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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 count, 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 these samples are representative of avid mountain bikers and nonbikers. a. Using a confidence level of \(95 \%,\) estimate the difference between the proportion of avid mountain bikers with low sperm count and the proportion for nonbikers. b. Is it reasonable to conclude that mountain biking 12 hours per week or more causes low sperm count? Explain.

Short Answer

Expert verified
a. We estimate, with a 95% confidence level, that the difference in the proportion of avid mountain bikers with low sperm count, and nonbikers, is between 0.5092 and 0.7708. b. Yet, it is not reasonable to conclude that mountain biking 12 hours or more per week causes low sperm count based solely on these data. Correlation does not imply causation, and other variables may also influence these results.

Step by step solution

01

Identify the Proportions

Identify the specific proportions given in the problem. Here, it's stated that 90% of the avid mountain bikers studied had low sperm count, compared to 26% of the nonbikers. These can be written as proportions \(p_1 = 0.90\) and \(p_2 = 0.26\).
02

Calculation of Difference

Calculate the difference between the two proportions. The difference \(d\) can be calculated as \(d = p_1 - p_2 = 0.90 - 0.26 = 0.64\). This value represents the observed difference in proportions.
03

Compute the Standard Error

We calculate the standard error (SE) using the formula: \(SE = \sqrt{[p_1(1 - p_1) / n_1] + [p_2(1 - p_2) / n_2]}\). Here, \(n_1\) and \(n_2\) represent the sample size for both groups, which in this case is 100 for both. This gives us: \(SE = \sqrt{(0.9 * 0.1 / 100) + (0.26 * 0.74 / 100) } = 0.067\).
04

Compute the Confidence Interval

We now use the standard normal value for 95% confidence, which is 1.96, to compute the confidence Interval (CI). Using the formula CI = \(d \pm\) Z * SE), where Z is the standard score, we have: CI = \(0.64 \pm 1.96 * 0.067\). This gives an interval from 0.5092 to 0.7708. This means we are 95% confident that the difference in proportions of low sperm count between avid mountain bikers and nonbikers is between 0.5092 and 0.7708.
05

Analyzing the Causation

When interpreting the results of statistical studies, it’s important to distinguish between correlation and causation. The data from the study shows a strong correlation between mountain biking for 12 hours or more per week and low sperm count. However, these results alone do not imply causation. While the observed effect might be due to mountain biking, it could also be the result of confounding variables not accounted for in this study. So, it's not reasonable to conclude from the study that mountain biking 12 hours per week or more causes low sperm count solely based on these results. Further studies, possibly experimental in nature, are needed to clearly establish causation.

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Key Concepts

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

Confidence Interval Calculation
Understanding confidence intervals is crucial when interpreting study results. A confidence interval gives a range of values within which we can expect the true population parameter to lie, given a certain level of confidence. For instance, in our mountain biking exercise, a 95% confidence interval was calculated to estimate the difference in low sperm count between avid mountain bikers and nonbikers.

The process involved five steps:
  • Identifying the sample proportions.
  • Calculating the observed difference in these proportions.
  • Computing the standard error (SE).
  • Using the standard normal distribution (Z-score) and the SE to calculate the interval.
  • Interpreting the interval, while understanding that it gives a range in which the true difference likely exists.
It's important to note that the wider the interval, the less precise our estimate is. Conversely, a narrower interval offers a more precise estimate but requires a larger sample size or a smaller level of confidence.
Difference in Proportions
In studies comparing two groups, analysts may be interested in the difference in proportions — essentially, how much the proportion of some characteristic varies from one group to another. In the mountain biking study, the proportion of low sperm count in avid mountain bikers (90%) was compared to that of nonbikers (26%).

The calculation of this difference is straightforward: substract one proportion from the other. The challenge often lies in interpreting this difference meaningfully. For instance, a big numerical divergence might sound alarming, but without a proper confidence interval, we cannot assess the precision of this estimated difference. When the interval is very wide, it means that the actual difference might be lesser or greater, and this uncertainty should be acknowledged when making conclusions.
Correlation vs Causation
A common dilemma in studies is distinguishing between correlation and causation. Correlation simply means that two variables are related, whereas causation implies that one variable directly affects another. In our exercise, the high proportion of low sperm count among avid mountain bikers relative to nonbikers does not conclusively establish that the mountain biking causes the condition.

Many factors come into play when deciding causation:
  • Temporal precedence - the cause must precede the effect.
  • There are no alternative explanations - other potential causes must be ruled out.
  • The relationship is not spurious - it is not due to a confounding variable.
Researchers often conduct controlled experiments to establish causality. Without such robust study designs, we can only infer that there is a correlation, but cannot assert causation.

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Most popular questions from this chapter

The report referenced in the previous exercise also stated that the proportion who thought their parents would help with buying a house or renting an apartment for the sample of young adults was 0.37 . For the sample of parents, the proportion who said they would help with buying a house or renting an apartment was 0.27 . Based on these data, can you conclude that the proportion of parents who say they would help with buying a house or renting an apartment is significantly less than the proportion of young adults who think that their parents would help?

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Common Sense Media surveyed 1,000 teens and 1,000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace ("Teens Show, Tell Too Much Online," San Francisco Chronicle, August 10,2009 ). The two samples were independently selected and were chosen to be representative of American teens and parents of American teens. When asked if they check online social networking sites more than 10 times a day, 220 of the teens surveyed said yes. When parents of teens were asked if their teen checked networking sites more than 10 times a day, 40 said yes. Use a significance level of 0.01 to determine if there is convincing evidence that the proportion of all parents who think their teen checks social networking sites more than 10 times a day is less than the proportion of all teens who check more than 10 times a day.

The article "Portable MP3 Player Ownership Reaches New High" (Ipsos Insight, June 29,2006 ) reported that in 2006 , \(20 \%\) of those in a random sample of 1,112 Americans ages 12 and older indicated that they owned an MP3 player. In a similar survey conducted in \(2005,\) only \(15 \%\) reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1,112 . a. Are the sample sizes large enough to use the largesample confidence interval for a difference in population proportions? b. Estimate the difference in the proportion of Americans ages 12 and older who owned an MP3 player in 2006 and the corresponding proportion for 2005 using a \(95 \%\) confidence interval. c. Is zero included in the confidence interval? What does this suggest about the change in this proportion from 2005 to \(2006 ?\) d. Interpret the confidence interval in the context of this problem.

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