A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1082 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.53 hours with a standard deviation of 0.71 hour. a. Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day
b. The nutritionist is 95% confident that the amount of time spent eating or drinking per day for any individual is betweenandhours.
c. There is a 95% probability that the mean amount of time spent eating or drinking per day is between and hours.
d. The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day is betweenand

a) The 95% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day is (1.49, 1.57).

d) The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day for any individual is between 1.49 and 1.57 hours.

(c and b can not be concluded from the confidence interval)

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=1.53.

The sample size is N=1082.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

The degrees of freedom for this sample size are:

The t-value for a 95% confidence interval and 1081 degrees of freedom is t=1.962.

The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the mean is (1.49, 1.57).

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step-by-step explanation:

c

step-by-step explanation: