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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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 =(80.54,84.62)
1. List the data: \(84, 86, 84, 86, 79, 82, 87, 81, 84, 78, 74, 86\).
2. Calculate the sample mean (\(\bar{x}\)):
3. Apply the standard deviation formula:
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:
D. Develop a 90% Confidence Interval:
1. Use the point estimate, margin of error, and sample mean:
2. Calculate the confidence interval:
Lower limit:
Upper limit:
3. Round each bound to 2 decimal places:
Answer: 90% Confidence Interval =
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