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In the July 29, 2001, issue of The Journal News (Hamilton, Ohio) Lynn Elber of the Associated Press reported on a study conduct...Asked by Anonymous
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.
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.
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Answered by
NIKI
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.
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.
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