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.
(a)What is your best estimate of the population mean?
(b)Develop a 99 percent confidence interval for the population mean wage (including benefits) for these employees.
(c)How large a sample is needed to assess the population mean with an allowable error of $2 at 95 percent confidence?
To answer these questions, we will use the concepts of estimation and confidence intervals. Let's go step by step:
(a) To estimate the population mean wage, we can use the sample mean as our best estimate. In this case, the sample mean is $56 per hour. Therefore, our best estimate for the population mean wage is $56 per hour.
(b) To develop a confidence interval for the population mean wage, we need to determine the margin of error and then construct the interval.
Step 1: Determining the margin of error
The margin of error is calculated using the sample standard deviation, sample size, and the desired level of confidence. The formula for the margin of error is:
Margin of Error = (Z-score) * (Standard Deviation / Square Root of Sample Size)
For a 99 percent confidence interval, the Z-score corresponding to the desired confidence level is approximately 2.576. In this case, the sample standard deviation is $6.35, and the sample size is 31.
Margin of Error = 2.576 * (6.35 / sqrt(31))
Step 2: Constructing the interval
The confidence interval formula is:
Confidence Interval = (Sample Mean - Margin of Error, Sample Mean + Margin of Error)
Substituting the values, we have:
Confidence Interval = ($56 - Margin of Error, $56 + Margin of Error)
Now, calculate the Margin of Error using the formula from Step 1 and substitute it into the Confidence Interval formula.
(c) To determine the sample size needed to assess the population mean with a specific allowable error at a certain confidence level, we can use the formula for sample size calculation. The formula is:
Sample Size = (Z-score)^2 * (Standard Deviation^2) / (Allowable Error)^2
In this case, the allowable error is $2, and the desired confidence level is 95 percent. The Z-score corresponding to a 95 percent confidence level is approximately 1.96. Substitute these values into the formula to calculate the required sample size.
Sample Size = (1.96^2 * 6.35^2) / (2^2)
Evaluate the result to find the sample size needed.
Note: These calculations assume that the data follows a normal distribution and that the sample is a simple random sample.
(a) The best estimate of the population mean is equal to the average wage of the sample, which is $56 per hour.
(b) To develop a 99 percent confidence interval, we can use the formula:
Confidence interval = X̄ ± (Z * (σ / √n))
Where:
X̄ is the sample mean, which is $56 per hour
Z is the Z-score for a 99 percent confidence level, which is approximately 2.576
σ is the sample standard deviation, which is $6.35 per hour
n is the sample size, which is 31
Substituting these values into the formula, we get:
Confidence interval = 56 ± (2.576 * (6.35 / √31))
Calculating this expression gives us a confidence interval of approximately $52.43 to $59.57 per hour. Therefore, we can be 99 percent confident that the true population mean wage for the Florida state authority employees falls within this range.
(c) To determine the sample size needed to estimate the population mean with an allowable error of $2 at 95 percent confidence, we can use the formula:
n = (Z^2 * σ^2) / E^2
Where:
Z is the Z-score for a 95 percent confidence level, which is approximately 1.96
σ is the estimated population standard deviation, which is $6.35 per hour
E is the maximum allowable error, which is $2
Substituting these values into the formula, we get:
n = (1.96^2 * 6.35^2) / 2^2
Calculating this expression gives us a sample size of approximately 61. Therefore, a sample size of at least 61 is needed to estimate the population mean with an allowable error of $2 at 95 percent confidence.