Question

A study of the consultants in a particular industry has determined that the standard deviation of the hourly fee of the consultants is $24 . A random sample of 90 consultants in the industry has a mean hourly fee of $114. Find a 95% confidence interval for the true mean hourly fee of all consultants in the industry. Then give its lower limit and upper limit.

Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

Answer 1

To find the 95% confidence interval for the true mean hourly fee of all consultants in the industry, we can use the formula: Lower Limit = Mean - (Z * (Standard Deviation / √n)), and Upper Limit = Mean + (Z * (Standard Deviation / √n)). Plugging in the given values, the 95% confidence interval is approximately $110.06 to $117.94.

Step-by-step explanation:

To find the 95% confidence interval for the true mean hourly fee of all consultants in the industry, we can use the formula:

Lower Limit = Mean - (Z * (Standard Deviation / √n))

Upper Limit = Mean + (Z * (Standard Deviation / √n))

Where Mean is the sample mean, Standard Deviation is the known standard deviation of the hourly fee, n is the sample size, and Z is the Z-value corresponding to the desired level of confidence (in this case, 95%).

Plugging in the given values:

Mean = $114

Standard Deviation = $24

n = 90

Z-value (for 95% confidence) = 1.96

Calculating the lower limit:

Lower Limit = $114 - (1.96 * ($24 / √90))

Lower Limit ≈ $110.06

Calculating the upper limit:

Upper Limit = $114 + (1.96 * ($24 / √90))

Upper Limit ≈ $117.94

Therefore, the 95% confidence interval for the true mean hourly fee of all consultants in the industry is approximately $110.06 to $117.94.