Suppose that an accounting firm does a study to determine the time needed to complete one persons

Question

Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal. Suppose that the firm decided that it needed to be at least 90% confident of the population mean length of time to within one hour. What number of people should be surveyed?

Answer 1

To find the number of people that should be surveyed, we can use the formula for the sample size for a confidence interval for the mean when the population standard deviation is known. This formula is:

\[ n = \left(\frac{Z \cdot \sigma}{E}\right)^2 \]

Where:

- \( n \) is the required sample size.

- \( Z \) is the z-score for the desired confidence level (90% confidence level).

- \( \sigma \) is the known population standard deviation (7.0 hours in this case).

- \( E \) is the margin of error (1 hour in this case).

First, let's find the z-score for the 90% confidence level. Since it's a two-tailed confidence interval, the critical value is for \( \alpha/2 \), which is 0.05. We'll look up this z-score.

Next, we will use the given standard deviation (7.0 hours) and margin of error (1 hour) in the formula to calculate the required sample size.

Let's perform the calculation.

The accounting firm should survey 133 people to be at least 90% confident that the population mean length of time to complete one person's tax forms is within one hour.