Question 257321: This is a question about V-chips. The Kaiser Family Foundation conducted a study regarding parents' use of V-chips for controlling their children's television viewing. The study asked parents who own TVs equipped with V-chips whether they use the V-chips to block programming with questionable content.
a. Suppose 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% of parents polled used their V-chips. If the poll surveyed 1,000 parents, and if for the sake of argument we assume that 20% 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)I tried to get the minus sign or underscore in there. Sorry.
What I have done so far is to subract .17 from .20 to get 0.03. I am now stuck.
Help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! This is a question about V-chips. The Kaiser Family Foundation conducted a study regarding parents' use of V-chips for controlling their children's television viewing. The study asked parents who own TVs equipped with V-chips whether they use the V-chips to block programming with questionable content.
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a. Suppose 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.
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The study actually found that 17% of parents polled used their V-chips. If the poll surveyed 1,000 parents, and if for the sake of argument we assume that 20% 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.
is, calculate P(p<_.17)
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The mean of the distribution is 0.20 or 20%
The standard deviation of the dist. is sqrt[0.20*0.80/1000] = 0.01265...
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z(0.17) = (0.17-0.20)/0.01265 = -0.03/0.01265 = -2.3715...
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P(p <- 0.17) = P(z <= -2.3715) = 0.008857...
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Cheers,
Stan H.
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