Katrina wants to estimate the proportion of adults who read at least 10 books last year. to do​ so, she obtains a simple random sample of 100 adults and constructs a​ 95% confidence interval. matthew also wants to estimate the proportion of adults who read at least 10 books last year. he obtains a simple random sample of 400 adults and constructs a​ 99% confidence interval. assuming both katrina and matthew obtained the same point​ estimate, whose estimate will have the smaller margin of​ error? justify your answer.

Answer:D. Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidenceStep-by-step explanation:In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]In whichz is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].The margin of error is the division by 2 of the upper level of the interval [tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex] by the lower level [tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex], so this is[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]Both Katrina and Matthew are working with the same population, so [tex]\pi[/tex] is equal for both.KatrinaKatrina wants to estimate the proportion of adults who read at least 10 books last year. To do​ so, she obtains a simple random sample of 100 adults and constructs a​ 95% confidence interval. This means that she has [tex]z = 1.96[/tex] and [tex]n = 100[/tex]. So her margin of error is[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex][tex]M = (1.96)\sqrt{\frac{\pi(1-\pi)}{100}}[/tex][tex]M = 0.196\sqrt{\pi(1-\pi)}[/tex]MatthewHe obtains a simple random sample of 400 adults and constructs a​ 99% confidence interval.This means that, for Matthew, [tex]z = 2.575, n = 400[/tex]. So his margin of error is:[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex][tex]M = (2.575)\sqrt{\frac{\pi(1-\pi)}{20}}[/tex][tex]M = 0.12875\sqrt{\pi(1-\pi)}[/tex]Higher confidence levels usually led to higher margins of error. However, Matthew's larger sample size compensates his higher confidence level.So the correct answer is:D. Matthew's estimate will have the smaller margin of error because the larger sample size more than compensates for the higher level of confidence