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A random sample of 85 adults found that the average calorie consumption was 2100 per day. Previous research has found a standard deviation of 450 calories, and you use this value for \sigma σ. A researcher wants to estimate a 95% confidence interval and is willing to accept a margin of error of \pm 50 ± 50 calories. She knows it will cost $50 to survey each member of the sample. Given this information, how much will it cost to survey the minimum number of people?

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Answer:

$15600

Step-by-step explanation:

Minimum number of samples can be calculated using the formula:

N≥[tex](\frac{z*s}{ME})^2[/tex] where

  • N is the sample size
  • z is the corresponding z-score for 95% confidence level (1.96)
  • s is the standard deviation of the calorie consumption (450 cal.)
  • ME is the margin of error that the researcher is willing to accept. (50 cal)

N≥[tex](\frac{1.96*450}{50})^2[/tex] =311.2

Therefore, minimum required sample size for 50 cal standard error in 95% confidence is 312

Since the survey costs $50 for each person then the survey costs minimum

50*312=$15600

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