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.

Part A:

Get the z score:

z = (phat-p)/sqrt(p*(1-p)/N)

z = (0.17-0.2)/sqrt(0.2*0.8/1000)

z = -2.3717

p(z < -2.3717)

= 0.0089





Part B:

Since that probability is extremely low, we can conclude that less than 20% actually use the chips.

To calculate the probability, we can use the binomial distribution formula.

a) The formula for the probability of observing a sample proportion of 0.17 or less is:

P( pˆ ≤ 0.17) = Σ [ nCk * p^k * (1-p)^(n-k) ] for k = 0 to 170

where n is the sample size (1,000 in this case), p is the assumed proportion (0.2 in this case), and pˆ is the observed sample proportion (0.17 in this case).

To calculate this probability, we need to evaluate the summation for each value of k from 0 to 170 and sum them up:

P( pˆ ≤ 0.17) = Σ [1000Ck * (0.2)^k * (0.8)^(1000-k) ] for k = 0 to 170

This calculation can be quite complex, but it can be approximated using statistical software or calculators. The result of this calculation is the probability of observing a sample proportion of 0.17 or less.

b) Based on the probability calculated in part a, if the probability is very small (let's say less than 0.05), then we can conclude that it is unlikely for fewer than 20 percent of parents who own TV sets with V-chips actually use the devices. If the probability is greater than 0.05, then we would not have enough evidence to conclude that fewer than 20 percent of parents use V-chips.

In this case, we would need to compare the probability calculated in part a to a significance level (usually denoted as α) to make a conclusion. If the probability is less than α, we reject the claim of fewer than 20 percent of parents using V-chips, otherwise, we fail to reject the claim.

To calculate the probability of observing a sample proportion of .17 or less (P(p̂ ≤ .17)), we can use the normal approximation to the binomial distribution.

a) In this case, the sample proportion (p̂) is .17, and we assume that 20% of parents who own V-chips actually use the devices (p = .2). We are given that the sample size (n) is 1,000 parents.

To apply the normal approximation, we need to check two conditions:
1) np = n * p = 1000 * .2 = 200 ≥ 10
2) n(1-p) = n * (1-.2) = 1000 * .8 = 800 ≥ 10

Both conditions are satisfied, so we can use the normal approximation.

To calculate the probability, we will standardize the sample proportion:
Z = (p̂ - p) / √((p * (1-p)) / n)

Z = (.17 - .2) / √((.2 * .8) / 1000)

Calculating this, we find that Z ≈ -2.12.

Next, we need to find the probability of Z being less than or equal to -2.12. This can be done using a standard normal distribution table or a calculator that provides the cumulative probability function.

From the table or calculator, we find that P(Z ≤ -2.12) ≈ 0.016.

So, the probability of observing a sample proportion of .17 or less is approximately 0.016 or 1.6%.

b) Based on the probability calculated in part a, we can conclude that fewer than 20 percent of parents who own TV sets equipped with V-chips actually use the devices. This is because the probability is very low (approximately 1.6%), suggesting that a sample proportion of .17 or less is unlikely to occur if the true proportion of parents using the devices is 20%. However, it's important to note that statistical significance does not necessarily imply practical significance. Further analysis and additional evidence may be needed to draw stronger conclusions about the actual usage of V-chips by parents.