# A research institute poll asked respondents if they felt vulnerable to identity theft. in the​ poll, and who said​ yes. use a confidence level.

Question: A research institute poll asked respondents if they felt vulnerable to identity theft. in the​ poll, and who said​ yes. use a confidence level.

Assuming that the poll asked a simple random sample of n respondents and that the responses were independent of each other, we can use a confidence level and the sample proportion to construct a confidence interval for the true proportion of people who feel vulnerable to identity theft in the population.

Let's say we want to construct a 95% confidence interval. We first need to calculate the sample proportion:

p̂ = x/n = 840/1200 = 0.7

where x is the number of respondents who said "yes" and n is the sample size.

Next, we can use the following formula to calculate the margin of error:

E = z*sqrt(p̂(1-p̂)/n)

where z is the z-score associated with the desired confidence level (in this case, 1.96 for a 95% confidence level), sqrt represents the square root function, and p̂ and (1-p̂) represent the proportion and complement of the proportion, respectively.

Plugging in the values, we get:

E = 1.96*sqrt(0.7(1-0.7)/1200) ≈ 0.032

Therefore, the 95% confidence interval for the true proportion of people who feel vulnerable to identity theft in the population is:

0.7 ± 0.032 or (0.668, 0.732)

This means that we are 95% confident that the true proportion of people in the population who feel vulnerable to identity theft falls within the range of 0.668 to 0.732. Rjwala is an educational platform, in which you get many information related to homework and studies. In this we also provide trending questions which come out of recent recent exams.