Answer:
Explanation:
Hello!
The variable of interest is
X: Number of people that feel vulnerable to identity theft in a sample of 929.
This variable is discrete and has a binomial distribution. X~Bi(n;p)
The parameter of interest is the population proportion of people that feel vulnerable to identity theft.
To calculate the 99% CI for the population proportion you have to use the approximate distribution to normal for the sample proportion p'≈N(p;
)
a. The best point estimate for p is the sample proportion p' you calculate it as:
p'= x/n= 523/929= 0.56
b. The formula for the confidence interval is
p' ±

Where
is the margin of error
In this case


c. Then the interval is
0.56 ± 0.04
[0.52;0.6]
d.
With a 99% confidence level, you can expect that the interval [0.52;0.6] will include the true value of the proportion of people that feel vulnerable to identity theft.
The correct answer is
3. there is a 99% chance that the true value of the population proportion will fall between the lower bound and the upper bound.
I hope this helps!