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February 28, 2015

February 28, 2015

Posted by **Anonymous** on Sunday, October 17, 2010 at 9:32pm.

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.

- statistics -
**NIKI**, Sunday, October 17, 2010 at 9:35pmPart 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.

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