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. 2 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 (pˆ �� .17).
B. Based on the probability you computed in part a, would you conclude that fewer than 20 per- cent 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 (P̂ ≤ .17), we can use the binomial distribution formula. The formula for calculating the probability of a binomial distribution is:
P(x) = (nCx)(p^x)(q^(n-x))
Where:
P(x) is the probability of getting exactly x successes in n trials
n is the number of trials (sample size)
x is the number of successes (in this case, the number of parents who use V-chips)
p is the probability of success in a single trial (the assumed proportion of parents who use V-chips)
q is the probability of failure in a single trial (1 - p)
Given that the study surveyed 1,000 parents (n = 1000) and for the sake of argument we assume that 20 percent of parents who own V-chips actually use the devices (p = 0.2), we can calculate the probability.
Using a statistical software or calculator, the probability can be calculated using the cumulative distribution function (CDF) of the binomial distribution.
P(p̂ ≤ 0.17) = P(x ≤ 0.17 * n) = P(x ≤ 0.17 * 1000)
Calculating this probability will give us the answer to part a of the question.
In part b, we need to interpret the probability calculated in part a to determine if we can conclude that fewer than 20 percent of parents who own TV sets equipped with V-chips actually use the devices.
If the probability calculated in part a is very small (close to 0), it means that observing a sample proportion of 0.17 or less is highly unlikely assuming that the true proportion is 0.2. In this case, it would be unlikely that fewer than 20 percent of parents use V-chips.
On the other hand, if the probability is relatively high (close to 1), it means that observing a sample proportion of 0.17 or less is likely, and it would support the claim that fewer than 20 percent of parents use V-chips.
So, based on the probability calculated in part a, we can make a conclusion about the claim.