15. On February 20, 2018 ncbnews.com ran a story titled "Poll: Support for gun control hits record high." This article references a survey performed by Quinnipiac University of 1,351 voters nationwide and of those sampled 883 stated that they wantstricter gun control laws. Assume that the survey utilized a random sample of all Americans.Calculate 90% confidence interval for the proportion of all Americans that want stricter gun control laws.a. (0.6323, 0.6749)Ob. (0.6282, 0.6790)Oc. (0.6202, 0.6869)Od. (0.6370, 0.6702)

Confidence interval is an interval such that the probability of the unknown parameter to be included in the interval is equal to a preassigned probability, termed as confidence coefficient. In simple language, a 100(1 - α)% confidence interval has the probability to include the unknown parameter with probability (1 - α).A 100(1 – α)% confidence interval for population proportion is given by, where n is the number of observations, p^ is the sample proportion, Zα is the value of the standard normal distribution for α level of significance. Here we are given, Also Now, from standard normal distribution table we have, Therefore, a 90% confidence interval for the proportion of all Americans that want stricter gun control laws is Answer: The correct option is: First Option. Therefore, all other options are incorrect.