Shargorodsky, Curhan, Curhan, and Eavey (2010) examined hearing loss data for 1771 participants from the National Health and Nutrition Examination Survey (NHANES, 2005-6), aged 12−19 years. (“NHANES provides nationally representative cross-sectional data on the health status of the civilian, non-institutionalized U.S. population.”) A standardized health exam was given along with the survey and through audiometric measurements, 333 teens were found to have at least some level of hearing loss (average hearing threshold level at least 15 dB in either ear). News of this study spread quickly, many blaming the prevalence of hearing loss on the higher use of ear buds by teens. At MSNBC.com (8/17/2010), Carla Johnson summarized the study with the headline “1 in 5 U.S. teens has hearing loss, study says.”

Suppose we want to test

H₀: π = 0.20

H_a: π ≠ 0.20

 

Because we are sampling without replacement from a finite population, technically the Binomial distribution is not the best model of the sampling process. Intead, we should use the hypergeometric distribution.

According to the 2008 U.S. Census, there were 21,469,780 teens in America between the ages of 15 and 19.

Let X represent the number of successes in a random sample of 1771 teens from a population with .20(21469780) = 4293956 successes.

Use JMP or R to find P(X ≤ 333)

• JMP: Open the Journal file and select Distribution Calculator. Use the Distribution pull-down menu to select Hypergeometric. Enter this population size, number of successes in the population ("Number of Items") and sample size.  Find P(X≤ 333).  
• R: iscamhyperprob(333, total = 21469780, succ = 4293956 , n = 1771, lower.tail = TRUE )

But this is a large population size, so compare this result to the one-proportion z-test.

Use R or JMP or Theory-Based Inference applet to find the two-sided p-value.  

Make sure in the output you upload it is clear how you found the p-value.

The American Veterinary Medical Association (AVMA) has conducted a Pet Demographics Survey about every five years since the early 1980s. In 2016, the sample of 50,347 households surveyed in Dec. 2015 reported 25.4% of households reporting owning at least one cat.  Or is that the cat(s) owning the home?

 

Suppose an alien lands on Earth, observes variations in gender among humans, and sets out to estimate the proportion of humans who are women. Fortunately, the alien took a good statistics course on its home planet and knows to take a sample and produce a confidence interval. Suppose the alien samples the members of the 2024 U.S. Senate and finds 25 women in the group.

• If the population size is large compared to the sample size (e.g., more than 20 times), then we can use the same binomial and normal methods as before. (In other words, don't worry about the hypergeometric distribution for now!)
• Keep in mind the difference between statistical significance and practical significance. With large sample sizes, sample proportions will vary little from sample to sample, and so even small differences (that may seem minor to most of us) will be statistically significant. Saying that a sample result is unlikely to happen by chance (and therefore is statistically significant) is not the same as saying the result is important or even noteworthy (practically significant), depending on the context involved.
• First, statistical tests and confidence intervals do not compensate for the problems of a biased sampling procedure. If the sample is collected from the population in a biased manner, the ensuing confidence interval method will be a biased, and potentially misleading, estimate of the population parameter of interest.
  • Keep in mind "bias" is a property of the method, not an individual sample.
• A second important point to remember is that confidence intervals and significance tests use sample statistics to estimate population or process parameters. When the data at hand constitute the entire population of interest (i.e., you have a census), then constructing a confidence interval from these data is meaningless.