Question
Determine the sample size required to estimate a population proportion to within 0.032 with 95.7% confidence, assuming that you have no knowledge of the approximate value of the sample proportion.
Sample Size =
B. Repeat part the previous problem, but now with the knowledge that the population proportion is approximately 0.33.
Sample Size =
i don't understand how you can do this without the margin or error.
Sample Size =
B. Repeat part the previous problem, but now with the knowledge that the population proportion is approximately 0.33.
Sample Size =
i don't understand how you can do this without the margin or error.
Answers
Try this formula:
n = [(z-value)^2 * p * q]/E^2
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is 0.032. Z-value is found using a z-table (for 95.7% confidence).
For B, use 0.33 for p and 0.67 for q.
Note: q = 1 - p
I'll let you take it from here.
n = [(z-value)^2 * p * q]/E^2
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is 0.032. Z-value is found using a z-table (for 95.7% confidence).
For B, use 0.33 for p and 0.67 for q.
Note: q = 1 - p
I'll let you take it from here.
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 11 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.30 gram.
Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.09 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.09 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
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