Question

In 2003, the Accreditation Council for Graduate Medical Education (ACGME) implemented new rules lin residents. A key component of these rules is that residents should work no more than 80 hours per we of weekly hours worked in 2022 by a sample of residents at the Tidelands Medical Center. (Use t Distri 84 86 84 86 79 82 87 81 84 78 74 86 B. What is the point estimate of the population standard deviation? Note: Round your answer to 2 decimal places. c. What is the margin of error for a 90% confidence interval estimate? Note: Round your answer to 2 decimal places. d. Develop a 90% confidence interval for the population mean. Note: Round your answers to 2 decimal places.

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Answer to a math question In 2003, the Accreditation Council for Graduate Medical Education (ACGME) implemented new rules lin residents. A key component of these rules is that residents should work no more than 80 hours per we of weekly hours worked in 2022 by a sample of residents at the Tidelands Medical Center. (Use t Distri 84 86 84 86 79 82 87 81 84 78 74 86 B. What is the point estimate of the population standard deviation? Note: Round your answer to 2 decimal places. c. What is the margin of error for a 90% confidence interval estimate? Note: Round your answer to 2 decimal places. d. Develop a 90% confidence interval for the population mean. Note: Round your answers to 2 decimal places.

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Eliseo

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111 Answers

B. Calculate the sample standard deviation ($s$):

1. List the data: $84, 86, 84, 86, 79, 82, 87, 81, 84, 78, 74, 86$.

2. Calculate the sample mean ($\bar{x}$):

\bar{x} = \frac{84 + 86 + 84 + 86 + 79 + 82 + 87 + 81 + 84 + 78 + 74 + 86}{12} = \frac{991}{12} = 82.58

3. Apply the standard deviation formula:

s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}

s = \sqrt{\frac{(84-82.58)^2 + (86-82.58)^2 + \ldots + (86-82.58)^2}{11}}

s = \sqrt{\frac{1.98 + 11.78 + 1.98 + 11.78 + 13.04 + 0.3364 + 20.57 + 2.48 + 1.98 + 21.02 + 73.73 + 11.78}{11}}

s = \sqrt{\frac{40.54 + 36.84 + 2.48 + 34.82}{11}}

s = \sqrt{\frac{151.46}{11}}

s=\sqrt{15.5379}

s\approx3.94

C. Margin of Error:

1. Determine the critical value for 90% confidence with $ n = 12 $ which results in degrees of freedom $ df = 11 $. Use the t-distribution table.

2. Critical value ($t^*$) for 90% confidence interval with $df = 11$ is approximately 1.796.

3. Calculate the margin of error:

E=t^*\times\frac{s}{\sqrt{n}}=1.796\times\frac{3.94}{\sqrt{12}}

E=1.796\times1.137

E\approx2.04

D. Develop a 90% Confidence Interval:

1. Use the point estimate, margin of error, and sample mean:

\bar{x}\pm E=82.58\pm2.04

2. Calculate the confidence interval:

Lower limit:

82.58-2.04=80.54

Upper limit:

82.58+2.04=84.62

3. Round each bound to 2 decimal places:

(80.54,84.62)

Answer: 90% Confidence Interval =