A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. Assuming a standard deviation of 3 hours, what is the required sample size if the error should be less than ½ hour with a 99% level of confidence?

Using a Z-value of 2.575, the answer is 238.7; or 240 if the answer is in multiple choice form.

Formula:

n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 2.58 using a z-table to represent the 99% confidence interval, sd = 3, E = 0.5, ^2 means squared, and * means to multiply.

Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.

To calculate the required sample size, 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 = standard deviation
E = desired margin of error

In this case, the desired margin of error is ½ hour, which is equal to 0.5.

First, we need to find the Z-score for a 99% level of confidence. The Z-score that corresponds to a 99% level of confidence is approximately 2.576.

Next, we can substitute the given values into the formula:

n = ((2.576 * 3) / 0.5) ^ 2

Simplifying this equation, we get:

n = (7.728 / 0.5) ^ 2
n = 15.456 ^ 2
n ≈ 238.881

Rounding up to the nearest whole number, we get:

n = 239

Therefore, the required sample size for the survey is 239.

To calculate the required sample size for a given level of confidence and desired margin of error, 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 level of confidence,
σ is the standard deviation of the population, and
E is the maximum desired margin of error.

In this case, the desired level of confidence is 99%, the standard deviation is given as 3 hours, and the maximum desired margin of error is 0.5 hour.

1. Determine the Z-score corresponding to a 99% level of confidence:
To find the Z-score for a 99% confidence level, we can consult a Z-table or use a statistical calculator. The Z-score for a 99% confidence level is approximately 2.58.

2. Substitute the values into the formula:
n = (Z * σ / E)^2
= (2.58 * 3 / 0.5)^2

3. Calculate the sample size:
n = (7.74 / 0.5)^2
= 15.48^2
= 239.5004

Since we cannot have a fractional sample size, we need to round up to the nearest whole number.

Therefore, the required sample size is 240 students.