An investigator wants to estimate caffeine consumption in high school students. How many students would be required to ensure that a 95% confidence interval estimate for the mean caffeine intake is within 15mg of the true mean? Assume the standard deviation in caffeine intake is 68mg. How many students would be required to estimate the proportion of students who consume coffee. Suppose we want the estimate to be within 5% of the true proportion with 95% confidence.
Use formulas to find sample size.
For the first part:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent the 95% confidence interval, sd = 68, E = 15, ^2 means squared, and * means to multiply.
For the second part:
n = [(z-value)^2 * p * q]/E^2
... where n = sample size, z-value is 1.96, p = .5 (when no value is stated in the problem), q = 1 - p, ^2 means squared, * means to multiply, and E = .05 (for 5%).
Plug the values into the formulas and finish the calculations. Round your answers to the next highest whole number.
Hope this helps.
To determine the sample size required to estimate the mean caffeine intake, we can use the following formula:
n = [(Z * σ) / E]^2
Where:
n = sample size
Z = Z-score (corresponding to the desired level of confidence, which is 95%)
σ = standard deviation
E = margin of error
Substituting the given values:
Z = 1.96 (the critical value for a 95% confidence level)
σ = 68mg (standard deviation)
E = 15mg (margin of error)
n = [(1.96 * 68) / 15]^2
n ≈ 166.65
Rounding up to the nearest whole number, you would need at least 167 high school students to estimate the mean caffeine intake within 15mg of the true mean, with a 95% confidence level.
To determine the sample size required to estimate the proportion of students who consume coffee, we can use the following formula:
n = [(Z^2 * p * q) / E^2]
Where:
n = sample size
Z = Z-score (corresponding to the desired level of confidence, which is 95%)
p = estimated proportion (also known as the expected proportion, you can use 0.5 if there is no prior estimate)
q = 1 - p
E = margin of error
Substituting the given values:
Z = 1.96 (the critical value for a 95% confidence level)
p = 0.5 (since we don't have an estimated proportion)
q = 1 - 0.5 = 0.5
E = 0.05 (5% margin of error)
n = [(1.96^2 * 0.5 * 0.5) / 0.05^2]
n ≈ 384.16
Rounding up to the nearest whole number, you would need at least 385 high school students to estimate the proportion of students who consume coffee within 5% of the true proportion, with a 95% confidence level.
To determine the number of students required to ensure a 95% confidence interval estimate for the mean caffeine intake is within 15mg of the true mean, we can use the formula for sample size calculation:
n = (Z * σ / E)^2
Where:
n = required sample size
Z = desired Z-score for the desired confidence level (for 95% confidence level, Z = 1.96)
σ = standard deviation of the population (68mg)
E = desired margin of error (15mg)
Plugging in the values:
n = (1.96 * 68 / 15)^2
n ≈ 103.17
To ensure a 95% confidence interval estimate for the mean caffeine intake within 15mg of the true mean, you would need a sample size of at least 104 high school students (rounded up to the nearest whole number).
To estimate the proportion of students who consume coffee, we can use the formula for sample size calculation:
n = (Z^2 * p * (1 - p)) / (E^2)
Where:
n = required sample size
Z = desired Z-score for the desired confidence level (for 95% confidence level, Z = 1.96)
p = estimated proportion of students who consume coffee (unknown, we'll use 0.5 for a conservative estimate)
E = desired margin of error (5%)
Plugging in the values:
n = (1.96^2 * 0.5 * (1 - 0.5)) / (0.05^2)
n ≈ 384.16
To estimate the proportion of students who consume coffee within 5% of the true proportion with 95% confidence, you would need a sample size of at least 385 high school students (rounded up to the nearest whole number).