Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 100 people. The sample average is 23.8 hours. There is a known population standard deviation of 6.0 hours. The population distribution is assumed to be normal.
cool. what's the question?
how do you Construct a 90% confidence interval for the population average time to complete the tax forms.
To answer this question, we need to calculate the margin of error for the survey's sample mean. The margin of error represents the range in which the true population mean is likely to fall based on the survey results.
To calculate the margin of error, we use the formula:
Margin of Error = Z * (Population Standard Deviation / √(Sample Size))
In this case, the sample size is 100, the population standard deviation is 6.0 hours, and the Z-value is based on the desired confidence level. Since you have not specified a confidence level, we will assume a 95% confidence level, which corresponds to a Z-value of 1.96 (approximate value).
Substituting these values into the formula, we get:
Margin of Error = 1.96 * (6.0 / √100)
Simplifying the equation:
Margin of Error = 1.96 * (6.0 / 10)
Margin of Error ≈ 1.176
Now, we can use the margin of error to estimate the confidence interval for the true population mean.
Confidence Interval = Sample Mean ± Margin of Error
Substituting the given sample mean of 23.8 hours and the calculated margin of error, we get:
Confidence Interval = 23.8 ± 1.176
Confidence Interval ≈ (22.624, 25.976)
Therefore, we can be 95% confident that the true population mean falls within the range of 22.624 to 25.976 hours.