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"Mountain Biking May Reduce Fertility in Men, Study Says" was the headline of an article appearing in the San Luis Obispo Tribune (December 3,2002 ). This conclusion was based on an Austrian study that compared sperm counts of avid mountain bikers (those who ride at least 12 hours per week) and nonbikers. Ninety percent of the avid mountain bikers studied had low sperm counts, as compared to \(26 \%\) of the nonbikers. Suppose that these percentages were based on independent samples of 100 avid mountain bikers and 100 nonbikers and that it is reasonable to view these samples as representative of Austrian avid mountain bikers and nonbikers. a. Do these data provide convincing evidence that the proportion of Austrian avid mountain bikers with low sperm count is higher than the proportion of Austrian nonbikers? b. Based on the outcome of the test in Part (a), is it reasonable to conclude that mountain biking 12 hours per week or more causes low sperm count? Explain.

Short Answer

Expert verified
a. The statistical test does provide evidence to suggest that mountain bikers have a greater proportion of low sperm counts compared to nonbikers. b. However, it's not reasonable to conclude that mountain biking causes low sperm count just based on this study. This is because correlation does not imply causation.

Step by step solution

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Step 1. Setting up the Hypothesis

First, create the null and alternative hypotheses. The null hypothesis \(H_0\) is that the proportion of mountain bikers with low sperm count \(P_{MB}\) and the proportion of nonbikers with a low sperm count \(P_{NB}\) is the same, written as: \(H_0: P_{MB} = P_{NB}\). The alternative hypothesis \(H_A\) is that the proportion of mountain bikers with low sperm count is greater than the proportion of nonbikers with low sperm count, written as: \(H_A: P_{MB} > P_{NB}\)
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Step 2. Calculation of z-statistic

The test statistic for a hypothesis test about a difference in proportions conducted using large independent samples, is a z-score. We need to calculate this. The formula for the z-statistic is: \[Z = \frac{(\hat{p}_{1} - \hat{p}_{2}) - 0}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_{1}} + \frac{1}{n_{2}})}}\] Where, The overall sample proportion \(\hat{p} = \frac{x_{1} + x_{2}}{n_{1} + n_{2}}\) Here, \(n_{1} = n_{2} = 100\), the number of mountain bikers and nonbikers; \(\hat{p}_{1} = 0.9\) and \(\hat{p}_{2} = 0.26\), the proportions of mountain bikers and nonbikers with low sperm count.
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Step 3. Calculation of p-value

The p-value is the probability that a test statistic being as extreme as, or more so, than what was observed under the assumption that the null hypothesis is true. Because the alternative hypothesis is that \(P_{MB} > P_{NB}\), the p-value is the area to the right of the computed z-score. Calculate this using a standard normal z-table or statistical software.
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Step 4. Conclusion of the Hypothesis Test

If the p-value is less than the chosen significance level (usually 0.05), then reject the null hypothesis and state that there is evidence to suggest that the proportion of mountain bikers with low sperm count is greater than the proportion of nonbikers.
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Step 5. Answering the Second Part

While the hypothesis test may show a significant difference in the proportion of low sperm counts between mountain bikers and nonbikers, this does not necessarily mean that mountain biking is the cause. It could be that mountain bikers have something else in common that leads to a low sperm count. Further studies would be needed to determine causality.

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Key Concepts

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

Statistical Significance
Understanding statistical significance is crucial when analyzing the results of a study. It is a measure of whether the findings from your data are likely due to chance or if they reflect a true effect in the population. When a result is statistically significant, it means that the observed pattern is unlikely to have occurred by random variations alone.

For instance, in hypothesis testing, statistical significance helps researchers decide if they should reject the null hypothesis. A common threshold for significance is a p-value less than 0.05. If a study's p-value falls beneath this level, the difference or relationship observed in the data is considered statistically significant, suggesting that the results are not simply due to random chance.
Proportion Comparison
Comparing proportions is a statistical method used to determine if there is a significant difference between the proportion of a particular outcome in two different groups. It involves calculating the proportion of each group experiencing the outcome and then quantifying the difference. This can be particularly useful in medical or social science research.

For example, in the case of the mountain bikers and nonbikers, the proportion comparison was used to assess whether a significantly larger proportion of mountain bikers had low sperm counts compared to nonbikers.
Null and Alternative Hypotheses
Hypotheses are the backbone of statistical testing. The null hypothesis (\(H_0\)) represents the default statement, usually reflecting no effect or no difference. It asserts that any observed difference is due to sampling error or chance. Alternatively, the alternative hypothesis (\(H_A\) or \(H_1\)) is what you aim to support, suggesting that there is a true effect or difference.

In our exercise, the null hypothesis posits no difference in low sperm count proportions between mountain bikers and nonbikers, while the alternative hypothesis posits that mountain bikers have a higher proportion of low sperm counts than nonbikers.
Z-Statistic Calculation
The z-statistic is a measure that helps researchers understand how far a data point is from the mean of a distribution, in standard deviation units. Calculating the z-statistic in proportion comparisons involves using the formula given in the solution steps. When the resulting z-value is large (positive or negative), it suggests that our sample proportion is much different from the hypothesized proportion under the null.

In the exercise, the z-statistic quantifies the difference between the avid mountain bikers and nonbikers' sperm count proportions, considering the variability in those proportions.
P-Value Interpretation
The p-value is one of the most important aspects of hypothesis testing as it helps us interpret the results of our statistical test. It represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated from our sample data, assuming the null hypothesis is true.

In the context of our exercise, a low p-value would suggest that it is very unlikely that the higher proportion of low sperm counts among mountain bikers is due to chance alone. This could lead to rejecting the null hypothesis in favor of the alternative.
Causal Inference
Causal inference is the process of drawing a conclusion about a causal connection based on the conditions of the occurrence of an effect. It is a complex task that requires more than just statistical evidence; it demands rigorous experimental or observational study design to control for confounding variables.

Referencing the exercise about mountain bikers, while statistical significance might indicate a higher proportion of low sperm counts compared to nonbikers, it doesn't prove that mountain biking causes the low sperm count. Other factors may be at play, and only a well-designed study can start to make causal determinations.

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Most popular questions from this chapter

The effect of loneliness among college students was examined in the article "The Importance of Perceived Duration: Loneliness and Its Relationship to Self-Esteem and Academic Performance" (Journal College Student Development \([1994]: 456-460\) ). Based on reported frequency and duration of loneliness, subjects were divided into two groups. The first group was the short-duration loneliness group, and the second was the long-duration loneliness group. A self-esteem inventory was administered to students in both groups. For the short-duration group, the reported mean self-esteem score was \(76.78\) and the standard deviation was \(17.80\). For the long-duration group, the mean and standard deviation were \(64.00\) and \(15.68\), respectively. Do the data support the researcher's claim that mean self-esteem is lower for students classified as having long-duration loneliness? Test the relevant hypotheses using \(\alpha=.01\). Be sure to state any assumptions that are necessary for your test to be valid.

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Two different underground pipe coatings for preventing corrosion are to be compared. The effect of a coating (as measured by maximum depth of corrosion penetration on a piece of pipe) may vary with depth, orientation, soil type, pipe composition, etc. Describe how an experiment that filters out the effects of these extraneous factors could be carried out.

O British health officials have expressed concern about problems associated with vitamin D deficiency among certain immigrants. Doctors have conjectured that such a deficiency could be related to the amount of fiber in a person's diet. An experiment was designed to compare the vitamin D plasma half-life for two groups of healthy individuals. One group was placed on a normal diet, whereas the second group was placed on a high-fiber diet. The accompanying table gives the resulting data (from "Reduced Plasma Half-Lives of Radio- Labeled \(25(\mathrm{OH}) \mathrm{D} 3\) in Subjects Receiving a High-Fibre Diet," British Journal of Nutrition \([1993]: 213-216\) ). \(\begin{array}{lllllll}\text { Normal diet } & 19.1 & 24.0 & 28.6 & 29.7 & 30.0 & 34.8\end{array}\) \(\begin{array}{llllllll}\text { High-fiber diet } & 12.0 & 13.0 & 13.6 & 20.5 & 22.7 & 23.7 & 24.8\end{array}\) Use the following MINITAB output to determine whether the data indicate that the mean half-life is higher for those on a normal diet than those on a high- fiber diet. Assume that treatments were assigned at random and the two plasma half-life distributions are normal. Test the appropriate hypotheses using \(\alpha=.01\). Two-sample T for normal ws high \(\begin{array}{lllrr} & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text { Normal } & 6 & 27.70 & 5.44 & 2.2 \\ \text { High } & 7 & 18.61 & 5.55 & 2.1 \\ 95 \% & \text { C.I. for mu normal }-m u \text { high: }(2.3,15.9) & \end{array}\) T-Test mu normal = mu high(vs >): \(\mathrm{T}=2.97 \mathrm{P}=0.0070 \mathrm{DF}=10\)

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