Week 4 Homework
Question $\mathbf{8}$ of $\mathbf{2 0}$ (1 point) I Question Attempt: 1 of Unlimited
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A study of the consultants in a particular industry has determined that the standard deviation of the hourly fee of the consultants is $\$ 19$. A random sample of 90 consultants in the industry has a mean hourly fee of $\$ 121$. Find a $99 \%$ 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. (If necessary, consult a list of formulas.)
Lower limit:
Upper limit:

Step 1 ：Given values are mean = 121, standard deviation = 19, sample size = 90, and z-score for 99% confidence interval = 2.576.

Step 2 ：First, calculate the margin of error using the formula: \( z \times \frac{standard deviation}{\sqrt{sample size}} \).

Step 3 ：Substitute the given values into the formula to get: \( 2.576 \times \frac{19}{\sqrt{90}} \approx 5.159 \).

Step 4 ：Next, calculate the lower and upper limits of the confidence interval using the formulas: \( mean - margin of error \) and \( mean + margin of error \) respectively.

Step 5 ：Substitute the values into the formulas to get: \( 121 - 5.159 \approx 115.8 \) and \( 121 + 5.159 \approx 126.2 \).

Step 6 ：Round the lower and upper limits to one decimal place to get: lower limit = 115.8 and upper limit = 126.2.

Step 7 ：Final Answer: The lower limit of the 99% confidence interval for the true mean hourly fee of all consultants in the industry is \( \boxed{115.8} \) and the upper limit is \( \boxed{126.2} \).