In developing patient appointment schedules, a medical center wants to estimate the mean time that a staff member spends with each patient. How large a sample should be taken if the desired margin of error is two minutes at a 95% level of confidence? How large a sample should be taken for a 99% level of confidence? Use a planning value for the population standard deviation of 7 minutes.
a. 95% Confidence (to the nearest whole number):
b. 99% Confidence (to the nearest whole number):
Formula:
n = {[(z-value)(sd)]/E}^2
a) n = [(1.96 * 7)/2]^2
b) n = [(2.575 * 7)/2)^2
Calculate. Round to the next highest whole number.
a. To estimate the sample size needed for a desired margin of error of two minutes at a 95% level of confidence, we can use the formula:
n = (Z * σ / E)^2
where:
- n is the sample size
- Z is the Z-score corresponding to the desired confidence level (95% confidence corresponds to a Z-score of 1.96)
- σ is the population standard deviation
- E is the desired margin of error
Plugging in the values, we have:
n = (1.96 * 7 / 2)^2 = (13.72 / 2)^2 = 6.86^2 ≈ 47.12
Rounding up to the nearest whole number, the sample size needed for a 95% confidence level is approximately 48.
b. To estimate the sample size needed for a desired margin of error of two minutes at a 99% level of confidence, we can use the same formula as above, but with a different Z-score corresponding to a 99% confidence level (2.576):
n = (2.576 * 7 / 2)^2 = (18.07 / 2)^2 = 9.04^2 ≈ 81.63
Rounding up to the nearest whole number, the sample size needed for a 99% confidence level is approximately 82.
To determine the necessary sample size for estimating the mean time with a desired margin of error at different levels of confidence, we can use the formula:
n = (Z * σ / E)^2
Where:
n = sample size
Z = z-score corresponding to the desired level of confidence
σ = population standard deviation
E = desired margin of error
a. 95% Confidence:
For a 95% confidence level, the z-score is approximately 1.96 (you can look this up in a standard normal distribution table). The desired margin of error is 2 minutes, and the population standard deviation is given as 7 minutes.
Substituting these values into the formula:
n = (1.96 * 7 / 2)^2
n = (13.72 / 2)^2
n = 6.86^2
n ≈ 47
Therefore, a sample size of approximately 47 should be taken for a 95% confidence level.
b. 99% Confidence:
For a 99% confidence level, the z-score is approximately 2.58. Using the same margin of error (2 minutes) and population standard deviation (7 minutes):
Substituting these values into the formula:
n = (2.58 * 7 / 2)^2
n = (18.06 / 2)^2
n = 9.03^2
n ≈ 82
Therefore, a sample size of approximately 82 should be taken for a 99% confidence level.