SOLUTION: You randomly select 650 home-accident deaths and find that 117 of them are caused by falls. Construct the 99% confidence interval for the true population proportion of all of the h

Question 1192816: You randomly select 650 home-accident deaths and find that 117 of them are caused by falls. Construct the 99% confidence interval for the true population proportion of all of the home-accident deaths caused by falls.

Found 2 solutions by Boreal, math_tutor2020:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
p hat=117/650=0.18
99% half-interval is z(0.995)*sqrt(0.18*0.82/650)
=2.576*0.0151
=0.0388
the interval is (0.1412, 0.2188)
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

n = 650 = sample size
x = 117 = number of falls
phat = x/n = 117/650 = 0.18

At 99% confidence, we have z = 2.576 as the approximate critical value.
This value is used often in confidence intervals that you should memorize it, or have it handy on a reference table.

E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 2.576*sqrt(0.18*(1-0.18)/650)
E = 0.03881792879981
E = 0.0388
which is approximate

L = lower boundary
L = phat - E
L = 0.18 - 0.0388
L = 0.1412

U = upper boundary
U = phat + E
U = 0.18 + 0.0388
U = 0.2188

Answer: (0.1412, 0.2188)
The boundary values are approximate because the value of E was approximate. Round however instructed.

An alternative way to express the confidence interval is to write 0.1412 < p < 0.2188
This shows that we're 99% confident the population proportion (p) is somewhere between 0.1412 and 0.2188
In other words, we're 99% confident the population percentage is between roughly 14.12% and roughly 21.88%

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