Suppose a certain species bird has an average weight of xbar 3.40 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with sigma 0.29 grams. Find the sample size necessary for a 98% confidence level with a maximal error of estimate E=0.09 for the mean weights of the hummingbirds. (Points: 5)
To find the sample size necessary for a 98% confidence level with a maximal error of estimate E=0.09 for the mean weights of the hummingbirds, you can use the formula:
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 98% confidence level)
σ = standard deviation of the population
E = maximal error of estimate (also known as the margin of error)
First, let's calculate the Z-score using the standard normal distribution table. The confidence level of 98% means that we need to find the Z-score corresponding to a cumulative probability of 0.99. Looking up this value in the standard normal distribution table, we find that the Z-score is approximately 2.33.
Next, plug the known values into the formula and solve for n:
n = (2.33 * 0.29 / 0.09)^2
n ≈ (0.6757 / 0.09)^2
n ≈ 7.507^2
n ≈ 56.308
n ≈ 57 (rounded up to the nearest whole number)
Therefore, the sample size necessary for a 98% confidence level with a maximal error of estimate of 0.09 for the mean weights of the hummingbirds is approximately 57.
To find the sample size necessary for a 98% confidence level with a maximal error of estimate (E) of 0.09 for the mean weights of the hummingbirds, we can use the formula:
n = (Z * σ / E)²
where:
- n is the required sample size
- Z is the Z-score corresponding to the desired confidence level (in this case, 98%)
- σ is the standard deviation of the population (0.29 grams)
- E is the maximal error of estimate (0.09 grams)
Step 1: Find the Z-score
For a 98% confidence level, the Z-score can be found using a standard normal distribution table or a Z-score calculator. The Z-score for a 98% confidence level corresponds to 2.33.
Step 2: Plug values into the formula
Substitute the values into the formula:
n = (2.33 * 0.29 / 0.09)²
Step 3: Calculate the sample size
Calculate the value of n:
n = (0.6777 / 0.09)²
n = 7.53²
n ≈ 56.70
Step 4: Round up to the nearest whole number
Since the sample size should be a whole number, round up the result to the nearest whole number:
n = 57
Therefore, the sample size necessary for a 98% confidence level with a maximal error of estimate of 0.09 grams for the mean weights of the hummingbirds is 57.
Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 2.33 using a z-table to represent the 98% confidence interval, sd = 0.29, E = 0.09, ^2 means squared, and * means to multiply.
Your answer looks correct! Good job.
I did 2.33x.29/.09 then sqared and got 56.3 this good?
(ZcO/E)^2 =n is this right?