SOLUTION: Suppose that a market research firm is hired to estimate the percent
of adults living in a large city who have cell phones. 600 randomly selected
adult residents in this city are
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-> SOLUTION: Suppose that a market research firm is hired to estimate the percent
of adults living in a large city who have cell phones. 600 randomly selected
adult residents in this city are
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Question 1150274: Suppose that a market research firm is hired to estimate the percent
of adults living in a large city who have cell phones. 600 randomly selected
adult residents in this city are surveyed to determine whether they have cell
phones. Of the 600 people surveyed, 451 responded yes - they own cell phones.
Using a 90% confidence level, compute a confidence interval estimate for the true proportion of adults residents of this city who have cell phones and interpret
the confidence interval. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! 90% CI has z=1.645
p hat is 451/600=0.752
half interval for a 1 sample proportion is z*sqrt(p*(1-p)/n); sqrt (0.751*0.249/600)=0.01765
z*se=0.029
90% CI=(0.723, 0.781)
We don't know the true value, but we can be 90% confident it is in that interval.
