In a poll conducted, 439 of 1007 adults in a certain country polled said they were "picky eaters." A) what proportion of the respondents said they were picky eaters? B) A 95% confidence interval for the population proportion is ( , ) Question Help ▼In a poll conducted, 439 of 1007 adults in a certain country polled said they were "picky eaters." Complete parts (a) through (d) below.(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 usedrandom sampling, but a more complex method than SRS.)A 95% confidence interval for the population proportion is ( ).(Round to two decimal places as needed.)c. Would a 90% confidence interval based on this sample be wider or narower than the 95% interval? Give a reason for your answer.O A. A 90% confidence interval would be wider than the 95% interval because a 90% confidence interval includes less of the sampling distribution.Click to select your answer(s).Next« PreviousMacBook AirFSraF10F4&ETUO1PGK8734

Name: In a poll conducted, 439 of 1007 adults in a certain country polled sa
Uploaded: 2022-12-08T14:54:05.000Z
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Description: In a poll conducted, 439 of 1007 adults in a certain country polled said they were "picky eaters." A) what proportion of the respondents said they were picky eaters? B) A 95% confidence interval for the population proportion is ( , ) Question Help ▼I

a. The proportion of the respondents said they were picky eaters is, p^=xn=4391007=0.4359≈0.44 Thus,…

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In a poll conducted, 439 of 1007 adults in a certain country polled said they were "picky eaters." A) what proportion of the respondents said they were picky eaters? B) A 95% confidence interval for the population proportion is ( , )