Final answer:
To estimate the average amount of beef eaten by Americans to within ±6 pounds at a 90% confidence level, a consulting firm would need at least 43 people in their sample, based on a preliminary sample standard deviation of 23.7 pounds.
Step-by-step explanation:
To determine the sample size needed to estimate the average amount of beef eaten by all adult Americans to within ±6 pounds at a 90% confidence level, we use the formula to calculate sample size for estimating a mean:
n = (z* ÷ E)^2 ÷ (s^2)
Where:
- n = sample size
- z* = z-value corresponding to the desired confidence level
- E = margin of error
- s = sample standard deviation
For a 90% confidence level, the z-value is typically about 1.645. The margin of error (E) is ±6 pounds, and the sample standard deviation (s), as given in part (a) of the question, is 23.7 pounds. Plugging in these values:
n = (1.645 ÷ 6)^2 ÷ (23.7^2)
n = (0.2741667)^2 ÷ (561.69)
n = 0.075169 ÷ 561.69
n ≈ 42.364
Since we cannot have a fraction of a person, we would round up to the next whole number. Therefore, the consulting company should expect to need a sample size of at least 43 people to estimate the average amount of beef eaten by all adult Americans to the desired margin of error at a 90% confidence level.