_1 = \mu_2\), where \(\mu_1\) is the population mean IQ for the group with fewer than 10 headers and \(\mu_2\) is the population mean IQ for the group with 10 or more headers. The alternative hypothesis will be: \(\mu_1 \neq \mu_2\).
02

Calculate the test statistic

Next, the test statistic must be calculated, which will follow a Student's t-distribution. The formula for the test statistic in this case is: \[ t = \frac{{(\bar{X}_1 - \bar{X}_2)}}{{\sqrt{\frac{{s_1^2}}{n_1} + \frac{{s_2^2}}{n_2}}}} \] Where \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means, \(s_1^2\) and \(s_2^2\) are the sample variances, and \(n_1\) and \(n_2\) are the sample sizes for the two groups. Substituting the given values from the table above, you will find the test statistic.
03

Find the p-value

After finding the test statistic, the next step is to find the p-value. This can be done using statistical software or a t-distribution table. The value corresponds to the probability of finding the observed, or more extreme, results when the null hypothesis of a study question is true.
04

Make a Conclusion

If the p-value is less than the significance level, \(\alpha=0.05\), reject the null hypothesis in favor of the alternative. This means there is evidence to suggest that there's a difference in means. In terms of this question, it means it might support the researcher's conclusion that heading the ball affects players' IQs. However, even if there is evidence to support the conclusion, it does not prove causation. Other factors might affect the players' IQs.

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Key Concepts

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

T-Test for Independent Means
When comparing the average scores or measurements of two different groups, researchers often use a statistical method known as the t-test for independent means. This method helps to determine if the differences in group means are statistically significant or if they could have occurred by chance. The t-test assumes that the two groups are independent of each other with different members in each group. The situation with the soccer players and their IQ scores is a suitable scenario for this test since we have two distinct groups 鈥� those who head the ball frequently and those who don't.

In the provided exercise, the calculation of the t-statistic is a critical step. It takes into account the means, standard deviations, and sample sizes from both groups. The resulting statistic is then used to gauge the likelihood of the null hypothesis 鈥� which states there is no effect or difference 鈥� being true. If the calculated t is sufficiently large, it may lead us to doubt the null hypothesis and consider the alternative that there indeed is a significant difference between the groups' mean IQ scores.
P-Value Interpretation
After calculating the t-statistic, obtaining the p-value is the next logical step. The p-value is a probability score that tells us how likely it is to get a result as extreme as, or more extreme than, what we've observed in our study if the null hypothesis were true. A low p-value (typically less than 0.05) suggests that the observed data is quite unlikely under the null hypothesis, and thus, we may reject the null hypothesis.

In the context of this exercise, if we found a p-value that is less than the alpha level of 0.05, it would mean there is less than a 5% chance that the difference in IQ scores between the two groups of soccer players occurred due to random variation alone. Therefore, we would reject the null hypothesis that there is no difference in mean IQs. However, interpreting the p-value should be done with care; a small p-value does not prove that one variable causes the difference in the other. It simply highlights a statistical association that warrants further investigation.
Causation vs Correlation
Determining causation is a complex matter in scientific research. A key distinction to make is that between correlation and causation. Correlation refers to a relationship where two variables move together, but it does not mean one causes the other to happen. Causation, on the other hand, implies that one event is the result of the occurrence of the other event; there is a cause and effect relationship.

In our example of soccer players and IQ scores, even if we find a statistically significant difference in IQ based on the frequency of heading the ball, we cannot immediately conclude that heading the ball causes a decrease in IQ. There could be other variables at play, such as the overall physical activity levels, dietary habits, education, or even socioeconomic status that might influence both soccer playing and IQ. To establish causation, further studies, possibly with experimental and longitudinal designs, are required to rule out alternative explanations and to track the relationship over time.

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Most popular questions from this chapter

The article "Fish Oil Staves Off Schizophrenia" (USA Today, February 2, \(2 \mathrm{O} 1 \mathrm{O}\) ) describes a study in which 81 patients age 13 to 25 who were considered atrisk for mental illness were randomly assigned to one of two groups. Those in one group took four fish oil capsules daily. The other group took a placebo. After 1 year, \(5 \%\) of those in the fish oil group and \(28 \%\) of those in the placebo group had become psychotic. Is it appropriate to use the two-sample \(z\) test of this section to test hypotheses about the difference in the proportions of patients receiving the fish oil and the placebo treatments who became psychotic? Explain why or why not.

The Insurance Institute for Highway Safety issued a press release titled "Teen Drivers Often Ignoring Bans on Using Cell Phones" (June 9,2008 ). The following quote is from the press release: Just \(1-2\) months prior to the ban's Dec. 1,2006 start, 11 percent of teen drivers were observed using cell phones as they left school in the afternoon. About 5 months after the ban took effect, \(12 \%\) of teen drivers were observed using cell phones. Suppose that the two samples of teen drivers (before the ban, after the ban) can be regarded as representative of these populations of teen drivers. Suppose also that 200 teen drivers were observed before the ban (so \(n_{1}=200\) and \(\hat{p}_{1}=.11\) ) and 150 teen drivers were observed after the ban. a. Construct and interpret a \(95 \%\) confidence interval for the difference in the proportion using a cell phone while driving before the ban and the proportion after the ban. b. Is zero included in the confidence interval of Part (c)? What does this imply about the difference in the population proportions?

An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route can be justified only if it reduces the mean travel time by more than 10 minutes. a. If \(\mu_{1}\) is the mean for the scenic route and \(\mu_{2}\) for the nonscenic route, what hypotheses should be tested? b. If \(\mu_{1}\) is the mean for the nonscenic route and \(\mu_{2}\) for the scenic route, what hypotheses should be tested?

Two proposed computer mouse designs were compared by recording wrist extension in degrees for 24 people who each used both mouse types ("Comparative Study of Two Computer Mouse Designs," Cornell Human Factors Laboratory Technical Report RP7992). The difference in wrist extension was computed by subtracting extension for mouse type \(\mathrm{B}\) from the wrist extension for mouse type A for each student. The mean difference was reported to be 8.82 degrees. Assume that it is reasonable to regard this sample of 24 people as 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 B? Use a .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 B? Use a .05 significance level. c. Briefly explain why a different conclusion was reached in the hypothesis tests of Parts (a) and (b).

The paper "Ready or Not? Criteria for Marriage Readiness among Emerging Adults" (Journal of \(\underline{\text { Ado- }}\) lescent Research [2009]: 349-375) surveyed emerging adults (defined as age 18 to 25 ) from five different colleges in the United States. Several questions on the survey were used to construct a scale designed to measure endorsement of cohabitation. The paper states that "on average, emerging adult men \((\mathrm{M}=3.75, \mathrm{SD}=1.21)\) reported higher levels of cohabitation endorsement than emerging adult women \((\mathrm{M}=3.39, \mathrm{SD}=1.17) . "\) The sample sizes were 481 for women and 307 for men. a. Carry out a hypothesis test to determine if the reported difference in sample means provides convincing evidence that the mean cohabitation endorsement for emerging adult women is significantly less than the mean for emerging adult men for students at these five colleges. b. What additional information would you want in order to determine whether it is reasonable to generalize the conclusion of the hypothesis test from Part (a) to all college students?
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