In a poll conducted, 444 of 1009 adults in a certain country polled said they were "picky eaters." Complete parts (a) through (d) below. a. What proportion of the respondents said they were picky eaters? 0.44 (Round to two decimal places as needed.) b. Find a 95% confidence interval for the population proportion of adults in the country who say they are picky eaters. Assume the poll used a simple random sample (SRS). (In fact, it used random sampling, but a more complex method than SRS.) A 95% confidence interval for the population proportion is ( 0.41, 0.47). (Round to two decimal places as needed.) c. Would a 90% confidence interval based on this sample be wider or narrower than the 95% interval? Give a reason for your answer. OA. A 90% confidence interval would be wider than the 95% interval because a 90% confidence interval includes less of the sampling distribution. B. A 90% confidence interval would be narrower than the 95% interval because a 90% confidence interval includes less of the sampling distribution. OC. A 90% confidence interval would be narrower than the 95% interval because a 95% confidence interval includes less of the sampling distribution. O D. A 90% confidence interval would be wider than the 95% interval because a 95% confidence interval includes less of the sampling distribution. d. Construct the 90% confidence interval. Was your conclusion in part (c) correct? A 90% confidence interval for the population proportion is ( (Round to three decimal places as needed.)