A paper summarizes data from a study of food waste in the United States. Data from a representative sample of adult
Americans resulted in a mean food waste per day of 417 grams. For purposes of this exercise, assume that the sample size was 500 and that the sample standard deviation was 533 grams
Construct a 98% confidence interval for the mean food waste weight (in grams) for adult Americans. (Enter your answer using interval notation. Round your answer to two decimal places.)
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To construct a 98% confidence interval, we can use the formula:
Confidence Interval = sample mean ± Z * (sample standard deviation / √n)
where:
- sample mean = 417 grams
- sample standard deviation = 533 grams
- n = sample size = 500
- Z is the Z score associated with the desired confidence level (98% or 0.98)
To find the Z score, we can use a Z-table or a calculator. For a confidence level of 98%, the Z score is approximately 2.33.
Confidence Interval = 417 ± 2.33 * (533 / √500)
Calculating the square root of 500:
√500 ≈ 22.36
Confidence Interval ≈ 417 ± 2.33 * (533 / 22.36)
Calculating the division:
≈ 417 ± 2.33 * 23.84
Calculating the multiplication:
≈ 417 ± 55.55
The lower bound of the confidence interval:
≈ 417 - 55.55 ≈ 361.45
The upper bound of the confidence interval:
≈ 417 + 55.55 ≈ 472.55
Thus, the 98% confidence interval for the mean food waste weight (in grams) for adult Americans is approximately [361.45, 472.55].
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