The Corporate Lawyer, a magazine for corporate lawyers, would like to report the mean amount earned by lawyers in their area of specialization. How large a sample is required if the 99 percent level of confidence is used and the estimate is to be within $1,200? The population standard deviation is $15,000.
Captain D’s tuna is sold in cans that have a net weight of 8.015 ounces. The weights are normally distributed with a mean of 8.015 ounces and a standard deviation of 0.11 ounces. You take a sample of 35 cans. Compute the probability that the sample would have a mean:
greater than 8.01 ounces?
less than 8.025 ounces?
between 7.995 and 8.04 ounces
Second problem:
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.
The Corporate Lawyer, a magazine for corporate lawyers, would like to report the mean amount earned by lawyers in their area of specialization. How large a sample is required if the 97 percent level of confidence is used and the estimate is to be within $2,500? The standard deviation is $16,000.
To find out the required sample size, you can use the formula for the sample size of a mean estimation.
The formula is given by:
n = (Z * σ / E)²
Where:
n = required sample size
Z = Z-score corresponding to the desired level of confidence
σ = population standard deviation
E = maximum margin of error (difference between the sample mean and the true mean)
In this case, we want a 99 percent level of confidence, which corresponds to a Z-score of 2.576 (obtained from a standard normal distribution table).
Now let's plug in the values:
n = (2.576 * 15,000 / 1,200)²
Calculating further:
n ≈ (38,640 / 1,200)²
n ≈ 32.2²
n ≈ 1,038.84
Therefore, a sample size of approximately 1,039 is required to estimate the mean amount earned by lawyers with a 99 percent level of confidence and a margin of error within $1,200.