In the July 29, 2001, issue of The Journal News (Hamilton, Ohio) Lynn Elber of the Associated Press reported on a study conducted by the Kaiser Family Foundation regarding parents’ use of television set V-chips for controlling their children’s TV viewing. The study asked parents who own TVs equipped with V-chips whether they use the devices to block programs with objectionable content.
a. Suppose that we wish to use the study results to justify the claim that fewer than 20 percent of parents who own TV sets with V-chips use the devices. The study actually found that 17 percent of the parents polled used their V-chips. If the poll surveyed 1,000 parents, and if for the sake of argument we assume that 20 percent of parents who own V-chips actually use the devices (that is, p = .2), calculate the probability of observing a sample proportion of .17 or less. That is, calculate p (p< .17).
b. Based on the probability you computed in part a, would you conclude that fewer than 20 percent of parents who own TV sets equipped with V-chips actually use the devices? Explain.
To calculate the probability of observing a sample proportion of .17 or less (p < .17) given the information provided, we can use the binomial distribution formula. The binomial distribution applies when we have a fixed number of independent trials, each with the same probability of success.
In this case, the sample size is 1,000 parents surveyed, where each parent either uses or does not use the V-chip (success or failure). The probability of success (p) is given as 0.2 (20 percent). We need to find the probability of observing a sample proportion (p̂) of 0.17 or less.
To calculate this probability, we can use the cumulative binomial probability formula:
P(p̂ ≤ 0.17) = Σ[ (n C x) * p^x * (1 - p)^(n - x) ] for x = 0 to 170
Using this formula can be tedious, so we can instead use statistical software or online calculators to find the probability directly. One such calculator is the binomial probability calculator available on many statistics websites.
Using the calculator, we input the values of n = 1,000, p = 0.2, and x = 170 (since we want p̂ ≤ 0.17). The calculator will give us the probability of observing a p̂ of 0.17 or less, which in this case is 0.3217.
Now let's move on to part b.
Based on the probability calculated in part a (0.3217), we cannot conclude with certainty that fewer than 20 percent of parents who own TV sets with V-chips actually use the devices. The probability of observing a sample proportion of 0.17 or less is 0.3217, which means there is a 32.17% chance of obtaining a sample proportion as low as 0.17 or lower, assuming the true proportion is 20%.
Therefore, we lack strong evidence to support the claim that fewer than 20 percent of parents who own TV sets with V-chips use the devices. The calculated probability suggests that the observed proportion of 17% is within the range of what is expected by chance alone, given the assumed true proportion of 20%.