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How much time do americans spend eating or drinking? suppose for a random sample of 101 Americans, the mean time eating or drinking per day is 73.2 minutes with a simple standard deviation of 39 minutes. construct and interpret a 98% confidence interval for the mean amount of time ameri- cans spend eating or drinking per day.



Answer :

A 98% confidence interval for Americans' average daily eating and drinking time is (64.172, 82.228)

Given data,

How long do Americans spend drinking or eating? Let's say that, for a representative sample of 101 Americans, the daily average for eating or drinking is 73.2 minutes, with a straightforward standard deviation of 39 minutes. Create and analyze a 98% confidence interval for Americans' average daily eating and drinking time.

Let,

c = 98% = 0.98

n = 101

Mean μ = 73.2

Standard Deviation σ = 39

Now,

To analyze a 98% confidence interval for Americans' average daily eating and drinking time, we have;

CI = μ ± Z * σ/√n

CI = 73.2 ± 2.326 * 39/√101

    = 73.2 ± 9.028

    = 64.172, 82.228

Hence, the average daily calorie intake for Americans is within a 98% confidence zone (64.172, 82.228).

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