A recent survey by the Gallup organization reported that 66% of U.S. adults view

Question

A recent survey by the Gallup organization reported that 66% of U.S. adults view the restaurant industry positively. A researcher for a public relations firm in a large metropolitan area hears this and decides to replicate the survey for a random sample of 250 adults in their city.
a. Assuming that the percentage in this city is the same as nationally, describe the mean, standard deviation, and shape of the sampling distribution for the sample proportion of adults who view the restaurant industry positively.
b. Calculate the probability that the sample proportion will be more than 0.72.
c. Suppose the sample proportion is more than 0.72. Does this mean more adults in this
city view the restaurant industry positively than the percentage reported by Gallup?
Explain why or why not using probabilities.

Explanation / Answer

a) mean of  sample proportion =0.66

for n=250 ; std deviation =(p(1-p)/n)1/2 =0.0300

as np >10 and n(1-p)>10 ; therefore shape of the sampling distribution for the sample proportion will be approximately normal

b) P(P>0.72) =1-P(P<0.72)= 1-P(Z<(0.72-0.66)/0.03)=1-P(Z<2 )=1-0.9772 =0.0228

c)

as above probability is less than 0.05 ; therefore above event comes under range of unusual events and therefore we an say that more adults in this city view the restaurant industry positively than the percentage reported by Gallup

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