The image of the public library is constantly changing, and their online services continue to grow. Usage of the library's home page grew by 17% during the past 12 months. It has been estimated that the current average length of a visit to the library's homepage is approximately 20 minutes. The library wants to take a sample to statistically estimate this mean. How large will the sample need to be to estimate the mean within 0.2 of 1 standard deviation with 0.95 confidence? (Round you answer up to the nearest whole number.)


Can someone help with formula for this and what do you use the percentage for on this one?

To determine the necessary sample size to estimate the mean within a given margin of error with a certain level of confidence, we can use the formula for sample size estimation:

n = (Z * σ / E)^2

Where:
- n is the required sample size
- Z is the Z score corresponding to the desired confidence level
- σ is the population standard deviation
- E is the margin of error

In this case, we are given that the library wants to estimate the mean length of a visit to their homepage. The margin of error is given as 0.2 of 1 standard deviation, so E = 0.2 * σ.

The confidence level is given as 0.95, which corresponds to a Z score of 1.96. This value can be obtained from standard Z score tables or calculated using statistical software.

By substituting these values into the formula, we can solve for the required sample size (n):

n = (1.96 * σ / (0.2 * σ))^2
n = 97.04

To round up to the nearest whole number, the library will need a sample size of 98 in order to estimate the mean within 0.2 of 1 standard deviation with 0.95 confidence.

The percentage you mentioned is the confidence level, which represents the level of certainty or confidence in the estimation. In this case, a confidence level of 0.95 means we are 95% confident that the true mean length of a visit to the library's homepage will fall within the estimated range.

To calculate the sample size needed to estimate the mean within a certain margin of error with a specific level of confidence, you can use the following formula:

n = (Z * σ / E)^2

Where:
n = sample size
Z = z-score corresponding to the desired confidence level (in this case, 0.95 confidence corresponds to a z-score of 1.96)
σ = standard deviation of the population (since we don't have information about the population standard deviation, we can use a conservative estimate, such as 0.5)
E = margin of error (0.2 * 1 standard deviation)

Using the formula, we can substitute the given values:

Z = 1.96
σ = 0.5
E = 0.2 * 1 = 0.2

n = (1.96 * 0.5 / 0.2)^2
n = 4.9^2
n ≈ 24.01

Since you cannot have a fraction of a sample, you need to round up to the nearest whole number. Therefore, the sample size needed is 25.