I believe the answer is c but i cant figure out the rest A n

I believe the answer is c but i cant figure out the rest

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1093 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.56 hours with a standard deviation of 0.59 hour. Complete parts (a) through (d) below. (b) In 2010, there were over 200 million people nationally age 15 or older. Explain why this, along with the fact that the data were obtained using a random sample, satisfies the requirements for constructing a confidence interval. A.The sample size is less than 5% of the population. B. The sample size is greater than 10% of the population. C. The sample size is less than 10% of the population. D. The sample size is greater than 5% of the population. (c) Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. Select the correct choice below and fill in the answer boxes, if applicable, in your choice. (Type integers or decimals rounded to two decimal places as needed. Use ascending order.) A. The nutritionist is 95% confident that the amount of time spent eating or drinking per day for any individual is between [ ] and [ ] hours. B. There is a 95% probability that the mean amount of time spent eating or drinking per day is between [ ] and [ ] hours. C. The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day is between [ ] and [ ] hours. D. The requirements for constructing a confidence interval are not satisfied.

Solution

The sample size is less than 5% of the population

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Option B is right

Std error for sample = std dev / rt n = 0.59/rt 200

= 0.04171

Margin of error for 95% = 1.96(0.04171)

= 0.08177

Conf interval lower bound = 1.56-0.08177 = -1.4782

Upper bound = 1.64177