In a random sample of 184 college students, 97 had part-time jobs. Find the margin of error for the 95% confidence interval used to estimate the population proportion. 0.06490.12600.07210.0027

Answer:0.0721Step-by-step explanation:A 95% confidence interval for the population proportion is calculated using the formula;[tex]p+/-1.96\sqrt{\frac{p(1-p)}{n} }[/tex]In this case p is the sample proportion, calculated as ;number of successes/sample size. It is the equivalent of the probability of success in a binomial distribution. This implies that the sample proportion in this case will be; 97/184The value 1.96 is the confidence coefficient associated with a 95% confidence  interval for the population proportion since the sample proportions under the central limit theorem are assumed o be normally distributed.n is simply the sample size, 184 college students.Now, the expression [tex]1.96\sqrt{\frac{p(1-p)}{n} }[/tex] is what is termed as the margin of error for a 95% confidence interval of the population proportion. We simply substitute the values given into the expression and simplify it.[tex]1.96\sqrt{\frac{\frac{97}{184} (1-\frac{97}{184} )}{184} }\\ =0.0721[/tex]