In: Statistics and Probability
Use the sample data and confidence level given below to complete parts (a) through (d).
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll,
n equals 931n=931
and
x equals 596x=596
who said "yes." Use a
99 %99%
confidence level.
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Click the icon to view a table of z scores.
a) Find the best point estimate of the population proportion p.
. 552.552
(Round to three decimal places as needed.)
b) Identify the value of the margin of error E.
Eequals=. 511.511
(Round to three decimal places as needed.)
c) Construct the confidence interval.
nothing less than p less than<p<nothing
(Round to three decimal places as needed.)
d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.
A.
One has
9999%
confidence that the sample proportion is equal to the population proportion.
B.
9999%
of sample proportions will fall between the lower bound and the upper bound.
C.
One has
9999%
confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
D.
There is a
9999%
chance that the true value of the population proportion will fall between the lower bound and the upper bound.
More
Given that x=596 n=931
a) Find the best point estimate of the population proportion p.
p = x / n = 596/ 931= 0.640
#b) Identify the value of the margin of error E.
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.990 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.58
Margin of error = E = Z / 2 * ((P * (1 - P)) / n)
= 2.58 * (((0.640 * 0.360) / 931) = 0.020
#Margin of error=0.020
c) Construct the confidence interval.
A 99 % confidence interval for population proportion p is ,
P - E < P <P + E
0.640 - 0.020 < p < 0.640+ 0.020
0.620 < p < 0.660
d)correctly interprets the confidence interva
c) One has 99%confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.