A simple random sample of 250 Americans revealed that 112 of them owned a DVD player. DVD companies, as an industry, brag that half of all Americans own a DVD player.
Assuming that the DVD companies are correct, what is the probability of a sample of 250 Americans producing a sample proportion as low as the one from this sample?
Assuming that the DVD companies are correct, what is the probability of a sample of 250 Americans producing a sample proportion as low as the one from this sample?
-
Kristin
The sample proportion is equal to 112/250 = 0.448
Now convert to z-values ...
P( p < 0.448) = P[ z < (0.448 - 0.5) / sqrt (0.5 x 0.5 / 250)] = P( z < -1.644)
Now using a Standard Normal table ...
P( z < -1.644) = 0.05 or 5%
Hope that helped
The sample proportion is equal to 112/250 = 0.448
Now convert to z-values ...
P( p < 0.448) = P[ z < (0.448 - 0.5) / sqrt (0.5 x 0.5 / 250)] = P( z < -1.644)
Now using a Standard Normal table ...
P( z < -1.644) = 0.05 or 5%
Hope that helped