Posted by Anonymous on Sunday, October 17, 2010 at 9:32pm.
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 Vchips for controlling their children’s TV viewing. The study asked parents who own TVs equipped with Vchips 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 Vchips use the devices. The study actually found that 17 percent of the parents polled used their Vchips.2 If the poll surveyed 1,000 parents, and if for the sake of argument we assume that 20 percent of parents who own Vchips 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 Vchips actually use the devices? Explain.

statistics  NIKI, Sunday, October 17, 2010 at 9:35pm
Part A:
Get the z score:
z = (phatp)/sqrt(p*(1p)/N)
z = (0.170.2)/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.