Answer:
Step-by-step 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; [tex]\frac{p(1-p)}{n}[/tex])
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' ± [tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex]
Where [tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }[/tex] is the margin of error
In this case [tex]Z_{1-\alpha /2}= Z_{0.995}= 2.586[/tex]
[tex]Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }= 2.586*\sqrt{\frac{0.56*0.44}{929} }= 0.04[/tex]
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!
