Answer:
a) The 95% CI for the mean surgery time is (133.05, 140.75).
b) The 99.5% CI for the mean surgery time is (131.37, 142.43).
c) The level of confidence of the interval (133.9, 139.9) is 69%.
d) The sample size should be 219 surgeries.
e) The sample size should be 377 surgeries.
Step-by-step explanation:
We have a sample, of size n=132, for which the mean time was 136.9 minutes with a standard deviation of 22.6 minutes.
a) We have to find a 95% CI for the mean surgery time.
The critical value of z for a 95% CI is z=1.96.
The margin of error of the CI can be calculated as:
![E=z\cdot s/\sqrt{n}=1.96*22.6/\sqrt{132}=44.296/11.489=3.85](https://tex.z-dn.net/?f=E%3Dz%5Ccdot%20s%2F%5Csqrt%7Bn%7D%3D1.96%2A22.6%2F%5Csqrt%7B132%7D%3D44.296%2F11.489%3D3.85)
Then, the lower and upper bounds of the confidence interval are:
![LL=\bar x-E=136.9-3.85=133.05\\\\UL=\bar x+E=136.9+3.85=140.75](https://tex.z-dn.net/?f=LL%3D%5Cbar%20x-E%3D136.9-3.85%3D133.05%5C%5C%5C%5CUL%3D%5Cbar%20x%2BE%3D136.9%2B3.85%3D140.75)
The 95% CI for the mean surgery time is (133.05, 140.75).
b) Now, we have to find a 99.5% CI for the mean surgery time.
The critical value of z for a 99.5% CI is z=2.81.
The margin of error of the CI can be calculated as:
![E=z\cdot s/\sqrt{n}=2.81*22.6/\sqrt{132}=63.506/11.489=5.53](https://tex.z-dn.net/?f=E%3Dz%5Ccdot%20s%2F%5Csqrt%7Bn%7D%3D2.81%2A22.6%2F%5Csqrt%7B132%7D%3D63.506%2F11.489%3D5.53)
Then, the lower and upper bounds of the confidence interval are:
![LL=\bar x-E=136.9-5.53=131.37\\\\UL=\bar x+E=136.9+5.53=142.43](https://tex.z-dn.net/?f=LL%3D%5Cbar%20x-E%3D136.9-5.53%3D131.37%5C%5C%5C%5CUL%3D%5Cbar%20x%2BE%3D136.9%2B5.53%3D142.43)
The 99.5% CI for the mean surgery time is (131.37, 142.43).
c) We can calculate the level of confidence, calculating the z-score for the margin of error in that interval.
We know that the difference between the upper bound and lower bound is 2 times the margin of error:
![UL-LL=2E\\\\E=\dfrac{UL-LL}{2}=\dfrac{139.9-133.9}{2}=\dfrac{6}{2}=3](https://tex.z-dn.net/?f=UL-LL%3D2E%5C%5C%5C%5CE%3D%5Cdfrac%7BUL-LL%7D%7B2%7D%3D%5Cdfrac%7B139.9-133.9%7D%7B2%7D%3D%5Cdfrac%7B6%7D%7B2%7D%3D3)
Then, we can write the equation for the margin of error to know the z-value.
![E=z \cdot s/\sqrt{n}\\\\z= E\cdot \sqrt{n}/s=2*\sqrt{132}/22.6=2*11.5/22.6=1.018](https://tex.z-dn.net/?f=E%3Dz%20%5Ccdot%20s%2F%5Csqrt%7Bn%7D%5C%5C%5C%5Cz%3D%20E%5Ccdot%20%5Csqrt%7Bn%7D%2Fs%3D2%2A%5Csqrt%7B132%7D%2F22.6%3D2%2A11.5%2F22.6%3D1.018)
The confidence level for this interval is then equal to the probability that the absolute value of z is bigger than 1.018:
![P(-|z|](https://tex.z-dn.net/?f=P%28-%7Cz%7C%3CZ%3C%7Cz%7C%29%3DP%28-1.018%3CZ%3C1.018%29%3D0.69)
The level of confidence of the interval (133.9, 139.9) is 69%.
d) We have to calculate the sample size n to have a margin of error, for a 95% CI, that is equal to 3.
The critical value for a 95% CI is z=1.96.
Then, the sample size can be calculated as:
![E=z\cdot s/\sqrt{n}\\\\n=(\dfrac{z\cdot s}{E})^2=(\dfrac{1.96*22.6}{3})^2=14.77^2=218.015\approx 219](https://tex.z-dn.net/?f=E%3Dz%5Ccdot%20s%2F%5Csqrt%7Bn%7D%5C%5C%5C%5Cn%3D%28%5Cdfrac%7Bz%5Ccdot%20s%7D%7BE%7D%29%5E2%3D%28%5Cdfrac%7B1.96%2A22.6%7D%7B3%7D%29%5E2%3D14.77%5E2%3D218.015%5Capprox%20219)
The sample size should be 219 surgeries.
e) We have to calculate the sample size n to have a margin of error, for a 99% CI, that is equal to 3.
The critical value for a 99% CI is z=2.576.
Then, the sample size can be calculated as:
![E=z\cdot s/\sqrt{n}\\\\n=(\dfrac{z\cdot s}{E})^2=(\dfrac{1.96*22.6}{3})^2=19.41^2=376.59\approx 377](https://tex.z-dn.net/?f=E%3Dz%5Ccdot%20s%2F%5Csqrt%7Bn%7D%5C%5C%5C%5Cn%3D%28%5Cdfrac%7Bz%5Ccdot%20s%7D%7BE%7D%29%5E2%3D%28%5Cdfrac%7B1.96%2A22.6%7D%7B3%7D%29%5E2%3D19.41%5E2%3D376.59%5Capprox%20377)
The sample size should be 377 surgeries.