A random sample of 31 people employed by the Florida state authority established they earned an average wage (including benefits) of $56 per hour. The sample standard deviation was $6.35 per hour.
Develop a 99 percent confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
To develop a confidence interval for the population mean wage, we can use the formula:
Confidence interval = sample mean ± (critical value * standard error)
1. Calculate the standard error: The formula for standard error is given by the sample standard deviation divided by the square root of the sample size.
Standard error = sample standard deviation / √(sample size)
Given that the sample standard deviation is $6.35 per hour and the sample size is 31, we can calculate the standard error:
Standard error = 6.35 / √(31) ≈ 1.1316
2. Determine the critical value: The critical value is based on the desired confidence level and the degrees of freedom. In this case, since we want a 99 percent confidence interval, the corresponding critical value is obtained from a t-distribution with (n-1) degrees of freedom.
Since the sample size is 31, the degrees of freedom is equal to (31 - 1) = 30.
Using a t-distribution table or a scientific calculator, the critical value for a 99 percent confidence interval with 30 degrees of freedom is approximately 2.756.
3. Calculate the confidence interval:
Confidence interval = sample mean ± (critical value * standard error)
Given that the sample mean is $56 per hour, the confidence interval can be calculated as follows:
Confidence interval = 56 ± (2.756 * 1.1316)
Lower limit = 56 - (2.756 * 1.1316) ≈ 52.0605
Upper limit = 56 + (2.756 * 1.1316) ≈ 59.9395
Therefore, the 99 percent confidence interval for the population mean wage (including benefits) for these employees is approximately $52.06 to $59.94 per hour.