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 0.22 of 1 standard deviation with 0.9 confidence? (Round you answer up to the nearest whole number.)

To determine the sample size needed to estimate the mean within 0.22 of 1 standard deviation with 0.9 confidence, you can use the formula for the required sample size, which is:

n = (Z * σ / E)^2

Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (0.9), which can be looked up from the standard normal distribution table. For a 0.9 confidence level, the corresponding Z-score is approximately 1.645.
σ = standard deviation of the population (unknown in this case, so we will use a conservative estimate)
E = maximum error (0.22 of 1 standard deviation)

Since the standard deviation of the population is unknown, we need to estimate it. Therefore, we can use a conservative estimate of 0.5 (which is typically larger than the actual standard deviation) to ensure a larger sample size, making the estimate more reliable.

Plugging in the values into the formula:

n = (1.645 * 0.5 / 0.22)^2

n ≈ (0.8225 / 0.22)^2

n ≈ 3.74^2

n ≈ 13.99

Rounding up to the nearest whole number:

n ≈ 14

Hence, the sample size needed to estimate the mean within 0.22 of 1 standard deviation with 0.9 confidence is 14.

To find the sample size needed to estimate the mean within a certain range with a desired level of confidence, we can use the formula:

n = (Z * σ / E)²

Where:
n = sample size
Z = Z-score corresponding to the desired confidence level (in this case, 0.9)
σ = standard deviation
E = margin of error (0.22 standard deviations in this case)

To calculate the sample size, we need to determine the standard deviation (σ). Given that the length of a visit to the library's homepage has not been provided, we cannot directly calculate the standard deviation. However, we can use the current estimate of the mean visit length to estimate the standard deviation.

Since the average visit length is approximately 20 minutes, we can assume that the data follows a normal distribution. To estimate the standard deviation, we can use the formula:

σ ≈ σ = range / 4

Where range is the difference between the maximum and minimum values of the data. However, since we don't have the range, we can use a conservative estimate of the range based on the given data.

Let's assume the maximum possible visit length is 60 minutes and the minimum possible visit length is 0 minutes. Therefore, the estimated range is 60 - 0 = 60 minutes.

Using this estimated range, we can calculate the estimated standard deviation:

σ ≈ σ = 60 / 4 = 15 minutes

Now that we have an estimate for the standard deviation, we can proceed to calculate the sample size:

n = (Z * σ / E)²
n = (Z * 15 / (0.22 * 15))²

Next, we need to find the Z-score corresponding to a confidence level of 0.9. We can use a Z-table or a calculator to find this value. The Z-score for a 0.9 confidence level is approximately 1.645.

Using this value, we can calculate the sample size:

n = (1.645 * 15 / (0.22 * 15))²
n = (1.645 / 0.22)²
n ≈ 56.01

Finally, rounding up to the nearest whole number:

n ≈ 57

Therefore, the sample size needed to estimate the mean within 0.22 of 1 standard deviation with 0.9 confidence is 57.