# 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.

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