A random sample of 85 adults found that average calorie consumption was 2,100 per day. previous research has found a standard deviation of 450 calories, and you assume this value for. a researcher has been given $10,000 to conduct a similar survey, which costs $50.00 per person surveyed. she must compute a 95% confidence interval. given her budget restriction, what would the minimum margin of error be for the confidence interval for the population mean?

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Business

Yates

19 February, 00:20

A random sample of 85 adults found that average calorie consumption was 2,100 per day. previous research has found a standard deviation of 450 calories, and you assume this value for. a researcher has been given $10,000 to conduct a similar survey, which costs $50.00 per person surveyed. she must compute a 95% confidence interval. given her budget restriction, what would the minimum margin of error be for the confidence interval for the population mean?

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Ada · Answer 1

Average consumption, p = 2,100

Standard deviation assumed from previous studies, sd = 450

Z (at 95% confidence interval) = 1.96

Amount of funds available = $10,000

Cost per person supplied = $50.00

Therefore,

Sample size, n = 10000/50 = 200

Minimum margin of error will be achieved when the sample size is bigger than 85 adults.

Margin error = Z*sd/Sqrt (n) = 1.96*450/Sqrt (200) = + / - 62.37

Additionally,

Lowest range = 2100-62.37 = 2,037.63

Highest range = 2100+62.37 = 2,162.37