# A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1065 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.62 hours with a standard deviation of 0.51 hour.

(1)Determine and interpret a 95% confidence interval for the mean amount of time Americans age 15 or older spend eating/drinking each day.

(A)The nutritionest is 95% confident that the mean amount of time spent eating or drinking per day for any is between ___, and____ hours.

## Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.025) and its Z score.

95% = mean ± 1.96 SEm

SEm = SD/√n

## To determine the 95% confidence interval for the mean amount of time Americans age 15 or older spend eating/drinking each day, we can use the formula:

Confidence Interval = Mean ± (Z * (Standard Deviation / √(Sample Size)))

Where:
- Mean is the sample mean amount of time spent eating or drinking per day (1.62 hours).
- Z is the Z-score corresponding to the desired level of confidence (in this case, 95% confidence, so Z = 1.96).
- Standard Deviation is the population standard deviation (0.51 hours).
- Sample Size is the number of people in the sample (1065).

Plugging in the values:

Confidence Interval = 1.62 ± (1.96 * (0.51 / √1065))

Calculating the square root of the sample size:

√1065 ≈ 32.6

Confidence Interval = 1.62 ± (1.96 * (0.51 / 32.6))

Calculating the value inside the parentheses:

(0.51 / 32.6) ≈ 0.0157

Confidence Interval = 1.62 ± (1.96 * 0.0157)

Calculating the multiplication:

(1.96 * 0.0157) ≈ 0.0308

Confidence Interval = 1.62 ± 0.0308

Calculating the lower and upper limits of the confidence interval:

Lower Limit = 1.62 - 0.0308 ≈ 1.5892
Upper Limit = 1.62 + 0.0308 ≈ 1.6508

Therefore, the nutritionist is 95% confident that the mean amount of time spent eating or drinking per day for Americans age 15 or older is between 1.5892 and 1.6508 hours.

## To determine the 95% confidence interval for the mean amount of time Americans age 15 or older spend eating or drinking each day, we can use the formula:

Confidence Interval = mean ± (critical value * standard deviation / sqrt(sample size))

Step 1: Find the critical value.
Since the sample size is relatively large (n ≥ 30), we can use the Z-table to find the critical value for a 95% confidence level. The critical value for a 95% confidence level is approximately 1.96.

Step 2: Calculate the confidence interval.
Mean = 1.62 hours
Standard Deviation = 0.51 hour
Sample Size = 1065

Confidence Interval = 1.62 ± (1.96 * 0.51 / sqrt(1065))

Calculating this expression will give us the lower and upper bounds of the confidence interval.

Confidence Interval = 1.62 ± (1.96 * 0.51 / sqrt(1065))
Confidence Interval = 1.62 ± (1.96 * 0.0156)

Lower Bound = 1.62 - (1.96 * 0.0156)
Upper Bound = 1.62 + (1.96 * 0.0156)

Lower Bound ≈ 1.59 hours
Upper Bound ≈ 1.65 hours

Therefore, the 95% confidence interval for the mean amount of time Americans age 15 or older spend eating/drinking each day is approximately 1.59 to 1.65 hours.

(A) The nutritionist is 95% confident that the mean amount of time spent eating or drinking per day for any individual is between 1.59 and 1.65 hours.